The seven step series

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Johnathan Corgan

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Jun 29, 2009, 11:22:59 AM6/29/09
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Bruno,

I think you were off to a good start with your planned series of posts
about the seven step argument. I believe your first installment was a
discussion of set theory as one of the mathematical preliminaries to the
actual argument.

I am looking forward to your next installment.

Regards,

Johnathan Corgan


Bruno Marchal

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Jun 30, 2009, 6:45:27 AM6/30/09
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Hi Johnathan,


Well, thanks. I am not sure Kim and Marty are there, but I can provide
a summary, and recall the motivation.

Marty, did you come back from holiday? Kim? still interested in
electronical summer's school on mathematics.

The goal of the seven step thread is to make clear the seventh step of
the UDA (Universal Dovetailer Argument). The purpose of the UDA is to
make clear that the mind-body problem (or the consciousness/reality
problem, or the first person/third person) problem is reduced, when we
do the computationalist assumption, to a pure body appearance or
discourse problem. UDA shows that if we assume the comp. hyp. then we
have to explain the appearance of matter from machine or number self-
reference only. The proof is constructive, it shows *how* the laws of
physics have to be extracted from self-reference.

Later, much later, I could explain, if everyone is OK with UDA, how we
can already extract from self-reference the general shape of physics,
so that we can already refute empirically, or confirm, the comp. hyp.
And it appears that the empirical quantum mechanics, currently,
confirms the comp. hyp. Quantum mechanics confirms the partial
indetermination of the outcomes of our possible experiences, and the
"high non booleanity" of the propositions describing those outcomes".

The object of the "seventh step thread' consists in making the seventh
step accessible to non mathematicians. So we have to start from zero.
I have decided to start from elementary "naive" set theory, without
which we cannot do anything in math. I will avoid all special
mathematical symbols, and use instead words with capital letters.

We have not yet done a lot. So I can sum up, with the new "notations".

Definition. A set is just a "many" considered, when clear enough, as a
"one". So a set is just a collection of objects, and those objects are
called the element, or the member, of the set. If some x is an element
of some set A, we write x BELONGS-TO A, or (x BELONGS-TO A).
A set can be described in extension or in intension. "in extension"
means that we give all elements of the set, enclosed in accolades.
When the set is not to complex (meaning big or infinite), we can use
the "...". We can give name to a set, to ease or talk about that set,
like we do all the times in mathematics. Most of the set we will
consider are set of mathematical object, mainly numbers in the
beginning, and then set of ... sets.

Example-exercise:

1°) Let A be the set {0, 1, 2, 3}. ("A" is said to be a local name for
the set {0, 1, 2, 3}. And local means that such a name is used in a
local context. One paragraph later "A" could designed another, so be
careful). If "A" names {0, 1, 2, 3}, we will write "A = {0, 1, 2, 3}".

OK, so with A = {0, 1, 2, 3}. Which of the following propositions are
true

1) the number 2 is a member of A
2) the number 12 is a member of A
3) the number 12 is not a member of A
4) (3 BELONGS-TO A)
5) all members of A are numbers
6) one element of A is not a number
7) A can be defined in intension in the following way A = {x SUCH-THAT
x is a positive integer little than 4}

2°) Same questions with the set A = {0, 1, 2, 3, ... , 61, 62, 63}

This makes 14 exercises, which should be easy. I intent to keep it
that way. I continue after I get either answers (correct or wrong), or
questions.

Everyone is welcome to participate. Yet, I ask those who are quick to
respect those who are slow. To be slow in the beginning usually help
for being deep in the sequel.

Best,

Bruno

http://iridia.ulb.ac.be/~marchal/

John Mikes

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Jun 30, 2009, 8:32:42 AM6/30/09
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Hi, Bruno
you know that I am in a different mindset, yet happy to read your train of thoughts. I consider a set a limited model of elements (and conclusions thereof are not applicable to wider domains) - when I read your
"A set can be described in extension or in intension. "in extension" means that we give all elements of the set, enclosed in accolades."
I was really happy with the next sentence:
"When the set is not to complex (meaning big or infinite), we can use the "...".  - "
(I missed here the exemption of the 'infinite' "set", really a contradiction, to which the 'set' considerations cannot apply - OR can they? if you have something on that...) "Many" cannot be infinite (by MY definition).
I loved your words on QM, the (linear) extension of the figment physical world as described in reductionist physical sciences.
I also cannot wait for something more about your approach on
the "self reference" - the basis of physics? - especially as to
'self' of what (who)? I hope the answer will not be "machine" or comp, because then I have to continue "and what is that?"
(in more than a utilitarian explanation of what it does). ('it?')
What boils down to my ignorance as to the originating and maintaining to ANY action we speak about. The 'theos' of a non-assumed and non-supernatural factor (system?) yet involved in conducting all we just find natural and proceeding.
You may substitute 'numbers' for such, but so far did not reply (to my satisfaction at least) WHAT those 'numbers' may be.
Sorry, I am not of the religious kind.
*
Maybe my error is in 'believeing' that a REALITY may exist and 'we' have only access to part of it. Inventing for our comfort (the D. Bohmian idea) 'numbers' at the human level of pre-Platonian thinking. If 'reality' exists only by 'comp' or 'consequences' then I may be in a reversed error, due to brainwashing by in college imprinted  natural sciences - what I try to exceed yet it still sits there.
Our 'perceived reality' (ColinH) may also provide the numbers.
Now that sounds heretical enough in this thread. Forgive me.
*
Waiting for the self-reference, (who's?) - with thanks so far
 
John Mikes


m.a.

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Jun 30, 2009, 12:57:17 PM6/30/09
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Bruno,
I'm still with you but I seriously wonder how far I can follow.
I have the sort of mind that groups of logical statement and propositions
cause to simply shut down. I am more than willing to accept that your proof
is consistent and I assume that others on the list will point out flaws if
there are any. What I would really appreciate would be a prose explanation
of the sequence of ideas that lead to the conclusion that physics can emerge
from and be (in some sense) actualized by math. If Kim and others wish to
continue the exposition of UDA, I'll try to keep up, but I won't ask you to
continue solely on my account. Best wishes,




marty a.

Bruno Marchal

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Jul 1, 2009, 10:20:51 AM7/1/09
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Hi John,

On 30 Jun 2009, at 14:32, John Mikes wrote:

Hi, Bruno
you know that I am in a different mindset, yet happy to read your train of thoughts. I consider a set a limited model of elements (and conclusions thereof are not applicable to wider domains) -

Well, as Thorgny illustrated correctly, the notion of SET does not apply to SETS. usually the collection of all sets is not considered as a set, for many good reasons, some of which appear already in the treatise on Numbers by Plotinus. To derive from this that infinite sets cannot exist, like Thorgny seems to believed is invalid.


when I read your
"A set can be described in extension or in intension. "in extension" means that we give all elements of the set, enclosed in accolades."
I was really happy with the next sentence:
"When the set is not to complex (meaning big or infinite), we can use the "...".  - "
(I missed here the exemption of the 'infinite' "set", really a contradiction, to which the 'set' considerations cannot apply - OR can they?

Of course they can. This is done in everyday mathematics all the time since Cantor discovered the notion of sets. It can be said that set have been exploited especially for the handling of infinities.
Typical infinite sets are the set of natural numbers {0, 1, 2, ...}, the set of odd numbers {1, 3, 5, 7, ...}, the set of prime numbers {2, 3, 5, 7, 11, ...} as already proved by Euclid, the set of decimal approximation of most real numbers, etc.



if you have something on that...)

Google on it. The notion of set (finite and especially infinite sets) is pervading all modern mathematics.



"Many" cannot be infinite (by MY definition).

Well, I prefer to use the word in their most used and standard sense. 



I loved your words on QM, the (linear) extension of the figment physical world as described in reductionist physical sciences.
I also cannot wait for something more about your approach on
the "self reference" - the basis of physics? -

This is the whole point of the UDA, and AUDA.



especially as to
'self' of what (who)? I hope the answer will not be "machine" or comp, because then I have to continue "and what is that?"

Of course it is comp, although I use comp because it is the simple way. Then, digging on mathematical logic, the same result can be retrieve from much weaker assumption. But comp is believed by all scientist and philosophers, except Penrose. Even John Searle is computationalist with my weak definition of it.
"and what is that"? It is the point of the present thread to explain that as slowly as possible so that good willing non mathematicians can understand the key points. 



(in more than a utilitarian explanation of what it does). ('it?')
What boils down to my ignorance as to the originating and maintaining to ANY action we speak about. The 'theos' of a non-assumed and non-supernatural factor (system?) yet involved in conducting all we just find natural and proceeding.
You may substitute 'numbers' for such, but so far did not reply (to my satisfaction at least) WHAT those 'numbers' may be.
Sorry, I am not of the religious kind.

If you can compute 34+89, or 65*87, then you know enough. My use of number in UDA is not religious, it is the same use as those who use number to send man on the moon. Too much and premature philosophizing makes it hard to proceed. "Religion" appears later, when you say "yes" to the digitalist surgeon. Comp asks for an act of faith.



*
Maybe my error is in 'believeing' that a REALITY may exist and 'we' have only access to part of it.


Neither science, nor philosophy, nor theology, could develop without such assumption. It is because we believe that there is a reality, that we can build theories to infer the part on which we have no access. science always consist in an attempt to see the invisible, be it atoms, far away galaxies, or mathematical constructions. So this is not an error.




Inventing for our comfort (the D. Bohmian idea) 'numbers' at the human level of pre-Platonian thinking. If 'reality' exists only by 'comp' or 'consequences' then I may be in a reversed error, due to brainwashing by in college imprinted  natural sciences - what I try to exceed yet it still sits there.
Our 'perceived reality' (ColinH) may also provide the numbers.

Sure, but this is independent of the fact that 17 is a prime numbers independently of me. That the set of prime numbers is infinite, independently of me, etc. 



Now that sounds heretical enough in this thread. Forgive me.
*
Waiting for the self-reference, (who's?)

Who's? But the universal number's self-reference of course. Even the Lobian one.



- with thanks so far

You are welcome,

Bruno

Bruno Marchal

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Jul 1, 2009, 12:00:20 PM7/1/09
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Hi Marty,


On 30 Jun 2009, at 18:57, m.a. wrote:

>
> Bruno,
> I'm still with you but I seriously wonder how far I can
> follow.


Well, honestly, if you don't try to answer to the exercises, we will
never know.
But I can imagine some shyness for doing so.



>
> I have the sort of mind that groups of logical statement and
> propositions
> cause to simply shut down.


I disagree. You have already prove to me that you can handle such
propositions. Your problem is that you don't memorize what you
understand, so, especially after some break, you feel like you could
shut down.
This just means that you have to work a little more, and to stop
building negative self-prejudices, which are most of the time self-
fulfilling.
Now, I obvioulsy cannot ask you to do such a work, as life is rich and
full of consuming time opportunities, noir can I really provide the
motivation for doing so.
I appreciate very much your honesty, even if I suspect the reasons are
bad.



> I am more than willing to accept that your proof
> is consistent and I assume that others on the list will point out
> flaws if
> there are any.

The problem is that we have discussed it before. The only things which
remains to be explained is what is a mathematical computation. This is
not easy, and ask for some familiarity with basic mathematics.


> What I would really appreciate would be a prose explanation
> of the sequence of ideas that lead to the conclusion that physics
> can emerge
> from and be (in some sense) actualized by math.

My problem is that UDA is exactly that. I get the AUDA in the early
seventies, before UDA. And I have developed UDA in the late seventies,
so as to provide some help for my friends. Since then I know that UDA
is not so simple, especially the seventh and eighth steps. I can think
about making a shorter attempt.



> If Kim and others wish to
> continue the exposition of UDA, I'll try to keep up, but I won't ask
> you to
> continue solely on my account. Best wishes,




No problem Marty, at least you have tried. Kim? how do you do?
Johnathan?

I know that to play the "candid" role in a public way, you need some
courage, and it is OK to remain silent. But people have to understand
than on those delicate matters, it is not really possible to proceed
without asking question/exercise. We can only continue "the seven step
series" thread if people provide answer to the exercises.
I will nevertheless try to explain things in a more "journalistic",
and shorter, way. The problem is that such a thing can easily be
misinterpreted. I will think about it. The "real thing" is more easy
to explain, but admittedly longer when we start from quasi zero.

Bruno
http://iridia.ulb.ac.be/~marchal/



m.a.

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Jul 1, 2009, 12:43:04 PM7/1/09
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Bruno,
Your faith in my ability to master the logic is encouraging. I
remain determined to wrestle with the steps in whichever form in which you
decide to present them. Tally ho,



m.a.


----- Original Message -----
From: "Bruno Marchal" <mar...@ulb.ac.be>
To: <everyth...@googlegroups.com>
Sent: Wednesday, July 01, 2009 12:00 PM
Subject: Re: The seven step series



Hi Marty,


I know that to play the "candid" role in a public way, you need some
courage, and it is OK to remain silent. But people have to understand
than on those delicate matters, it is not really possible to proceed
without asking question/exercise. We can only continue "the seven step
series" thread if people provide answer to the exercises.
I will nevertheless try to explain things in a more "journalistic",
and shorter, way. The problem is that such a thing can easily be
misinterpreted. I will think about it. The "real thing" is more easy
to explain, but admittedly longer when we start from quasi zero.

Bruno







http://iridia.ulb.ac.be/~marchal/





m.a.

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Jul 1, 2009, 12:57:33 PM7/1/09
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Hi Bruno,
                I'm responding to the quiz (see below). What does "high non booleanity" mean in the context of para.2?
> ----- Original Message -----
> From: "Bruno Marchal" <
mar...@ulb.ac.be>
> To: <
everyth...@googlegroups.com
>

> Sent: Tuesday, June 30, 2009 6:45 AM
> Subject: Re: The seven step series
>
>
>
> Hi Johnathan,
>
>
> 1) the number 2 is a member of A   True
> 2) the number 12 is a member of A  False
> 3) the number 12 is not a member of A  True
> 4) (3 BELONGS-TO A)    True: but you haven't told us whether the parenthesis cancels the locality of brackets.
> 5) all members of A are numbers  True
> 6) one element of A is not a number  False: we've established that zero is a number.

> 7) A can be defined in intension in the following way A = {x SUCH-THAT
> x is a positive integer little than 4}    True...if zero is considered a positive integer.

>
> 2°) Same questions with the set A = {0, 1, 2, 3, ... , 61, 62, 63}
1. True
2. True
3. False
4. True: same question as 4 above.
5. True
6. False: zero is a number
7. False

John Mikes

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Jul 1, 2009, 6:22:43 PM7/1/09
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Hi, Bruno,
and thanks for taking the time to eviscerate my ignorance. I asked for it <G>. Just a bit in detail:
I did not pay enough attention to Torgny's sets vs set, I see now.
Question: There are 'rules' applicable how to handle 'a' set, operations and values to obtain etc. Are these rules also valid and applicable to Torgny's "sets", the infinite set of all sets? Can you do with that anything (practical in some sense)? even 'think' of it?  (BTW I did not mean that 'infinite' sets do not exist, only that we need a different consideration for them from the finite bunch usually understood as 'a' set - maybe also 'naming' them differently). 
 
I did not find the 'intension' sets (description), only the 'extension' part.  
 
(JM)"Many" cannot be infinite (by MY definition).(BM):Well, I prefer to use the word in their most used and standard sense. 
-----which one is that? Can, or cannot?-----
 
I did not identify "self-referenced" with computationalist. And that 'many' think so is no satisfying evidence in my view: scientific thinking is not a democratic voting formula. Even the "BIG" names... Al Gore and Jimmy Carter received Nobels.
There is no "scientific" statement which could not be contrasted by two opposite ones of 'other' scientists.
Now the delicate part:
BM:
"If you can compute 34+89, or 65*87, then you know enough."
IF!! I accept that a number means a math-book and writing a number is not 'that' number without applying 'rules'. Like:
34 is a 3 and a 4 unless you defined the space and comma if applicable. (I have something on that 'comma' I forgot to write about. Maybe now I will resort to it at the end). To make the 3 !!! and the 4 !!!! a !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! is far fetched.
Then again your + and * are NOT numbers. It looks to me (in a personalized overwhelming exaggeration, of course) that by 'numbers' you mean the 'science of math expressed in symbols of a huge vocabulary' (maybe books ->library!) Ph.D.
I don't deny the practicality of applying 'numbers-based' science in sending a man to Mars, but it is NOT the numbers that does the job. It is the complexity of the state of the art we reached, which includes science, technology, skills, ideas AND of course numbers-application. Bohm's idea - as I understood it - was that searching nature, you do not bounce into numbers,  you can observe 3-leaf or 4legged and manyshaped things, big and small, YOU (the human) can 'count them' if you invented the symbols 1 2 3 4 etc. but these refer to quantities and it required lots of abstracting in mental evolution to arrive in a numbers-based math - how humans think about nature.
That's what I referred to as the pre-Platonistic times.
*
Let me postpone the 'comma' part in the sets for next time.
Thanks again and my mind works in crooked ways, if you can excuse me for that. It seems I need too much learning to catch up.
 
John M

--------------------------------------------------------------------------

Bruno Marchal

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Jul 2, 2009, 10:34:18 AM7/2/09
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Hi Marty,

On 01 Jul 2009, at 18:57, m.a. wrote:

Hi Bruno,
                I'm responding to the quiz (see below). What does "high non booleanity" mean in the context of para.2?


We will need more math for the details, but "boolean" refer to classical or even platonist logic, when applied in a frame where we can agree that propositions are either true or false.Like arithmetic: you surely agree that a postive integer is either even, or not even.
 Classical logic is the one used implicitly or explicitly in most "scientific" discourses, especially in (classical) mathematics. "Boolean" comes from Boole who wrote a book "The laws of thought", and which plays some role in the birth of mathematical logic and computer science. But classical logic has a very long story, both in East and West.




>
> Example-exercise:
>
> 1°) Let A be the set {0, 1, 2, 3}. ("A" is said to be a local name for
> the set {0, 1, 2, 3}. And local means that such a name is used in a
> local context. One paragraph later "A" could designed another, so be
> careful). If "A" names {0, 1, 2, 3}, we will write "A = {0, 1, 2, 3}".
>
> OK, so with A = {0, 1, 2, 3}. Which of the following propositions are
> true
>
> 1) the number 2 is a member of A   True

OK.


> 2) the number 12 is a member of A  False

OK.


> 3) the number 12 is not a member of A  True

OK.


> 4) (3 BELONGS-TO A)    True: but you haven't told us whether the parenthesis cancels the locality of brackets.

OK.
You can suppress anything in the notations as far as you, and those who will read the text, can figure out what you mean. Here I could have simply 
4) 3 BELONGS-TO A.




> 5) all members of A are numbers  True

OK.



> 6) one element of A is not a number  False: we've established that zero is a number.

OK. (I am not sure that we have established that, nor even what you mean by "established", but we surely welcome 0 in the numbers. Note that in the beginning 0 was rejected. And both the numbers 1 and 2 takes time to be accepted as number, for the reason that "number" means originally numerous. The "num" of "numerous" has the same origin as "number".




> 7) A can be defined in intension in the following way A = {x SUCH-THAT
> x is a positive integer little than 4}    True...if zero is considered a positive integer.

OK.



>
> 2°) Same questions with the set A = {0, 1, 2, 3, ... , 61, 62, 63}
1. True
2. True
3. False
4. True: same question as 4 above.
5. True
6. False: zero is a number
7. False

OK. 
OK.
OK. 
OK.
OK. 
OK.
OK.

Very good. A "sans faute" we would say in french. In cyclism, you would be suspected taking drug! Bravo!

next lesson: (but take your time)


Could you tell me if you understand and/or remember those definitions (where a and b denoting arbitrary sets):

(a INTERSECTION b) = {x SUCH-THAT (x BELONGS-TO a) and (x BELONGS-TO b)}

(a UNION b) = {x SUCH THAT (x BELONGS-TO a) or (x BELONGS-TO b)}

Can you compute

{1, 2, 7, 789} UNION {1, 2, 7, 5678} = ?
{1, 2, 7, 789} INTERSECTION {1, 2, 7, 5678} = ?

Do you remember the empty set? Can you compute:

{1, 2} UNION { } = ?
{1} UNION { } = ?
{1, 2, 3} UNION {1, 2, 3} = ?
{ } UNION { } = ?
{1, 2} INTERSECTION { } = ?
{1} INTERSECTION { } = ?
{1, 2, 3} INTERSECTION {1, 2, 3} = ?
{ } INTERSECTION { } = ?


Now, an important distinction which will follow us through ... forever.  I suggest you read attentively the next two paragraphs two times before breakfast, every day for one week. :), Really take all your time. It concerns the notion of operation, and relation.

INTERSECTION and UNION, are operations on sets, like addition (+, or PLUS) and multiplication (*, or TIMES) are operation on numbers. This means, typically, that, if x and y denote numbers, then x + y, and x * y, will denote, or are equal to, numbers. For example 3 + 4 is equal to 7.
Similarly, if x and y denotes, or are equal, to sets, then x INTERSECTION y denotes, or is equal to, some set. For example {1,2} INTERSECTION {2, 7} is equal to some set, actually the set {2}. OK?

Operations are important, as you can guess, but relations are as well important. Operations lead to new elements, new objects. From the numbers 2 and 3, you get the element 5. Relations pertains or does not pertain, or equivalently, leads to true or false. 

Example. The relation LESS-THAN, among the numbers. (x LESS-THAN y) is true if x is less than y. So (3 LESS-THAN 56) is true, and (56 LESS-THAN 3) is false. An important relation pertaining on sets is the relation of inclusion, or of being a subset of a set.

By definition a set x will be said included in y (or be said subset of y), when all the elements of x are among the elements of y. We will write (x INCLUDED-IN y) when the set x is included in the set y.
For example, the set {1, 2} is included in the set {3, 2, 1}, but is not included in the set {3, 1}.

Exercise: in the following, what is true or false?

45 LESS-THAN 67
0 LESS-THAN 1
999 LESS-THAN 4
{1, 2, 3} INCLUDED-IN {4, 1, 5, 2, 3, 8}
{1} INCLUDED-IN {1, 2}


oops, I must go. You are lucky ;) 

Bruno






m.a.

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Jul 2, 2009, 12:42:49 PM7/2/09
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Could you tell me if you understand and/or remember those definitions (where a and b denoting arbitrary sets):

(a INTERSECTION b) = {x SUCH-THAT (x BELONGS-TO a) and (x BELONGS-TO b)}

(a UNION b) = {x SUCH THAT (x BELONGS-TO a) or (x BELONGS-TO b)}

Can you compute

{1, 2, 7, 789} UNION {1, 2, 7, 5678} = ?  1,2,7,789, 5678
{1, 2, 7, 789} INTERSECTION {1, 2, 7, 5678} = ? 1, 2, 7, 789
 
Do you remember the empty set? Can you compute:

{1, 2} UNION { } = ?  1,2
{1} UNION { } = ? { }
{1, 2, 3} UNION {1, 2, 3} = ? 1,2,3
{ } UNION { } = ? { }
{1, 2} INTERSECTION { } = ? { }
{1} INTERSECTION { } = ? { }
{1, 2, 3} INTERSECTION {1, 2, 3} = ? 1, 2 3
{ } INTERSECTION { } = ?  { }


Now, an important distinction which will follow us through ... forever.  I suggest you read attentively the next two paragraphs two times before breakfast, every day for one week. :), Really take all your time. It concerns the notion of operation, and relation.

INTERSECTION and UNION, are operations on sets, like addition (+, or PLUS) and multiplication (*, or TIMES) are operation on numbers. This means, typically, that, if x and y denote numbers, then x + y, and x * y, will denote, or are equal to, numbers. For example 3 + 4 is equal to 7.
Similarly, if x and y denotes, or are equal, to sets, then x INTERSECTION y denotes, or is equal to, some set. For example {1,2} INTERSECTION {2, 7} is equal to some set, actually the set {2}. OK?

Operations are important, as you can guess, but relations are as well important. Operations lead to new elements, new objects. From the numbers 2 and 3, you get the element 5. Relations pertains or does not pertain, or equivalently, leads to true or false. 

Example. The relation LESS-THAN, among the numbers. (x LESS-THAN y) is true if x is less than y. So (3 LESS-THAN 56) is true, and (56 LESS-THAN 3) is false. An important relation pertaining on sets is the relation of inclusion, or of being a subset of a set.

By definition a set x will be said included in y (or be said subset of y), when all the elements of x are among the elements of y. We will write (x INCLUDED-IN y) when the set x is included in the set y.
For example, the set {1, 2} is included in the set {3, 2, 1}, but is not included in the set {3, 1}.

Exercise: in the following, what is true or false?

45 LESS-THAN 67  true
0 LESS-THAN 1   true
999 LESS-THAN 4  false
{1, 2, 3} INCLUDED-IN {4, 1, 5, 2, 3, 8} true
{1} INCLUDED-IN {1, 2} true

Bruno Marchal

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Jul 2, 2009, 1:44:46 PM7/2/09
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You are quick!


On 02 Jul 2009, at 18:42, m.a. wrote:

 
 
Could you tell me if you understand and/or remember those definitions (where a and b denoting arbitrary sets):

(a INTERSECTION b) = {x SUCH-THAT (x BELONGS-TO a) and (x BELONGS-TO b)}

(a UNION b) = {x SUCH THAT (x BELONGS-TO a) or (x BELONGS-TO b)}

Can you compute

{1, 2, 7, 789} UNION {1, 2, 7, 5678} = ?  1,2,7,789, 5678
Almost OK. 
{1, 2, 7, 789} UNION {1, 2, 7, 5678} = {1,2,7,789, 5678}. 
Don't forget the accolades, which means that you have as result the SET {1,2,7,789, 5678}

{1, 2, 7, 789} INTERSECTION {1, 2, 7, 5678} = ? 1, 2, 7, 789

Not correct. To belong to A INTERSECTION B, the element must belong to A, *and* must belong to B. 1, 2 and 7 does belong indeed to A and to B, in this case, with A = {1, 2, 7, 789}, and B = {1, 2, 7, 5678}), but neither 789, nor 5678 do belong to both A and B.
So {1, 2, 7, 789} INTERSECTION {1, 2, 7, 5678} = {1, 2, 7}

Just tell me if you agree.

 
Do you remember the empty set? Can you compute:
{1, 2} UNION { } = ?  1,2


OK, but don't forget the accolades. 
{1, 2} UNION { } = ?  {1,2}


{1} UNION { } =  { }
You are too quick here, you forget to type the 1. 
{1} UNION { } =  {1 }


{1, 2, 3} UNION {1, 2, 3} = ? 1,2,3
OK (my mind adds the accolades)

{ } UNION { } = ? { }
Very good. You could eliminate the "?".

{1, 2} INTERSECTION { } = ? { }

Excellent.


{1} INTERSECTION { } = ? { }

Bravo.


{1, 2, 3} INTERSECTION {1, 2, 3} = ? 1, 2 3

Exact. (well, I continue to add the accolades, and eliminate the "?")

{ } INTERSECTION { } = ?  { }

Exact. In this case you see how much it is important to not forget the accolades!




Now, an important distinction which will follow us through ... forever.  I suggest you read attentively the next two paragraphs two times before breakfast, every day for one week. :), Really take all your time. It concerns the notion of operation, and relation.

INTERSECTION and UNION, are operations on sets, like addition (+, or PLUS) and multiplication (*, or TIMES) are operation on numbers. This means, typically, that, if x and y denote numbers, then x + y, and x * y, will denote, or are equal to, numbers. For example 3 + 4 is equal to 7.
Similarly, if x and y denotes, or are equal, to sets, then x INTERSECTION y denotes, or is equal to, some set. For example {1,2} INTERSECTION {2, 7} is equal to some set, actually the set {2}. OK?

Operations are important, as you can guess, but relations are as well important. Operations lead to new elements, new objects. From the numbers 2 and 3, you get the element 5. Relations pertains or does not pertain, or equivalently, leads to true or false. 

Example. The relation LESS-THAN, among the numbers. (x LESS-THAN y) is true if x is less than y. So (3 LESS-THAN 56) is true, and (56 LESS-THAN 3) is false. An important relation pertaining on sets is the relation of inclusion, or of being a subset of a set.

By definition a set x will be said included in y (or be said subset of y), when all the elements of x are among the elements of y. We will write (x INCLUDED-IN y) when the set x is included in the set y.
For example, the set {1, 2} is included in the set {3, 2, 1}, but is not included in the set {3, 1}.

Exercise: in the following, what is true or false?

45 LESS-THAN 67  true

OK.


0 LESS-THAN 1   true

OK.


999 LESS-THAN 4  false

OK.

{1, 2, 3} INCLUDED-IN {4, 1, 5, 2, 3, 8} true

OK.


{1} INCLUDED-IN {1, 2} true

OK.



oops, I must go. You are lucky ;) 

I'm back!  I give you two last exercises to ponder about, just  in case of insomnia. Again, take your time. I hope Kim follows, and does not look at the solution ! 


1°) In the two relational formula below, one is true, the other is false. Which one are what?

a)    { } INCLUDED-IN { }
b)    { } BELONGS-TO { } 

2°) And I give you a slightly longer exercise. Can you give me all the subsets of the set {1, 2} ?. That is, can you give me all the sets which are included in the set {1, 2} ? In case of doubt, reread the definitions, reread the examples, and never panic! I give you a hint: the set {1, 2} has four subsets. Can you find them?

Good job, Marty. 

Bruno


Bruno Marchal

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Jul 2, 2009, 1:52:55 PM7/2/09
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On 02 Jul 2009, at 00:22, John Mikes wrote:

> I don't deny the practicality of applying 'numbers-based' science in
> sending a man to Mars, but it is NOT the numbers that does the job.
> It is the complexity of the state of the art we reached, which
> includes science, technology, skills, ideas AND of course numbers-
> application. Bohm's idea - as I understood it - was that searching
> nature, you do not bounce into numbers, you can observe 3-leaf or
> 4legged and manyshaped things, big and small, YOU (the human) can
> 'count them' if you invented the symbols 1 2 3 4 etc. but these
> refer to quantities and it required lots of abstracting in mental
> evolution to arrive in a numbers-based math - how humans think about
> nature.


I know well that theory. It is based on the idea that some primary
Nature exists. A common "superstition" among christians and atheists.
Which could be true, actually. I don't know.

But what I am almost completely sure, is that if comp is true, then it
is has to be supersitution. And that is what I try to explain.


> Thanks again and my mind works in crooked ways, if you can excuse me
> for that. It seems I need too much learning to catch up.


You are welcome. If you have the time and courage, I really encourage
you to follow the thread. You may be surprised ... soon!

Bruno

http://iridia.ulb.ac.be/~marchal/

m.a.

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Jul 2, 2009, 6:18:48 PM7/2/09
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Bruno,
              Comments and questions are interspersed below.  
                                                                                                      marty
----- Original Message -----
Sent: Thursday, July 02, 2009 1:44 PM
Subject: Re: The seven step series

You are quick!


On 02 Jul 2009, at 18:42, m.a. wrote:

 
 
Could you tell me if you understand and/or remember those definitions (where a and b denoting arbitrary sets):

(a INTERSECTION b) = {x SUCH-THAT (x BELONGS-TO a) and (x BELONGS-TO b)}

(a UNION b) = {x SUCH THAT (x BELONGS-TO a) or (x BELONGS-TO b)}

Can you compute

{1, 2, 7, 789} UNION {1, 2, 7, 5678} = ?  1,2,7,789, 5678
Almost OK. 
{1, 2, 7, 789} UNION {1, 2, 7, 5678} = {1,2,7,789, 5678}. 
Don't forget the accolades, which means that you have as result the SET {1,2,7,789, 5678}

{1, 2, 7, 789} INTERSECTION {1, 2, 7, 5678} = ? 1, 2, 7, 789

Not correct. To belong to A INTERSECTION B, the element must belong to A, *and* must belong to B. 1, 2 and 7 does belong indeed to A and to B, in this case, with A = {1, 2, 7, 789}, and B = {1, 2, 7, 5678}), but neither 789, nor 5678 do belong to both A and B.
So {1, 2, 7, 789} INTERSECTION {1, 2, 7, 5678} = {1, 2, 7}

Just tell me if you agree.     I agree and can't understand how I could have been so careless.


 
Do you remember the empty set? Can you compute:
{1, 2} UNION { } = ?  1,2


OK, but don't forget the accolades.    Are accolades brackets?
{1, 2} UNION { } = ?  {1,2}


{1} UNION { } =  { }
You are too quick here, you forget to type the 1. 
{1} UNION { } =  {1 }             Yes, I mistook the {1} for the number of the question...not part of the equation. I tend to overlook the fine points.


{1, 2, 3} UNION {1, 2, 3} = ? 1,2,3
OK (my mind adds the accolades)

{ } UNION { } = ? { }
Very good. You could eliminate the "?".

{1, 2} INTERSECTION { } = ? { }

Excellent.


{1} INTERSECTION { } = ? { }

Bravo.


{1, 2, 3} INTERSECTION {1, 2, 3} = ? 1, 2 3

Exact. (well, I continue to add the accolades, and eliminate the "?")

{ } INTERSECTION { } = ?  { }

Exact. In this case you see how much it is important to not forget the accolades!




Now, an important distinction which will follow us through ... forever.  I suggest you read attentively the next two paragraphs two times before breakfast, every day for one week. :), Really take all your time. It concerns the notion of operation, and relation.

INTERSECTION and UNION, are operations on sets, like addition (+, or PLUS) and multiplication (*, or TIMES) are operation on numbers. This means, typically, that, if x and y denote numbers, then x + y, and x * y, will denote, or are equal to, numbers. For example 3 + 4 is equal to 7.
Similarly, if x and y denotes, or are equal, to sets, then x INTERSECTION y denotes, or is equal to, some set. For example {1,2} INTERSECTION {2, 7} is equal to some set, actually the set {2}. OK?......No!
 
                                                                                                                                                                                                                          Why not the sets {1,2,7} if INTERSECTION means BOTH?

Operations are important, as you can guess, but relations are as well important. Operations lead to new elements, new objects. From the numbers 2 and 3, you get the element 5. Relations pertains or does not pertain, or equivalently, leads to true or false. 

Example. The relation LESS-THAN, among the numbers. (x LESS-THAN y) is true if x is less than y. So (3 LESS-THAN 56) is true, and (56 LESS-THAN 3) is false. An important relation pertaining on sets is the relation of inclusion, or of being a subset of a set.

By definition a set x will be said included in y (or be said subset of y), when all the elements of x are among the elements of y. We will write (x INCLUDED-IN y) when the set x is included in the set y.
For example, the set {1, 2} is included in the set {3, 2, 1}, but is not included in the set {3, 1}.

Exercise: in the following, what is true or false?

45 LESS-THAN 67  true

OK.


0 LESS-THAN 1   true

OK.


999 LESS-THAN 4  false

OK.

{1, 2, 3} INCLUDED-IN {4, 1, 5, 2, 3, 8} true

OK.


{1} INCLUDED-IN {1, 2} true

OK.



oops, I must go. You are lucky ;) 

I'm back!  I give you two last exercises to ponder about, just  in case of insomnia. Again, take your time. I hope Kim follows, and does not look at the solution ! 


1°) In the two relational formula below, one is true, the other is false. Which one are what?

a)    { } INCLUDED-IN { }True
b)    { } BELONGS-TO { } True

2°) And I give you a slightly longer exercise. Can you give me all the subsets of the set {1, 2} ?. That is, can you give me all the sets which are included in the set {1, 2} ? In case of doubt, reread the definitions, reread the examples, and never panic! I give you a hint: the set {1, 2} has four subsets. Can you find them?
 
                                                                         {1} {2} {1,2} {2,1}     why not {3} ?

Bruno Marchal

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Jul 3, 2009, 4:07:14 AM7/3/09
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Hi Marty,


On 03 Jul 2009, at 00:18, m.a. wrote:

Bruno,
              Comments and questions are interspersed below.  
                                                                                                      marty

<snip>

Just tell me if you agree.     I agree and can't understand how I could have been so careless.

So you have to introspect a little bit. 




 
Do you remember the empty set? Can you compute:
{1, 2} UNION { } = ?  1,2


OK, but don't forget the accolades.    Are accolades brackets?

Due to Dirac, in Quantum Mechanics, I tend to believe that brackets are "<" and ">". parentheses are "(" and ")". I call "{" and "}" accolades, but perhaps they are called bracket. The terms are not important as far as we understand each other. How would you call "[" and "]" ?


You are too quick here, you forget to type the 1. 
{1} UNION { } =  {1 }             Yes, I mistook the {1} for the number of the question...not part of the equation. I tend to overlook the fine points.

It is good that you are aware of that facts. It is what matter. It is what will make it possible to develop the familiarity with the important fine points which can arise from time to time.




Now, an important distinction which will follow us through ... forever.  I suggest you read attentively the next two paragraphs two times before breakfast, every day for one week. :), Really take all your time. It concerns the notion of operation, and relation.

INTERSECTION and UNION, are operations on sets, like addition (+, or PLUS) and multiplication (*, or TIMES) are operation on numbers. This means, typically, that, if x and y denote numbers, then x + y, and x * y, will denote, or are equal to, numbers. For example 3 + 4 is equal to 7.
Similarly, if x and y denotes, or are equal, to sets, then x INTERSECTION y denotes, or is equal to, some set. For example {1,2} INTERSECTION {2, 7} is equal to some set, actually the set {2}. OK?......No!
 
                                                                                                                                                                                                                          Why not the sets {1,2,7} if INTERSECTION means BOTH?

Ah, but the word "both" alone is ambiguous. You could say that the UNION of two sets is the merging of BOTH set, and the intersection is the given of the elements which are in both set. So the union of {1, 2} and {2, 7} is {1, 2, 7}, which indeed merges BOTH sets. But for computing the intersection, you must ask yourself, does this *element* belongs to BOTH set? So, for the intersection of {1, 2} and {2, 7}, you have to ask yourself the following question: does 1 belong to both set? well, the answer is NO. the 1 belongs to the first set but not to the second, and so 1 does not belong to the intersection. Does 2 belongs to both sets? The answer is yes. 2 belongs to {1, 2} and 2 belongs to {2, 7}. Does 7 belongs to both sets, the answer is no, 7 belongs to the second set, but does not belong  to the first set, so 7 is not in the intersection.
Tell me if you are OK with this.



I'm back!  I give you two last exercises to ponder about, just  in case of insomnia. Again, take your time. I hope Kim follows, and does not look at the solution ! 


1°) In the two relational formula below, one is true, the other is false. Which one are what?

a)    { } INCLUDED-IN { }True


Very good. All elements of { } are among the elements of { }. This is sometimes said to be "trivially" true, because { } has no elements at all.
This is an example of an important "fine point".  Examples:

To verify if the set {1, 2, 3} is included in {34, 56, 7, 2, 100, 1, 45, 3, 4}, you have to check THREE things: does 1 belongs to the second set, does 2 belongs to the second set, does 3 belongs to the second set.

To verify if the set {1, 2} is included in {34, 56, 7, 2, 100, 1, 45, 3, 4}, you have to check TWO things: does 1 belongs to the second set, does 2 belongs to the second set.

To verify if {1} is included in {34, 56, 7, 2, 100, 1, 45, 3, 4}, you have to check ONE things: does 1 belongs to the second set.

To verify if { } is included in {34, 56, 7, 2, 100, 1, 45, 3, 4}, you have to verify ZERO thing! So it is automatically true. That is why logicians say it is trivially true. 

From this you should understand that the empty set, { }, is included to any set.

So { } is included in all the sets:  { }, {1}, {1, 2}, ... {0, 1, 2, ...}, ....

In particular, as you said correctly, { } is included in { }. Put in another way, ({ } INCLUDED-IN { }) = true.


b)    { } BELONGS-TO { } True

NOT correct. Remember that the empty set is empty, so nothing belongs to it. All formula like (x belongs to { }) will be false. You can conceive a set as an empty box { }, in which you can fill elements. So the set {a, b} is the empty set in which you put the elements a, and then, the element b. The accolades "{" and "}" represents the box itself, and what is in between the accolades represents the elements of the set. You could have guessed the solution because I was helping you when saying that one of the proposition is true and the other is false, and this means that, like many beginners, you read the enunciation of the problem too much quickly. That is why I suggest you take your time, and read often, at different time, the enunciation of the problems, and actually all explanations as well.

The moral is:

"x belongs to { }"     is never true, or is always false, whatever x represents.
"{ } included in x"    is always true, or never false, whatever x represents.



2°) And I give you a slightly longer exercise. Can you give me all the subsets of the set {1, 2} ?. That is, can you give me all the sets which are included in the set {1, 2} ? In case of doubt, reread the definitions, reread the examples, and never panic! I give you a hint: the set {1, 2} has four subsets. Can you find them?
 
                                                                         {1} {2} {1,2} {2,1}     why not {3} ?

Not too bad. 3/4 correct:  

{1} is included in {1, 2}.  Indeed.
{2} is included in {1, 2}. Indeed.
{1, 2} is included in {1, 2}. Indeed.

{2, 1} is included in {1, 2}. Indeed, that is true, but you have to remember what you have already agree on: the set {1, 2} is equal to the set {2, 1}, so this is not a new solution. It is the preceding one in disguised!

Why not {3}? {3} is not included in {1, 2} just because 3 does not belong to {1, 2}. Reread the definition of inclusion. A is included in B if all the elements of A belongs to B. OK?

So you have found three subsets, among the four. Reading today's explanations I think you could find the missing subset. I let you search a little bit.

So just one exercise: what is the missing subset?

I apologize if all of this is a bit boring, but soon enough it will be highly rewarding. You will see.

Bruno









Brent Meeker

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Jul 3, 2009, 12:59:52 PM7/3/09
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Bruno Marchal wrote:
> ...

> Due to Dirac, in Quantum Mechanics, I tend to believe that brackets
> are "<" and ">". parentheses are "(" and ")". I call "{" and "}"
> accolades, but perhaps they are called bracket. The terms are not
> important as far as we understand each other. How would you call "["
> and "]" ?

I've never seen "{" and "}" denoted "accolades" but I like it; they are
more commonly called "braces". I don't know a specific term for "["
and "]", I generally refer to them as "square brackets".

Brent

m.a.

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Jul 3, 2009, 10:31:34 PM7/3/09
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New comments in italics.
 For example {1,2} INTERSECTION {2, 7} is equal to some set, actually the set {2}. OK?......No!
 
                                                                                                      Why not the sets {1,2,7} if INTERSECTION means BOTH?

Ah, but the word "both" alone is ambiguous. You could say that the UNION of two sets is the merging of BOTH set, and the intersection is the given of the elements which are in both set. So the union of {1, 2} and {2, 7} is {1, 2, 7}, which indeed merges BOTH sets. But for computing the intersection, you must ask yourself, does this *element* belongs to BOTH set? So, for the intersection of {1, 2} and {2, 7}, you have to ask yourself the following question: does 1 belong to both set? well, the answer is NO. the 1 belongs to the first set but not to the second, and so 1 does not belong to the intersection. Does 2 belongs to both sets? The answer is yes. 2 belongs to {1, 2} and 2 belongs to {2, 7}. Does 7 belongs to both sets, the answer is no, 7 belongs to the second set, but does not belong  to the first set, so 7 is not in the intersection.
Tell me if you are OK with this.
                                                 Not OK. You previously defined UNION as one OR the other. Now you seem to be giving me the same definition for INTERSECTION.
Is the missing subset   { }    ?

Bruno Marchal

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Jul 4, 2009, 7:59:44 AM7/4/09
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On 04 Jul 2009, at 04:31, m.a. wrote:

New comments in italics.
 For example {1,2} INTERSECTION {2, 7} is equal to some set, actually the set {2}. OK?......No!
 
                                                                                                      Why not the sets {1,2,7} if INTERSECTION means BOTH?

Ah, but the word "both" alone is ambiguous. You could say that the UNION of two sets is the merging of BOTH set, and the intersection is the given of the elements which are in both set. So the union of {1, 2} and {2, 7} is {1, 2, 7}, which indeed merges BOTH sets. But for computing the intersection, you must ask yourself, does this *element* belongs to BOTH set? So, for the intersection of {1, 2} and {2, 7}, you have to ask yourself the following question: does 1 belong to both set? well, the answer is NO. the 1 belongs to the first set but not to the second, and so 1 does not belong to the intersection. Does 2 belongs to both sets? The answer is yes. 2 belongs to {1, 2} and 2 belongs to {2, 7}. Does 7 belongs to both sets, the answer is no, 7 belongs to the second set, but does not belong  to the first set, so 7 is not in the intersection.
Tell me if you are OK with this.
                                                 Not OK. You previously defined UNION as one OR the other. Now you seem to be giving me the same definition for INTERSECTION.

Let us take the set a = {1, 2} and the set b = {2, 3}.

Let us recall the definition (I abandon the capital letters because they are ugly, and I feel talking louder!)

1) intersection

(a intersection b) = {x such-that (x belongs-to a) AND (x belongs-to b) }.

So, some x, to belong to the intersection, has to belong simultaneously to the two sets involved. Only when x is equal to 2, is that condition verified.

2 belongs to (a intersection b) because 2 belongs to a, AND, 2 belongs to b.

That condition is not verify for x = 1, nor for x = 3. 3 belongs to b, but not to a. So 3 is not in the intersection. Nor 1, because it does not belong to a.

So (a intersection b) =  ( {1, 2} intersection {2, 3} ) = {2}.

2) Union

(a union b) = {x such-that (x belongs-to a) OR (x belongs-to b).

Does 1 belong to the union of a and b?  That is do we have that 1 belongs-to (a union b)? With same a and b as above.

Let us see.

Does 1 belongs to a union b?  Does 1 verify the condition written in the definition?  Do we have that (1 belongs-to a) OR (1 belongs to b))? 
A proposition shaped like P OR Q is true in the case one or both of P and Q is true. It is true that 1 belongs to {1, 2} OR to {2, 3}. A bit like "any number is odd or is not odd" is always true. So 1 is in the union.

2 is in the union, because it is true that 2 belongs to a or 2 belongs to b. Indeed 2 belongs to both of them. And 3 is in the union too, because iit belongs to one of them again, actually {2, 3}.

So (union b) = {1, 2, 3}.

OK?

Don't hesitate to tell me if it is not OK.




2°) And I give you a slightly longer exercise. Can you give me all the subsets of the set {1, 2} ?. That is, can you give me all the sets which are included in the set {1, 2} ? In case of doubt, reread the definitions, reread the examples, and never panic! I give you a hint: the set {1, 2} has four subsets. Can you find them?
 
                                                                         {1} {2} {1,2} {2,1}     why not {3} ?

Not too bad. 3/4 correct:  

{1} is included in {1, 2}.  Indeed.
{2} is included in {1, 2}. Indeed.
{1, 2} is included in {1, 2}. Indeed.

{2, 1} is included in {1, 2}. Indeed, that is true, but you have to remember what you have already agree on: the set {1, 2} is equal to the set {2, 1}, so this is not a new solution. It is the preceding one in disguised!

Why not {3}? {3} is not included in {1, 2} just because 3 does not belong to {1, 2}. Reread the definition of inclusion. A is included in B if all the elements of A belongs to B. OK?

So you have found three subsets, among the four. Reading today's explanations I think you could find the missing subset. I let you search a little bit.

So just one exercise: what is the missing subset?
 
Is the missing subset   { }    ?


Correct.

So the subsets of {1, 2} are { }, {1}, {2}, {1, 2}. 

Could you find all subsets of {1, 2, 3}?

And now I give you an exercise which is so much easy that you could panic, and so I will provide the solution. I have seen often that too much easy question can make a student panic, and then the prey of out-of place mockery, and useless loss of confidence.

The easy exercise: could you give me the set of subsets of {1, 2} ?

Solution: You already told me that the subsets of {1, 2} are { }, {1}, {2}, {1, 2}.  So, the set of subsets of {1, 2} is 

{ { }, {1}, {2}, {1, 2} }


OK? It is just the solutions you give me, enclosed by braces (accolades) "{",  "}". Look at the expression with a spectacle.
If we except the set of books on Brent Meeker's shell, up to now, we have met mainly set of numbers, like

{1}
{1, 2}
{0, 2, 4, 6, ...}
{0, 1, 2, 3, ...}

Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets. The set {2, 21, 14} is a set of numbers. The set { { }, {4, 78, 56} } is a set of sets. It has two elements: the empty set {}, and the set of numbers {4, 78, 56}. Do not confuse a number, like 24, and a set, like {24}, which is a set having a number has elements. In particular it is the case that  {4, 78, 56} belongs to { { }, {4, 78, 56} }. Take it easy, and meditate on the following exercise:

Which of the following are true

{3, 5} included-in {3, 5}
{3, 5} belongs-to {3, 5}
{3, 5} included-in { {3, 5} }
{3, 5} belongs-to { {3, 5} }

Take your time, 

Bruno




m.a.

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Jul 4, 2009, 9:17:06 AM7/4/09
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Bruno,
          Can you provide definitions of  "belongs-to"  and  "included-in" that distinguish them from "union" and "intersection"?
 
 
 
 
 
 
 

Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets. The set {2, 21, 14} is a set of numbers. The set { { }, {4, 78, 56} } is a set of sets. It has two elements: the empty set {}, and the set of numbers {4, 78, 56}. Do not confuse a number, like 24, and a set, like {24}, which is a set having a number has elements. In particular it is the case that  {4, 78, 56} belongs to { { }, {4, 78, 56} }. Take it easy, and meditate on the following exercise:

Which of the following are true

{3, 5} included-in {3, 5} True

Bruno Marchal

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Jul 4, 2009, 3:14:55 PM7/4/09
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On 04 Jul 2009, at 15:17, m.a. wrote:

Bruno,
          Can you provide definitions of  "belongs-to"  and  "included-in" that distinguish them from "union" and "intersection"?


"Belongs-to" and "included-in" are relations. Their value are true or false.


1) (x belongs-to A) means that the object x belongs to the set A.

Examples: 3 belongs-to {3, 4}, because 3 is an element of the set {3, 4}, or, put in another way (which can be useful for later):

 (3 belongs-to {3, 4}) = true


2) Similarly (x included-in y) is a relation bearing on sets, and (x included-in y) = true, means that x is included-in y, and this means that all elements of x are elements of y. You don't need to know more, but if you want you can define (x included-in y) by

For any z ((z belongs-to x) -> (z belongs-to y)). But I intended to introduce "->" later, so don't worry.

Example {3, 4} is included in {3,5,4} because all elements of {3, 4} are in {3, 5, 4}. We can write ({3,4} included-in {3,5,4}) = true. Another example is ({3, 4} included-in {3,6,7,9,567}) is false, because not all elements of {3,4} are in {3,6,7,9,567} (indeed 4 belongs to {3,4} and not to  {3,6,7,9,567}.

Union and intersection are not relation, but operation.

({3,4} union {4,5}) is not equal to true or to false, it is equal to a set: actually the set you get by the union of {3,4} with {4,5}, and this is {3,4,5}.
Likewise, the intersection of {3,4} with {4,5}, that is ({3,4} union {4,5}) is not true or false, but is equal to {4}. 

So:

({3,4} included-in {3,4,5}) = true
({3,4} union {3,4,5}) = {3,4,5}

You see the difference. It is the difference between "the brother of Paul", which denotes a human. and "Paul is greater than Julia", which is true or false.

Or, on the numbers, less-than (<)  is a relation, and addition and multiplication are operation:

(3 <  7) = true
(3+4) = 7

(7 <  3) = false
(7*3) = 21

({ } included-in {3,4}) = true
({1,2} intersection {2,7}) = {2}.

Does this help?

Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets. The set {2, 21, 14} is a set of numbers. The set { { }, {4, 78, 56} } is a set of sets. It has two elements: the empty set {}, and the set of numbers {4, 78, 56}. Do not confuse a number, like 24, and a set, like {24}, which is a set having a number has elements. In particular it is the case that  {4, 78, 56} belongs to { { }, {4, 78, 56} }. Take it easy, and meditate on the following exercise:

Which of the following are true

{3, 5} included-in {3, 5} True

This is correct.


{3, 5} belongs-to {3, 5} 
{3, 5} included-in { {3, 5} }
{3, 5} belongs-to { {3, 5} }

Take your time, 

Bruno








m.a.

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Jul 6, 2009, 10:12:43 AM7/6/09
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My answers.    m.a.
 
Bruno,
          Can you provide definitions of  "belongs-to"  and  "included-in" that distinguish them from "union" and "intersection"?
 
 
 
 
 
 
 

Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets. The set {2, 21, 14} is a set of numbers. The set { { }, {4, 78, 56} } is a set of sets. It has two elements: the empty set {}, and the set of numbers {4, 78, 56}. Do not confuse a number, like 24, and a set, like {24}, which is a set having a number has elements. In particular it is the case that  {4, 78, 56} belongs to { { }, {4, 78, 56} }. Take it easy, and meditate on the following exercise:

Which of the following are true

{3, 5} included-in {3, 5} True
 
{3, 5} belongs-to {3, 5} True
{3, 5} included-in { {3, 5} } False
{3, 5} belongs-to { {3, 5} } True

Bruno Marchal

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Jul 6, 2009, 12:14:20 PM7/6/09
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On 06 Jul 2009, at 16:12, m.a. wrote (in bold):

My answers.    m.a.
Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets. The set {2, 21, 14} is a set of numbers. The set { { }, {4, 78, 56} } is a set of sets. It has two elements: the empty set {}, and the set of numbers {4, 78, 56}. Do not confuse a number, like 24, and a set, like {24}, which is a set having a number has elements. In particular it is the case that  {4, 78, 56} belongs to { { }, {4, 78, 56} }. Take it easy, and meditate on the following exercise:

Which of the following are true

{3, 5} included-in {3, 5} True

OK.



 
{3, 5} belongs-to {3, 5} True

Not OK. The elements of {3, 5} are 3 and 5. {3, 5} is not an *element* of {3, 5}. 
Ask in case you are not OK with this, of course.




{3, 5} included-in { {3, 5} } False

OK. Very good.


{3, 5} belongs-to { {3, 5} } True

OK. {3, 5} is even the *only* element of  { {3, 5} }


No exercise today. Just a question, a suggestion, and a plan.

The question is: have you the feeling to learn something?

The suggestion: I think the best way to answer the preceding question consists in trying to explain what you learn to someone else. It is the best way to see if you remember and understand the definition. You could try to explain what you learn to some gentle "victim" in your neighborhood (wife, friend, child, parent, ...).

I give you a plan, and some more motivation. To get the seventh step in some proper way, there is a need to understand the mathematical notion of "universal machine". For this I need to explain what is a computable function. For this I need to explain what is a function, and for this I need to explain what is a set, given that functions can more easily be explained through sets relating sets. Once you will have a good grip of what is a universal machine, or what is a universal number, and what really means "universal", we will be able to tackle the notion of universal dovetailing, and especially the "mathematical universal dovetailing" (which is really important for the whole approach, and for the step eight). I am hesitating to work quickly on the notion of function, or to do some pieces of number theory and geometry to provide examples before.

As I said recently to John, the discovery of the notion universal machine is one of the most astonishing and gigantic discovery made by the humans, and what I do is just an exploitation of that discovery. Universes, cells, brains and computers are example of universal machine, and the notion of universal machine are a key to understand why eventually, once we say "yes to the doctor", and believe we can survive "qua computatio", we have to redefine physics as an invariant for the permutation of all possible observers, and how physics can be recovered from an invariant among all universal machines point-of-views ...

Feel free to slow me down, or to accelerate me, and to ask any question at whichever level of details you want. Feel free to ask any question that you have already asked.

Have a good day, and thanks for your effort and seriousness,

Bruno

PS. It should be obvious for everyone that if there are still questions, critics, objections, problems, feeling of dizziness, whatever, with the first six steps of the UDA, please, feel free to ask. And people should not hesitate to discuss other everything-like subject, I don't want to monopolize the list of course. But the UDA reasoning really changes the perspective on all possible TOEs, so I will feel free myself to point on UDA on each discussion where I find it relevant (of course also).




m.a.

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Jul 6, 2009, 10:03:41 PM7/6/09
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Questions and comments interspersed below (in bold).
 
----- Original Message -----
Sent: Monday, July 06, 2009 12:14 PM
Subject: Re: The seven step series


On 06 Jul 2009, at 16:12, m.a. wrote (in bold):

My answers.    m.a.
Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets. The set {2, 21, 14} is a set of numbers. The set { { }, {4, 78, 56} } is a set of sets. It has two elements: the empty set {}, and the set of numbers {4, 78, 56}. Do not confuse a number, like 24, and a set, like {24}, which is a set having a number has elements. In particular it is the case that  {4, 78, 56} belongs to { { }, {4, 78, 56} }. Take it easy, and meditate on the following exercise:

Which of the following are true

{3, 5} included-in {3, 5} True

OK.



 
{3, 5} belongs-to {3, 5} True

Not OK. The elements of {3, 5} are 3 and 5. {3, 5} is not an *element* of {3, 5}.Why not? They look like elements to me. Please define "elements" as applies to this example.. 
Ask in case you are not OK with this, of course.




{3, 5} included-in { {3, 5} } False

OK. Very good.


{3, 5} belongs-to { {3, 5} } True

OK. {3, 5} is even the *only* element of  { {3, 5} }


No exercise today. Just a question, a suggestion, and a plan.

The question is: have you the feeling to learn something?

The suggestion: I think the best way to answer the preceding question consists in trying to explain what you learn to someone else. It is the best way to see if you remember and understand the definition. You could try to explain what you learn to some gentle "victim" in your neighborhood (wife, friend, child, parent, ...).

I give you a plan, and some more motivation. To get the seventh step in some proper way, there is a need to understand the mathematical notion of "universal machine".
 
I've read about Turing machines if that's what you're referring to.
 
For this I need to explain what is a computable function. For this I need to explain what is a function, and for this I need to explain what is a set, given that functions can more easily be explained through sets relating sets. Once you will have a good grip of what is a universal machine, or what is a universal number, and what really means "universal", we will be able to tackle the notion of universal dovetailing, and especially the "mathematical universal dovetailing" (which is really important for the whole approach, and for the step eight). I am hesitating to work quickly on the notion of function, or to do some pieces of number theory and geometry to provide examples before.

As I said recently to John, the discovery of the notion universal machine is one of the most astonishing and gigantic discovery made by the humans, and what I do is just an exploitation of that discovery. Universes, cells, brains and computers are example of universal machine, and the notion of universal machine are a key to understand why eventually, once we say "yes to the doctor", and believe we can survive "qua computatio", we have to redefine physics as an invariant for the permutation of all possible observers, and how physics can be recovered from an invariant among all universal machines point-of-views ...

Feel free to slow me down, or to accelerate me, and to ask any question at whichever level of details you want. Feel free to ask any question that you have already asked.

Have a good day, and thanks for your effort and seriousness,

Bruno

PS. It should be obvious for everyone that if there are still questions, critics, objections, problems, feeling of dizziness, whatever, with the first six steps of the UDA, please, feel free to ask. And people should not hesitate to discuss other everything-like subject, I don't want to monopolize the list of course. But the UDA reasoning really changes the perspective on all possible TOEs, so I will feel free myself to point on UDA on each discussion where I find it relevant (of course also).


Bruno Marchal

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Jul 7, 2009, 3:30:12 AM7/7/09
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On 07 Jul 2009, at 04:03, m.a. wrote:

Questions and comments interspersed below (in bold)
{3, 5} belongs-to {3, 5} True

Not OK. The elements of {3, 5} are 3 and 5. {3, 5} is not an *element* of {3, 5}.Why not? They look like elements to me. Please define "elements" as applies to this example.. 

Think about a set as it is a sort of box. For example the set {3, 5} can be seen as the empty box { } in which you place the object 3, and then the object 5. In this case the set {3, 5} contains two elements, 3 and 5, which appears to be natural numbers. In particular the set {3, 5} contains only numbers.

So if I ask you if 3 and 5 are elements of {3, 5}, the answer is TRUE, and I guess that this is how you have interpret the question.

But the question was not "is 3 and 5 elements of {3, 5}?". The question was "is {3, 5} an element of {3, 5}"? This is really the question: "is the mathematical object {3, 5}, which is a set,  an element of {3, 5}?"; But just above we have seen that {3, 5} contains only numbers, and the object {3, 5} is not a number (indeed it is a set), and there is not set in {3,5}, only numbers.

Look: {3, 5} is a box which contains two numbers, 3 and 5, and nothing else.
A set in which {3, 5} would be itself an element would be, for example {7, 8, {3, 5}}, which can be seen as a box which contains three things, the number 3, the number 5, and the box {3, 5}.   {7, 8, {3, 5}} is an hybrid set which contains two numbers and a set. 

Do you see the difference between { }, the empty box, and {{ }}, which is a box which contains the empty box. If you put an empty box in a box, that box is no more empty: it contains an empty box. OK? All the interest of the notion of set, is that it makes a "many" into a "one". {3, 5} is the mathematical unique object, a set, which has 3 and 5 as element. And it can itself be an element of another set, like {{3, 5}}, or {{3, 5}, 7}.

You were confusing the question:

- Are the numbers 3, 5 elements of {3, 5}?  (answer: yes)
- Is the set  {3, 5} an element of {3, 5}?  (answer: no).

I give you more examples:

3 belongs-to {0, 1, 2, 3, 4}   TRUE.
{3} belongs to {0, 1, 2, 3, 4}  FALSE
{3} belongs-to {{0}, {1}, {2}, {3}, {4}}  TRUE
{3} belongs-to {0, 1, 2, {3}, 4} TRUE
{3} belongs-to {0, 1, 2, {3, 4}} FALSE
{3,4} belongs-to {0, 1, 2, {3, 4}} TRUE
{3, 4} belongs-to {{0, 1, 2} {3, 4}} TRUE

Tell me if you are OK with those examples. Keep in mind typical situation, like:

 {2, 3} is a set with two elements: the number 2, and 3.
{{2, 3}} is a set with one element: the set {2, 3}.



I give you a plan, and some more motivation. To get the seventh step in some proper way, there is a need to understand the mathematical notion of "universal machine".
 
I've read about Turing machines if that's what you're referring to.
Not exactly.  Turing machines are indeed "mathematical machine", and "universal Turing machine" do exist. But most Turing machine are not universal. And not all "universal machine" are Turing machine.

So the set of universal machines has a non empty intersection with the set of Turing machines, but that is the most we can say. Some Turing machine are not universal, and some universal machine are not Turing machine. But here we are anticipating.

Bruno



m.a.

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Jul 7, 2009, 10:18:32 AM7/7/09
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Thanks, Bruno. I think I've got it now. Sorry to be such a slow learner.       
                                                                                                                   marty
----- Original Message -----
Sent: Tuesday, July 07, 2009 3:30 AM
Subject: Re: The seven step series


Bruno Marchal

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On 07 Jul 2009, at 16:18, m.a. wrote:

Thanks, Bruno. I think I've got it now. Sorry to be such a slow learner.       
                                                                                                                   marty


Slow? It seems to me that you are rather quick. A scientist friend on mine took 4 years to get the difference between "belongs-to" and "included-in". 
Of course it will take more time you get friendly familiar with such matter, but that is completely normal. 
You are not slow at all, and I appreciate your seriousness and your courage, because you can say publicly "I don't understand, explain again". It is the big virtue which will help you to proceed.

Now, understanding is but one phase. You will have to remember what you learn. For this I suggest you do by yourself some summaries, and as I said it could help to try to explain to some others. But there is no problem to ask any question, or to suggest me to recall definitions and examples.
At some point I will sum up myself.

Oh, let me sum up a few bit before I introduce a new notion, and three exercise (but you can take some holiday before, take your time).

---

What are sets? Sets are sort of boxes which can contains anything, like numbers, or sets. Most of the set we have encounter were set of numbers, or set of sets. 

You can perhaps intuit some use of set in logic. For example saying that being a human makes you mortal can be analysed by the statement that the set of humans is included in the set of mortal beings. The proposition "Julia is a human" is equivalent with the proposition that Julia belongs to the set of humans. If I let H be a name for the set of humans, M be the name of the set of mortal beings, and j be a name for Julia, the fact that Julia is human, can be translated in "set theory" by 

(j belongs-to H), 

and the fact that being a human makes you mortal, can be translated by

(H included-in M);

Remember that (H included-in M) means that all elements of H are element of  M, and so it means that if j belongs-to H then j belongs-to M.

A logician would say that with the axioms (j belongs-to H) and (H included-in M), you can deduce that (j belongs-to M).

A logician never care if the axioms are true or false, he cares only on the validity of the reasoning. 

Remark. Personally, I don't believe that in "real life" there are sets, like those we can meet in math. Take the set of humans. Do we have a really a set ? An anti-computationalist could classify Julia as an inhuman zombie the day she got her artificial brain, so H is already different for a computationalist and an anti-computationalist!  In real life, sets can be locally useful, but it would be a sort of occamization, to quote John, (inspired by Russell) to apply the notion of set so straightforwardly. I have the same opinion for the use of set in mathematics, concerning the long run, but then I understand how much they make thinks far easier and clearer. Indeed they pervade all branches of math: topology, probability, algebra, logic, and computer science is no exception.  (I think they will disappear, but this will take millenia!)

------

Now it is time to do some exercise.

Do you remember, I asked you to give me all the subsets of {1, 2}. That is, all the sets which are included in {1, 2}. You gave me the correct answer: those subsets are { }, {1}, {2}, {1, 2}. You see that the set {1, 2} has 2 elements, and 4 subsets. But then I asked to give me the set of all subsets of {1, 2}. 
{1, 2} has four subsets, and it is natural to make that many a one, by considering *the* set of all subsets of {1, 2}. The answer is:

{{ }, {1}, {2}, {1, 2}} 

Considering all subsets of a set is a rather important operation, which we will meet more than one times in the sequel. Given its importance mathematicians gave it a name. It is the power operation. Later I will be able to explain why it is called power. 
It is an UNARY operation, which means it applies on ONE set. (Intersection, and union are BINARY operations, they need two sets to work on).

So (power x) = {y such-that y is included in x}, by definition.

For example:
(power {1, 2}) = {{ }, {1}, {2}, {1, 2}} 

Here are the three promised exercises. Compute


(power {1}) = ?
(power {1, 2, 3}) = ?
(power { }) = ?

Take your time,

Bruno





m.a.

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Jul 7, 2009, 8:28:17 PM7/7/09
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----- Original Message -----
Sent: Tuesday, July 07, 2009 2:07 PM
Subject: Re: The seven step series


On 07 Jul 2009, at 16:18, m.a. wrote:

 
Bruno,
 
I'm not entirely sure of these answers, but I think I learn more from your corrections than from pondering the rules to the point where I confuse myself.   m.a.

Do you remember, I asked you to give me all the subsets of {1, 2}. That is, all the sets which are included in {1, 2}. You gave me the correct answer: those subsets are { }, {1}, {2}, {1, 2}. You see that the set {1, 2} has 2 elements, and 4 subsets. But then I asked to give me the set of all subsets of {1, 2}. 
{1, 2} has four subsets, and it is natural to make that many a one, by considering *the* set of all subsets of {1, 2}. The answer is:

{{ }, {1}, {2}, {1, 2}} 

Considering all subsets of a set is a rather important operation, which we will meet more than one times in the sequel. Given its importance mathematicians gave it a name. It is the power operation. Later I will be able to explain why it is called power. 
It is an UNARY operation, which means it applies on ONE set. (Intersection, and union are BINARY operations, they need two sets to work on).

So (power x) = {y such-that y is included in x}, by definition.

For example:
(power {1, 2}) = {{ }, {1}, {2}, {1, 2}} 

Here are the three promised exercises. Compute


(power {1}) = ? {{ }, {1}}
(power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}}

Bruno Marchal

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Jul 8, 2009, 3:25:23 AM7/8/09
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Marty,


On 08 Jul 2009, at 02:28, m.a. wrote:

 
----- Original Message -----
Sent: Tuesday, July 07, 2009 2:07 PM
Subject: Re: The seven step series


On 07 Jul 2009, at 16:18, m.a. wrote:

 
Bruno,
 
I'm not entirely sure of these answers, but I think I learn more from your corrections than from pondering the rules to the point where I confuse myself.   m.a.

Hmm... You are not slow, but you may be a bit lazy :)
Pondering the rules to the point where you confuse yourself is necessary to develop ... well, the art of pondering the rules to the point where you confuse yourself, which is part of the work of the researcher.
But that is OK Marty, given that the goal here is just to give the necessary passive understanding of math so as making you able to grasp the seventh step of UDA. 



Do you remember, I asked you to give me all the subsets of {1, 2}. That is, all the sets which are included in {1, 2}. You gave me the correct answer: those subsets are { }, {1}, {2}, {1, 2}. You see that the set {1, 2} has 2 elements, and 4 subsets. But then I asked to give me the set of all subsets of {1, 2}. 
{1, 2} has four subsets, and it is natural to make that many a one, by considering *the* set of all subsets of {1, 2}. The answer is:

{{ }, {1}, {2}, {1, 2}} 

Considering all subsets of a set is a rather important operation, which we will meet more than one times in the sequel. Given its importance mathematicians gave it a name. It is the power operation. Later I will be able to explain why it is called power. 
It is an UNARY operation, which means it applies on ONE set. (Intersection, and union are BINARY operations, they need two sets to work on).

So (power x) = {y such-that y is included in x}, by definition.

For example:
(power {1, 2}) = {{ }, {1}, {2}, {1, 2}} 

Here are the three promised exercises. Compute


(power {1}) = ? {{ }, {1}}

Excellent.


(power { }) = ?   {{ }}

Excellent. People are often wrong on this one!


(power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}}


Here you are a bit lazy, as I said. You miss many subsets. Is not {1, 2} a subset of {1, 2, 3}? Is not {2, 3} a subset, and where is {1, 2, 3}?

I have to go right now, so I let you search, meanwhile, for the complete solution by yourself. I give you a hint (power {1,2,3}) has 8 elements.

And I give you a little subject research: if a set x has n elements, how many elements are in (power x)?

Bruno


m.a.

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Jul 8, 2009, 9:43:04 AM7/8/09
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Second try:
(power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}}


.

And I give you a little subject research: if a set x has n elements, how many elements are in (power x)? 
 
How's this for a wild guess? I have a feeling that it's missing the accolades, but I have no idea where to put them.
 
 
(power {x}) =  n(n-1) (n-2)...(n-x+1)
                                 x!

Bruno



Bruno Marchal

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Jul 8, 2009, 1:31:00 PM7/8/09
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On 08 Jul 2009, at 15:43, m.a. wrote:

Second try:
(power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}}

This is far better!  Not yet correct though.

I gave you the hint that there are 8 elements. Let us count: 

The empty set { }  ..................................1
Three singletons {1}, {2}, {3}................3
Two doubletons {1,2 }, {2,3 }................2
The biggest subset  {1,2,3}..................1

1 + 3 + 2 + 1 = 7

A subset is missing! Can you see which one?


And I give you a little subject research: if a set x has n elements, how many elements are in (power x)? 
 
How's this for a wild guess? I have a feeling that it's missing the accolades, but I have no idea where to put them 
(power {x}) =  n(n-1) (n-2)...(n-x+1)
                                 x!


Good intuition that there is a problem with the accolades. Although your expression is not *missing* accolades. Actually it has *too much* accolades. 

If x represents a set, for example the set {1,2}, it means that in the formula x can be substituted by {1,2}. we could write that x = {1, 2}, the accolades are in x, if you want. 
So on the left, you should have written (power x), like in the enunciation of the subject research, actually. 

If x is the set {1,2}, (power x) is (power {1, 2}). But (power {x}) is (power {{1,2,3}}), i.e. the powerset of {{1,2,3}}, which is {{ }, {{1,2,3}}}.
(Power {x}) looks like the powerset of an indeterminate singleton, a set with only one element.

You  could have written this:

When x has n elements, then (power x) = n(n-1) ... (n-x+1) / x!

Let us see. After all you already computes the powerset of { }, which is the set with 0 element, and you told me (power { })  = {{ }}. So it has one element, and your formula should confirm this, and ... well, your formula begins by n multiplied by something, if n = 0 then we will get 0,  because 0 times any number gives 0. But we have just seen that (power { }) = {{ }}, which is a set with 1 element. So your formula is already contradicted by the first example. Hmm...

May be that was bad luck, and sometimes in math the first example is also the trickiest, so let us look for n = 1. Let us take a set with one element, like {24}. Its power has 2 elements: {{ } {24}}, and you can guess that all singletons (set with two elements) have the same number of elements in their power set. So the answer is 2, in this case. If x has 1 element, the powerset of x, (power x) has two elements. Your formula should give 2, when n is equal to 1.
Let us see ...

it gives 1(1-1) ... (1 -

... but now, what could you mean by (n - x ...). ???


n is supposed to represent a number, x is supposed to represent a set, how could I, or you, subtract a set from a number?

So I'm afraid that your formula is senseless, although I will perhaps take the time, in some future, to explain why there *is* an atom of truth in it! The correct formula is much simpler, though.

Morality: if you have a theory, or a formula, test it on what you already know before submitting to publication!

To find the formula, you could try first the tedious brut force (if this is english). (Few mathematicians admit to do that, but all mathematicians do it!)

The number of sets included in { } = 1   (you have seen that). If x has 0 elements, (power x) = 1.
The number of sets included in  {a} = 2    (you have seen that). If x has 1 elements, (power x) = 2.
The number of sets included in {a, b} = 4  (you have seen that). If x has 2 elements (power x) = 4.
The number of sets included in {a,b,c} = 8 (cf the hint, exercise above). If x has 3 elements (power x) = 8

Could you compute and/or guess the number of sets included in {a,b,c,d} ?

And what about {a,b,c,d,e}?, and {a,b,c,d,e,f} ?, and ... ?

Bruno

PS Did I say that (power x) is called the powerset of x? It could be better to write (powerset x) instead of (power x), to emphasize that the powerset of x is the *set* of the sets included in x. OK? (Sorry).




m.a.

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Jul 8, 2009, 9:05:24 PM7/8/09
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Here's my third try. I'll continue working on the (power x) problem.   m.a.
----- Original Message -----
Sent: Wednesday, July 08, 2009 1:31 PM
Subject: Re: The seven step series


On 08 Jul 2009, at 15:43, m.a. wrote:

Second try:
(power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}}
 
           Third try:
                                         =   {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}, {{ },1,2,3}}

Bruno Marchal

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Jul 9, 2009, 4:06:36 AM7/9/09
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On 09 Jul 2009, at 03:05, m.a. wrote:

Here's my third try. I'll continue working on the (power x) problem.   m.a.
----- Original Message -----
Sent: Wednesday, July 08, 2009 1:31 PM
Subject: Re: The seven step series


On 08 Jul 2009, at 15:43, m.a. wrote:

Second try:
(power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}}
 
           Third try:
                                         =   {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}, {{ },1,2,3}}



You may be a bit saturated, because this is less correct than your preceding answer. You could, after some good night sleep, see this by yourself. Indeed if { { } 1, 2, 3} was a subset of { 1,2,3}, this would mean, by the definition of subset, that 1, 2, 3 and { } are elements of {1, 2, 3}. But { } is not an element of {1, 2, 3}.

The missing subset is just {1, 3}.

So instead of ...

I gave you the hint that there are 8 elements. Let us count: 

The empty set { }  ..................................1
Three singletons {1}, {2}, {3}................3
Two doubletons {1,2 }, {2,3 }................2
The biggest subset  {1,2,3}..................1

1 + 3 + 2 + 1 = 7

A subset is missing! Can you see which one?

... we have the much more symmetrical:

The empty set { }  .........................................1
Three singletons {1}, {2}, {3}.......................3
Three doubletons {1,2 }, {2,3 }, {1,3}..........3
The biggest subset  {1,2,3}..........................1

"1 3 3 1" is a row in the so called "Pascal triangle", and I have put those decomposition because the formula you gave me, was the formula giving the value of the number appearing in the Pascal triangle. 
In french we would say that you are searching the "little beast", making things more complex than they really are.

Let me do your, rather complex, reasoning:

I have a set A having n elements. A subset of A will have at most n elements. To find how many subsets are in A, I have to count the subset having 0 elements, 1 element, 2 elements; 3 elements, ... i elements, ... up to i = n. 
Searching in my memory, book or the net, I recall that the number of subsets having i elements taken in a set of n elements, let us write this (i;n),  is
n(n-1)(n-2) ... (n-i+1)/i!
So the number of elements in the powerset of A is the sum from i = 0 to n of (i; n) =

 sum from i = 0 to n of n(n-1)(n-2) ... (n-i+1)/i!

This is a very complex reasoning leading to a rather complex formula. It would have been rather not pedagogical from my part to give you a so difficult exercise, so you could have guessed a simpler reasoning, leading to a simpler formula.

Let us to do the "simpler" reasoning.

We have already seen that the powerset of a set with 0 element has 1 element. cf (powerset { }) = {{ }}
We have already seen that the powerset of a set with 1 element has 2 elements. cf (powerset {a}) = {{ }, {a}}
We have already seen that the powerset of a set with 2 element has 4 elements. cf (powerset {a, b}) = {{ }, {a} {b} {a, b}}
We have just seen that the powerset of a set with 3 element has 8 elements (cf above).

We see that the sequence of numbers on the right grows like

 1, 2, 4, 8, ....

And we have to guess the sequel, and find a general formula. Of course, if we don't see it yet we can still compute, tediously, the number of subsets of a set with four elements, like S = {a, b, c, d}. Let us do it:

The subset of S with 0 element { } .................................. 1
The subsets of S with 1 element {a}, {b}, {c}, {d} ...........4
The subsets of S with 2 elements {a,b} {a,c},{a,d} {b,c} {b,d}, {c,d} ...... 6 
The subsets of S with 3 elements {a,b,c}, {a,b,d}, {a,c,d}, {b,c,d} .......4
The subset of S with 4 elements {a,b,c,d} ..........................................1

Thus there are 1+4+6+4+1 subsets of {a,b,c,d}, and this gives 16.

Now our sequence of answers is a bit longer:

 1, 2, 4, 8, 16, ....

Exercise: can you guess how many subsets there are in a set with 5 elements, 6 elements, ...?

After that I will help you to get the formula, and we will be able to soon approach a far reaching question: how many subsets are included in the infinite set N = {0, 1, 2, 3, ...}. But some vocabulary will have to introduced, some generalization will have to be done. After that we will come to the machines, and the question of computability, but let us go easy and slowly. We have already done a lot, and you can take some rest when you want. Bravo for the work you have already done. 

I just give you another little, not so obvious but very pleasant exercise: look at many Mandelbrot set videos on youtube, and try to discover the recurring sequence 1, 2, 4, 8, 16, ... appearing in the zooms. This is experimental mathematics! This is for your enjoyment only. Sort of dessert.
BTW, the sequence 1, 2, 4, 8, 16, ... appears also in some of the thought experiments in the comp setting. Perhaps you remember?

Bruno




Bruno Marchal

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Jul 10, 2009, 4:24:55 PM7/10/09
to everything-list List
Hi all,

I suddenly feel sorry putting too much burden on just one
correspondent in the list, and I would appreciate if someone else
could propose answers or any remarks to the exercises.

I am also a bit anxious about Kim, who is the one who suggested me the
initial explanations, but who seems to have disappear right now.

There is also some sort of burden onto me, because it is hard to
explain "the real thing" concerning the seventh step, without
explaining or just illustrating at least some relevant portion of the
mathematical reality: mainly the unexpected mathematical discovery of
the universal functions, sets, numbers, systems, language, machine ...
I don't mention the absence of drawing ability which does not help.

Given that the list raised from a critical approach toward Tegmark and
Schmidhuber, I was usually assuming some knowledge in math and
physics. What was harder for me in the beginning was to motivate the
use of "philosophy of mind" notions, notably the key distinction
between the first and third person point of view. Then UDA should make
you realize how non obvious the relation between the first person and
the third person can be once we assume comp (= work in the theory
comp). My original goal was to illustrate that once we assume digital
mechanism, we can build a "scientific formulation" of the mind-body
problem or the consciousness-reality problem. We probably depart from
Tegmark and Schmidhuber, or Wolfram, by taking into account that
making comp explicit entails a delocalisation of the 1-person
relatively to the third person computations, and makes the identity
thesis, a most complex equivalence relations.

The knowledge of most people participating to the discussion is very
varied, due to the extreme transdiciplinarity of the subject, and the
interest it can evidently have for the layman (and indeed, for any
universal machine).
Marty asked me to make an attempt toward a "journalistic" description
of "how physics has to become part of number theory". This is very
difficult, and risky due to inevitable misunderstanding.
And I feel like I have to explain in what deep sense the mathematical
discovery of the "universal machine", made by Post, Turing, ... is
already a quite utterly astonishing, yet subtle, discovery. Gödel
himself took time to swallow it and he described Church thesis as an
epistemological miracle.
My intention was to derive properly Cantor theorem, and then Kleene
theorem, which was the object of my old "diagonalization posts".
I feel important that people understand how unbelievable Church thesis
is, and why most startling propositions, including incompleteness, are
easy consequences of it.

Typically I am happy to share my enthusiasm about all theorems in
computer science which leads to the reversal, but knowing myself I
know that I could accelerate too much and makes too much burden for
the correspondent especially if he is alone.
So before becoming an harasser myself I invite Marty to let other
people trying to answer the exercises.
Marty has fully agreed to this proposal and is happy the pressure is
off him to represent all those who are following anonymously.
Eventually I can show the solution and proceed in addressing the post
to everyone.
Note that this is what I have done with the combinators, feedback were
made out-of-line, then. But this lead to difficulties too. I cannot
solve all the exercise out-of-line. By experience this ends up with
finding myself writing too many posts with almost the same info to
different people.
Yet I can imagine how much it is to be the only public target of what
could look like an perpetual exam, and I really want to proceed in a
cooler way.

Some people have encouraged me, out-of-line, to proceed, but now I
think they should participate a little bit, if only to witness they
are following the thread. I will probably stop to propose "easy" (a
quite relative notion) exercise, but then it is important to stop me
once anything is unclear. This is a problem with math, if you miss a
piece, everything becomes senseless.

Understanding implies some self-implication in the reasoning. So,
either someone else try to participate, or I continue impersonally and
eventually I will try some "non technical summary". I recall that one
of the goal consists in explaining the difference between a
computation and a description of a computation (beyond just doing the
step 7).

Any remark to improve the communication or to design a better
methodology is welcome,

Best regards to all of you, and thanks for letting me know your
interests,

Bruno

http://iridia.ulb.ac.be/~marchal/

Johnathan Corgan

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Jul 10, 2009, 4:52:22 PM7/10/09
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On Fri, 2009-07-10 at 22:24 +0200, Bruno Marchal wrote:

> I suddenly feel sorry putting too much burden on just one
> correspondent in the list, and I would appreciate if someone else
> could propose answers or any remarks to the exercises.

Bruno--you're doing great. I think it is the case where silence means
"I understand, continue", rather than disinterest.

> There is also some sort of burden onto me, because it is hard to
> explain "the real thing" concerning the seventh step, without
> explaining or just illustrating at least some relevant portion of the
> mathematical reality: mainly the unexpected mathematical discovery of
> the universal functions, sets, numbers, systems, language, machine ...
> I don't mention the absence of drawing ability which does not help.

The derivation of your thesis from first principles is a very compelling
idea. The somewhat startling and unorthodox conclusions you espouse are
bound to cause confusion unless all their underpinnings are well
understood. The arguments from others then can have a much more
specific target than the top-level conclusions; instead they will come
out earlier in the derivation process and at the time of introduction of
the controversial subject.

> The knowledge of most people participating to the discussion is very
> varied, due to the extreme transdiciplinarity of the subject, and the
> interest it can evidently have for the layman (and indeed, for any
> universal machine).

While I do have training in math and physics, I still benefit from your
targeting the motivated layman. Personally, I'm not interested in doing
the exercises on the list, but they are still useful to check my
understanding.

> Best regards to all of you, and thanks for letting me know your
> interests,

By all means, proceed. Personally, if I don't understand something or
have an objection, you'll hear about it on the list, but I think you
should take silence as assent.

Johnathan Corgan


John Mikes

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Jul 12, 2009, 11:28:28 AM7/12/09
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Dear Bruno,
when I looked at the set-analysis it immediately popped up that {1,3} was missing, - YET - this fantastic<G> discovery of mine did not bring me closer to the idea "what are numbers".
It seems I can win the battle and still lose the war.
John

Bruno Marchal

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Jul 12, 2009, 11:39:46 AM7/12/09
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On 10 Jul 2009, at 22:52, Johnathan Corgan wrote:

>
> On Fri, 2009-07-10 at 22:24 +0200, Bruno Marchal wrote:
>
>> I suddenly feel sorry putting too much burden on just one
>> correspondent in the list, and I would appreciate if someone else
>> could propose answers or any remarks to the exercises.
>
> Bruno--you're doing great. I think it is the case where silence means
> "I understand, continue", rather than disinterest.

Well, thanks, OK, perhaps.


>
>
>> There is also some sort of burden onto me, because it is hard to
>> explain "the real thing" concerning the seventh step, without
>> explaining or just illustrating at least some relevant portion of the
>> mathematical reality: mainly the unexpected mathematical discovery of
>> the universal functions, sets, numbers, systems, language,
>> machine ...
>> I don't mention the absence of drawing ability which does not help.
>
> The derivation of your thesis from first principles is a very
> compelling
> idea. The somewhat startling and unorthodox conclusions you espouse
> are
> bound to cause confusion unless all their underpinnings are well
> understood.

There are two things. Understanding the conclusions, and understanding
how we get to them.
Many variations are possible in between are possible for varied
audience.


> The arguments from others then can have a much more
> specific target than the top-level conclusions; instead they will come
> out earlier in the derivation process and at the time of
> introduction of
> the controversial subject.

But what is controversial? I have never heard about something
controversial seen in the reasoning. The conclusion are astonishing,
and certainly annoying for someone who believes "religiously" in both
physicalism and digital mechanism.
The subject matter was controversial a long time ago, but today, it is
no more, I think. Well, it depends on which circle. That something
appears in the academy (like studies on consciousness, does not mean
that all academicians understand the questioning there, alas).
I have heard that the first person indeterminacy, which is my mean
early contribution, is controversial, but I have never seen any
controversy on it, just sometimes, some discussion on the vocabulary
or definition, which does not change any conclusion.
The subject matter is difficult, so it easier for the "religious"
people (like convinced atheists, to be clear) to speculate about some
difficulties they don't even try to single out.
I proceed in the scientific way, which means that I just ask
questions, and anyone can verify what follows from what, or interrupt
and present an objection. Up to now, none of the "real" objections
presented were fatal, and eventually those reduce also to a problem of
vocabulary.

>
>
>> The knowledge of most people participating to the discussion is very
>> varied, due to the extreme transdiciplinarity of the subject, and the
>> interest it can evidently have for the layman (and indeed, for any
>> universal machine).
>
> While I do have training in math and physics, I still benefit from
> your
> targeting the motivated layman. Personally, I'm not interested in
> doing
> the exercises on the list, but they are still useful to check my
> understanding.

OK.


>
>
>> Best regards to all of you, and thanks for letting me know your
>> interests,
>
> By all means, proceed. Personally, if I don't understand something or
> have an objection, you'll hear about it on the list, but I think you
> should take silence as assent.

If only silence could be assent!
But I am willing to take yours as such and I will proceed.

Best,

Bruno

http://iridia.ulb.ac.be/~marchal/

Bruno Marchal

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Jul 12, 2009, 12:26:01 PM7/12/09
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On 12 Jul 2009, at 17:28, John Mikes wrote:

Dear Bruno,
when I looked at the set-analysis it immediately popped up that {1,3} was missing, -


Good!




YET - this fantastic<G> discovery of mine did not bring me closer to the idea "what are numbers".


And, assuming comp, you never will. None of us will. Sets can provide a nice representation of the numbers, which I have already given some times ago, but for now, I prefer to take the numbers as basic entities. Sets are more complex entities than numbers, in general.

What are the numbers? I can give you only some hints, like zero is the number of matches in a empty box, one is the number of matches in a box having one match in, and so on ... 

About the "and so on ... " I can only hope you have been enough trained in high school to have an idea.

I cannot explain really what are numbers, but, if you assume the comp theory,  I can explain "completely", from it,  why numbers develop beliefs in galaxies, get conscious, makes a big variety of dreams, and eventually realize why they will never knows what numbers are, and what they are capable of.

Number are the necessary mystery, from which we start. What I like in comp, is that it explains why the numbers have to be a mystery.

Bruno




It seems I can win the battle and still lose the war.
John

On Wed, Jul 8, 2009 at 9:05 PM, m.a. <mart...@bellsouth.net> wrote:
Here's my third try. I'll continue working on the (power x) problem.   m.a.
----- Original Message -----
Sent: Wednesday, July 08, 2009 1:31 PM
Subject: Re: The seven step series


On 08 Jul 2009, at 15:43, m.a. wrote:

Second try:
(power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}}
 
           Third try:
                                         =   {{ }, {1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}, {{ },1,2,3}}

This is far better!  Not yet correct though.

I gave you the hint that there are 8 elements. Let us count: 

The empty set { }  ..................................1
Three singletons {1}, {2}, {3}................3
Two doubletons {1,2 }, {2,3 }................2
The biggest subset  {1,2,3}..................1

1 + 3 + 2 + 1 = 7

A subset is missing! Can you see which one?





Mirek Dobsicek

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Jul 12, 2009, 6:46:08 PM7/12/09
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Hi Bruno,

I'd like to let you know that I'm following the serie of your letters.
While I have the background you are covering right now, I still enjoy
your insights.

I joined the list like two years ago and from that time I've read most
of your key papers. Honestly, it is not the easiest stuff to read
style-wise. You try to precise, define well, etc. yet it cannot really
be compared to the quality of, let us say, Physical Review Letters and
alike articles. In my opinion, that is why it is hard to either agree or
disagree with your thesis.

I can imagine that right now you are tempted to write something along
the lines
a\ I just propose to take Church thesis seriously
b\ All I ask you is 'Do you say yes to the doctor?
While valid proposal and question, there is really not much to agree
with/disagree with/critize unless one is willing to undertake long
discussions, clarifications and position adjustments.


Anyway, your papers and letters are really a great source of ideas to
think about and that is exactly what I do. From the day one on the list
I keep myself busy with the question of "Why should I believe in the
Church thesis" (you see, I don't write "Why do I ..."). I've got into
the writing of Bernard Bolzano (I consider his work cruicial in order to
keep an open mind about the Cantor diagonal argument) ..
- and now back to the beginning of my letter -
Bolzano (Cantor), your insights and thinking about alternatives at any
moment make me pretty happy. Thanks!

Mirek

PS: I'd love to read a book by Bruno Marchal.

Kim Jones

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Jul 13, 2009, 3:38:47 AM7/13/09
to everyth...@googlegroups.com

On 11/07/2009, at 6:24 AM, Bruno Marchal wrote:

>
> I am also a bit anxious about Kim, who is the one who suggested me the
> initial explanations, but who seems to have disappear right now.

OK - I'm back. Since May 27 to two days ago I have been without
Internet access.

I made the mistake of upgrading my Broadband plan to add Internet
phone. It took two telcos a month to complete this ridiculously basic
operation with mistakes made and attendant extra waiting times.

Then, just as the connection was restored at the beginning of July,
the plumbing in this block of apartments fell apart and a major
excavation work went ahead and this time the plumbers cut the phone
cable and didn't realise it which meant I wasted another week trying
to get the problem diagnosed.

So now finally everything is back to normal. I have just started
reading this thread and can see that the class is a very exclusive
one! I will try my best to follow through on the exercises and the
comments, corrections. I feel I have access to the correct
mathematical symbols on my Mac now but *time* is the thing that I
don't have much of anymore, so I feel a bit depressed about the level
of effort I can devote to it. If only we didn't have to work for a
living things would be vastly easier.

The notion of sets is indeed a tricky one. I am just now going over
the initial exercises again. Do not wait for me. I am also trying to
catch up on about 4,000 emails.

Bruno - my sincerest apologies for this hiatus. You seem eager to get
to the seventh and eighth steps. Why wouldn't you be.

regards,

Kim


Bruno Marchal

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Jul 13, 2009, 9:37:56 AM7/13/09
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On 13 Jul 2009, at 00:46, Mirek Dobsicek wrote:

>
> I'd like to let you know that I'm following the serie of your letters.
> While I have the background you are covering right now, I still enjoy
> your insights.

Thanks for letting me know.

>
>
> I joined the list like two years ago and from that time I've read most
> of your key papers. Honestly, it is not the easiest stuff to read
> style-wise. You try to precise, define well, etc. yet it cannot really
> be compared to the quality of, let us say, Physical Review Letters and
> alike articles.

I work on a subject which is not usually approach in any reviews of
physics.
And then my english is sometimes a bit hazardous. But I have never get
referee or any feedback about the rigors, which in my field (machine
theology) is far more developed than usual. It is part of the problem
for some people I think. It is just unusual.

> In my opinion, that is why it is hard to either agree or
> disagree with your thesis.

I disagree. It is simple. Just say a number between 1 and 8, with a
justification of what you don't understand.
Perhaps between 0 and 8, if you have a problem with the definition of
comp.


>
>
> I can imagine that right now you are tempted to write something along
> the lines
> a\ I just propose to take Church thesis seriously
> b\ All I ask you is 'Do you say yes to the doctor?

yes, for the sake of the argument. A non computationalist can just
consider someone else saying yes to the doctor. A bit more is needed,
and it is necessary to recall the definition of comp: it exists a
level of description of my (generalized) such that I survive through a
digital functional substitution made at that level.


>
> While valid proposal and question, there is really not much to agree
> with/disagree with/critize unless one is willing to undertake long
> discussions, clarifications and position adjustments.

Indeed, there is nothing to disagree. Only to understand, and who
knows? to clarify. But then it is up to those who try to understand to
say what they don't understand, besides the intrinsic difficulty with
the subject. In the seventies some people argued that any sentence
containing the word "consciousness" was automatically crackpot. Of
course this is an illustration of the complete absence of
understanding of the axiomatic method. We never know what we are
talking about, we can only agree on starting propositions and method
of reasoning, and then see if the conclusion follows from what we have
admitted.


>
>
>
> Anyway, your papers and letters are really a great source of ideas to
> think about and that is exactly what I do.


I am happy with that.


> From the day one on the list
> I keep myself busy with the question of "Why should I believe in the
> Church thesis" (you see, I don't write "Why do I ...").

Good question. A lot of my work consists in showing that CT is a very
strong principle. It is far stronger than most computer scientist
imagine.

> I've got into
> the writing of Bernard Bolzano (I consider his work cruicial in
> order to
> keep an open mind about the Cantor diagonal argument) ..
> - and now back to the beginning of my letter -
> Bolzano (Cantor), your insights and thinking about alternatives at any
> moment make me pretty happy. Thanks!


You are welcome.

> PS: I'd love to read a book by Bruno Marchal.

I have already written three books, and one was ordered by a publisher
after getting a price. The two other one were disputed by different
publishers, and then suddenly, without explanation, all those projects
were abandoned. I have lost my trust in that kind of world I'm afraid.
I don't think the reason of that abandon has any relationship with my
work which is really of the type: "find the error". I will surely
write one paper and one book. Recently I have submitted a paper, and
the referees were quite enthusiast, but the paper has been refused for
being out of the topic, which it was not (unless you don't believe
that observers are person).
We will see. My work is simple in two senses: UDA is simple because
you need nothing more than a very tiny amount of understanding on
numbers, set, computable functions and consciousness/kowledge. AUDA is
relatively "simple" because you need only to understand Solovay's
theorem and the Theaetetical definitions of knowledge. It makes comp
hard to believe, no doubt, but here the work of Everett in quantum
mechanics can provide a big help. Ah, ok, all this at once needs works
and time, but nothing more.

http://iridia.ulb.ac.be/~marchal/

Bruno Marchal

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Jul 13, 2009, 9:50:27 AM7/13/09
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On 13 Jul 2009, at 09:38, Kim Jones wrote:

>
>
> On 11/07/2009, at 6:24 AM, Bruno Marchal wrote:
>
>>
>> I am also a bit anxious about Kim, who is the one who suggested me
>> the
>> initial explanations, but who seems to have disappear right now.
>
>
>
> OK - I'm back.


I am quite glad you are back, and alive!

> Since May 27 to two days ago I have been without
> Internet access.

Well, that is form of death nowadays. I wish you an happy resurrection.


>
>
> I made the mistake of upgrading my Broadband plan to add Internet
> phone. It took two telcos a month to complete this ridiculously basic
> operation with mistakes made and attendant extra waiting times.
>
> Then, just as the connection was restored at the beginning of July,
> the plumbing in this block of apartments fell apart and a major
> excavation work went ahead and this time the plumbers cut the phone
> cable and didn't realise it which meant I wasted another week trying
> to get the problem diagnosed.

Matter kicks back, even with comp!


>
>
> So now finally everything is back to normal. I have just started
> reading this thread and can see that the class is a very exclusive
> one! I will try my best to follow through on the exercises and the
> comments, corrections. I feel I have access to the correct
> mathematical symbols on my Mac now but *time* is the thing that I
> don't have much of anymore, so I feel a bit depressed about the level
> of effort I can devote to it. If only we didn't have to work for a
> living things would be vastly easier.

You are right. Take your time and ask any question. I will try to sum
up through the next posts.


>
>
> The notion of sets is indeed a tricky one. I am just now going over
> the initial exercises again. Do not wait for me. I am also trying to
> catch up on about 4,000 emails.

Gosh!


>
>
> Bruno - my sincerest apologies for this hiatus. You seem eager to get
> to the seventh and eighth steps. Why wouldn't you be.


No, we have all the time. I just answered to posts, and made
elementary recalls of math. But I realize that some people have not
follow any course in set theory. In Belgium it was taught as "modern
math" during 10 years in high school, and then it has disappeared
(alas 1000X). The discrepancy between the math level of different
people is very big, but the shortcut I am pursuing now, should not be
insurmountable. I will make the next posts as self-contained as
possible, but please interrupt when you feel I am esoteric.

Have a good day,

Bruno

http://iridia.ulb.ac.be/~marchal/

Bruno Marchal

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Jul 14, 2009, 4:40:23 AM7/14/09
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Hi Kim, Marty, Johnathan, John, Mirek, and all...


We were studying a bit of elementary set theory to prepare ourself to Cantor's theorem, and then Kleene's theorem, which are keys to a good understanding of the universal numbers, and to Church thesis, which are the keys of the seven steps. 
I intend to bring you to the comp enlightenment :)

But first some revision. Read the following with attention!

A set is a collection of things, which in general can themselves be anything. Its use consists in making a many into a one.
If something, say x,  belongs to a set S, it is usually called "element" of S. We abbreviate this by (x \belongs-to S).

Example:

A = {1, 2, 56}. A is a set with three elements which are the numbers 1, 2 and 56.

We write:

(1 \belongs-to {1, 2, 56}), or (1 \belongs-to A), or simply 1 \belongs-to A, when no confusions exist. The parentheses "(" and ")" are just delimiters for easing the reading. I write \belongs-to the relation "belongs to" to remind it is a mathematical symbol.

B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.

C = {34, 54, Paul, {3, 4}}

For this one, you may be in need of spectacles. In case of doubt, you can expand it a little bit:

C = {    34,     54,    Paul,   {3, 4}     }

You see that C is a sort of hybrid set which has 4 elements:

   - the number 34
   - the number 54
   - the human person Paul
   - the set {3, 4}

Two key remarks:
1) the number 3 is NOT an element of C. Nor is the number 4 an element of C. 3 and 4 are elements of {3, 4}, which is an element of C. But, generally, elements of elements are not elements! It could happen that element of element are element, like in D = {3, 4, {3, 4}}, the number 3 is both an element of D and element of an element of D ({3, 4}), but this is a special circumstance due to the way D is defined.
2) How do I know that "Paul" is a human, and not a dog. How do I know that "Paul" does not refer just to the string "paul". Obvioulsy the expression "paul" is ambiguous, and will usually be understood only in some context. This will not been a problem because the context will be clear. Actually we will consider only set of numbers, or set of mathematical objects which have already been defined. Here I have use the person Paul just to remind that typically set can have as elements any object you can conceive.

What is the set of even prime number strictly bigger than 2. Well, to solve this just recall that ALL prime numbers are odd, except 2. So this set is empty. The empty set { } is the set which has no elements. It plays the role of 0 in the world of sets.


We have seen some operations defined on sets.

We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))
8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Of course some sets can be disjoint, that is, can have an empty intersection:

{1, 2, 3} \inter {4, 5, 6} = { }.

Similarly we can define (S1 \union S2) by the set of the elements belonging to S1 or belonging to S2:

(S1 \union S2) = {x such-that ((x belongs-to S1) or (x belongs-to S2))}

We have, with S1 and S2 the same as above (S1 = {1, 2, 3} and S2 = {2, 3, 7, 8}):

(S1 \union S2) = {1, 2, 3, 7, 8}.

OK. I suggest you reread the preceding post, and let me know in case you have a problem.


We have seen also a key relation defined on sets: the relation of inclusion.

We say that (A \included-in B) is true when all elements of A are also elements of B.

Example:
The set of ferocious dogs is included in the set of ferocious animals.
The set of even numbers is included in the set of natural numbers.
The set {2, 6, 8} is included in the set {2, 3, 4, 5, 6, 7, 8}
The set {2, 6, 8} is NOT included in the set  {2, 3, 4, 5, 7, 8}.

When a set A is included in a set B, A is called a subset of B.

We were interested in looking to all subsets of a some set.

What are the subsets of {a, b} ?

They are { }, {a}, {b}, {a, b}. Why?

{a, b} is included in {a, b}. This is obvious. All elements of {a,b} are elements of {a, b}.
{a} is included in {a, b}, because all elements of {a} are elements of {a, b}
The same for {b}.
You see that to verify that a set with n elements is a subset of some set, you have to make n verifications. 
So, to see that the empty set is a subset of some set, you have to verify 0 things. So the empty set is a subset of any set.

proposition: { } is included-in any set.

So the subsets of {a, b} are { }, {a}, {b}, {a, b}.

But set have been invented to make a ONE from a MANY, and it is natural to consider THE set of all subsets of a set. It is called the powerset of that set.

So the powerset of {a, b} is THE set {{ }, {a}, {b}, {a, b}}. OK?

Train yourself on the following exercises:

What is the powerset of { }
What is the powerset of {a}
What is the powerset of {a, b, c}

Any question?

This was a bit of revision, to let Kim catch up.

The sequel will appear asap. Be sure everything is OK, and please, ask question if it is not. 
You can also ask any question on the first sixth steps of UDA ('course).

Bruno

Mirek Dobsicek

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Jul 14, 2009, 7:04:31 AM7/14/09
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Hi Bruno,

I am in a good mood and a bit picky :-) Do you know how many entries
google gave me upon entering
Theaetetical -marchal -bruno

Mirek

> for some people I think. It is just unusual.

Bruno Marchal

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Jul 14, 2009, 12:51:50 PM7/14/09
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Hi Mirek,

>
> Hi Bruno,
>
> I am in a good mood and a bit picky :-) Do you know how many entries
> google gave me upon entering
> Theaetetical -marchal -bruno


Well 144?

Good way to find my papers on that. The pages refer quickly to this
list or the FOR list.

Tomorrow I will not be there.

I let those interested to meditate on two questions (N is {0, 1, 2, 3,
4, ...}):

1) What is common between the set of all subsets of a set with n
elements, and the set of all finite sequences of "0" and "1" of length
n.
2) What is common between the set of all subsets of N, and the set of
all infinite sequences of "0" and "1".

Just some (finite and infinite) bread for surviving the day :)

Bruno


http://iridia.ulb.ac.be/~marchal/

John Mikes

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Jul 14, 2009, 6:50:37 PM7/14/09
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Bruno,
I appreciate your grade-school teaching. We (I for one) can use it.
I still find that whatever you explain is an 'extract' of what can be thought of a 'set' (a one representing a many).
Your 'powerset' is my example.
All those elements you put into { }s are the same as were the physical objects to Aristotle in his 'total' - the SUM of which was always MORE than the additives of those objects.
Relations!
The set is not an inordinate heap (correct me please, if I am off) of the elements, the elements are in SOME relation to each other and the "set"-idea of their ensemble, to form a SET.
You stop short at the naked elements together, as I see.
They wear cloths and hold hands. Mortar is among them.
Maybe your math-idea can tolerate any sequence and hiatus concerning to the 'set', and it still stays the same, as far as the "math-idea you need" goes,  but if I go further (and you indicated that ANYTHING can form a set) the relations of the set-partners comes into play. Not only those which WE choose for 'interesting' to such set, but ALL OF THEM influencing the character of that "ONE".
Just musing.
 
John

m.a.

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Jul 14, 2009, 10:15:35 PM7/14/09
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----- Original Message -----
Sent: Tuesday, July 14, 2009 4:40 AM
Subject: Re: The seven step series

Hi Kim, Marty, Johnathan, John, Mirek, and all...


Bruno: May I advise you about an instance of English usage? The word "supposed" in the next sentence is often used as sarcasm to imply serious doubt about the statement. In this context it can be interpreted as a slight. I think you meant to say "assumed" which implies an evident fact. Please don't apologize, we are most grateful for your efforts in using English and are happy to make allowances for minor slips.
B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.
 
 
 
 

I also have a question: see below:
 
We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))
8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Doesn't the statement in bold (above) contradict the statement immediately preceding (also in bold)?

Kim Jones

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Jul 15, 2009, 3:09:25 AM7/15/09
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On 14/07/2009, at 6:40 PM, Bruno Marchal wrote:

> The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7,
> 8} will be written (S1 \inter S2), and is equal to the set of
> elements which belongs to both S1 and S2. We have
>
> (S1 \inter S2) = {2, 3}
>
> We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and
> (x belongs-to S2))}
>
> 2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2
> belongs-to S2))
> 8 does not belongs to (S1 \inter S2) because it is false that ((2
> belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.
>

Quick (silly) questions:

1.

why do you have to write "\inter" ? Why not just write "inter" ?

Typing "\" causes me to make use of a key on my keyboard I have never
used before which is scary ;-)


2.

"such-that" is surely "such that" but the hyphen might just mean
something
(this is mathematics; there are dots and dashes and slashes all over
the place so you have to know what they all mean)

likewise

"belongs-to" would still mean the same thing if we wrote "belongs to"
would it not?


best

K

Bruno Marchal

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Jul 16, 2009, 4:13:24 AM7/16/09
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On 15 Jul 2009, at 00:50, John Mikes wrote:

Bruno,
I appreciate your grade-school teaching. We (I for one) can use it.
I still find that whatever you explain is an 'extract' of what can be thought of a 'set' (a one representing a many).
Your 'powerset' is my example.
All those elements you put into { }s are the same as were the physical objects to Aristotle in his 'total' - the SUM of which was always MORE than the additives of those objects.
Relations!
The set is not an inordinate heap (correct me please, if I am off) of the elements, the elements are in SOME relation to each other and the "set"-idea of their ensemble, to form a SET.
You stop short at the naked elements together, as I see.

You get the idea. 
We can add structure to sets, by explicitly endowing them with operations and relations.




They wear cloths and hold hands. Mortar is among them.
Maybe your math-idea can tolerate any sequence and hiatus concerning to the 'set', and it still stays the same, as far as the "math-idea you need" goes,  

Yes, it is the methodology.



but if I go further (and you indicated that ANYTHING can form a set)

More precisily, we can form a set of multiple thing we can conceive or defined.


the relations of the set-partners comes into play. Not only those which WE choose for 'interesting' to such set, but ALL OF THEM influencing the character of that "ONE".
Just musing.

It is OK. The idea consists in simplifying the things as much as possible, and then to realize that despite such simplification we are quickly driven to the unprovable, unnameable, un-reductible, far sooner than we could have imagine. 

Bruno

Bruno Marchal

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Jul 16, 2009, 4:28:06 AM7/16/09
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On 15 Jul 2009, at 04:15, m.a. wrote:

 
----- Original Message -----
Sent: Tuesday, July 14, 2009 4:40 AM
Subject: Re: The seven step series

Hi Kim, Marty, Johnathan, John, Mirek, and all...


Bruno: May I advise you about an instance of English usage? The word "supposed" in the next sentence is often used as sarcasm to imply serious doubt about the statement. In this context it can be interpreted as a slight. I think you meant to say "assumed" which implies an evident fact. Please don't apologize, we are most grateful for your efforts in using English and are happy to make allowances for minor slips.
B = {Kim, Marty, Russell, Bruno, George, Jurgen} is  a set with 5 elements which are supposed to be humans.

Thanks for letting me know. In french "I assume" is the same as "I suppose". I'm afraid it will take time for me not doing that error again. But don't hesitate to remind me of the "false friend" behavior. Sorry for the unintended sarcasm.
To be sure it is not really an assumption, and a "supposition" means more like an "obvious implicit fact we should take into account without mentioning", as opposed to an "assumption" which is more akin to "a key hypothesis". Here I was referring to conventions only, but then, as yopu point out, an non intended sarcasm could be see. Difficult. That is why I prefer to stick on less ambiguous, purely mathematical examples of sets.



I also have a question: see below:
 
We have seen INTERSECTION, and UNION.

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7, 8} will be written (S1 \inter S2), and is equal to the set of elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and (x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))
8 does not belongs to (S1 \inter S2) because it is false that ((2 belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.

Doesn't the statement in bold (above) contradict the statement immediately preceding (also in bold)?



You are completely right. I just did a copy and past, and forget to substitute the "2" by the "8".

So the statements are:

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2 belongs-to S2))
8 does not belongs to (S1 \inter S2) because it is false that ((8 belongs-to S1) and (8 belongs-to S2)).

Bruno



Bruno Marchal

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Jul 16, 2009, 4:48:45 AM7/16/09
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On 15 Jul 2009, at 09:09, Kim Jones wrote:



On 14/07/2009, at 6:40 PM, Bruno Marchal wrote:

The intersection of the two sets S1 = {1, 2, 3} and S2 = {2, 3, 7,  
8} will be written (S1 \inter S2), and is equal to the set of  
elements which belongs to both S1 and S2. We have

(S1 \inter S2) = {2, 3}

We can define (S1 \inter S2) = {x such-that ((x belongs-to S1) and  
(x belongs-to S2))}

2 belongs to (S1 \inter S2) because ((2 belongs-to S1) and (2  
belongs-to S2))
8 does not belongs to (S1 \inter S2) because it is false that ((2  
belongs-to S1) and (2 belongs-to S2)). Indeed 8 does not belong to S1.




Quick (silly) questions:

1.

why do you have to write "\inter"  ? Why not just write "inter"  ?

Typing "\" causes me to make use of a key on my keyboard I have never  
used before which is scary ;-)

For the intersection of two sets S1 and S1,  I have used  

1)  S1  S2

But the math symbol ""  did not go through all emailing system, so, I have used

2) S1 INTERSECTION S2

But then I recall that in mails, capital letters seems aggressive, loudly ..., so I have used

3) S1 intersection S2

But then the difference between what is supposed to be represent a mathematical symbols, and a word in english, disappears, so I have used

4) S1 \intersection S2

But this, on the last post, seems to me to be a little too long, and so I am using now

5) S1 \inter S2

Only God knows what I will use tomorrow. You should learn that there is no standard of mathematical notations, and no two mathematicians use the same symbols, and not one mathematician use the same symbols in two different texts. What is nice, is that, usually, mathematicians quickly redefine what they mean by any symbols at the beginning of their books and papers.
Of course doing math on mails aggravates apparently this search for the symbols which could satisfy everyone.

Sorry to scare you,


2.

"such-that" is surely "such that" but the hyphen might just mean  
something
(this is mathematics; there are dots and dashes and slashes all over  
the place so you have to know what they all mean)

likewise

"belongs-to" would still mean the same thing if we wrote "belongs to"  
would it not?

Same remark. I should have use \such-that, and \belongs-to.

Note that at this stage it is really not important to be aware of the difference between a symbol and what the symbols refer too, but in logic such differences acquire some importance at some point, and I just try to prepare you for such nuances, having the sequel of this introduction in my mind.

You question are not silly and makes sense, as you see,

Bruno



John Mikes

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Jul 16, 2009, 9:17:23 AM7/16/09
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Please read between your lines included in bold letters
John

On Thu, Jul 16, 2009 at 4:13 AM, Bruno Marchal <mar...@ulb.ac.be> wrote:
On 15 Jul 2009, at 00:50, John Mikes wrote:

Bruno,
I appreciate your grade-school teaching. We (I for one) can use it.
I still find that whatever you explain is an 'extract' of what can be thought of a 'set' (a one representing a many).
Your 'powerset' is my example.
All those elements you put into { }s are the same as were the physical objects to Aristotle in his 'total' - the SUM of which was always MORE than the additives of those objects.
Relations!
The set is not an inordinate heap (correct me please, if I am off) of the elements, the elements are in SOME relation to each other and the "set"-idea of their ensemble, to form a SET.
You stop short at the naked elements together, as I see.

You get the idea. 
We can add structure to sets, by explicitly endowing them with operations and relations.


Furhter below you also expose the contrary (to simplify) - I am afraid your "operations and relations" are restricted to the numbers-based (math?) domain, which is not what I mean by 'totality'.


They wear cloths and hold hands. Mortar is among them.
Maybe your math-idea can tolerate any sequence and hiatus concerning to the 'set', and it still stays the same, as far as the "math-idea you need" goes,  

Yes, it is the methodology.



but if I go further (and you indicated that ANYTHING can form a set)

More precisily, we can form a set of multiple thing we can conceive or defined.
 
I would not restrict 'a set' to what WE can conceive, or define now. (Not even within the 'math'-related domain).

the relations of the set-partners comes into play. Not only those which WE choose for 'interesting' to such set, but ALL OF THEM influencing the character of that "ONE".
Just musing.

It is OK. The idea consists in simplifying the things as much as possible, and then to realize that despite such simplification we are quickly driven to the unprovable, unnameable, un-reductible, far sooner than we could have imagine. 
 
I may suggest (or: assume?) that instead of "despite" it would make more sense to write: "AS A CONSEQUENCE"
- think about it.
 

Bruno
John

Bruno Marchal

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Jul 16, 2009, 9:19:31 AM7/16/09
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On 14 Jul 2009, at 10:40, Bruno Marchal wrote:

<snip>

> So the subsets of {a, b} are { }, {a}, {b}, {a, b}.
>
> But set have been invented to make a ONE from a MANY, and it is
> natural to consider THE set of all subsets of a set. It is called
> the powerset of that set.
>
> So the powerset of {a, b} is THE set {{ }, {a}, {b}, {a, b}}. OK?
>
> Train yourself on the following exercises:
>
> What is the powerset of { }
> What is the powerset of {a}
> What is the powerset of {a, b, c}


I give the answer, and I continue slowly.

1) What is the powerset of {a, b, c}?

By definition, the powerset of {a, b, c} is the set of all subsets of
{a, b, c}.
I go slowly.

Is the set {d, e, f} a subset of {a, b, d}? No. None of the elements
of {d, e, f} are elements of {a, b, c}. The question was ridiculous.

Is the set {a, b, d} a subset of {a, b, c}? No. One element of {a, b,
d}, indeed, d, does not belong to {a, b, c}, so {a, b, d} cannot be a
subset of {a, b, c}. The question was ridiculous again, but less
obviously so.

Is the set {a, b, c} a subset of {a, b, c}. Yes. All elements of {a,
b, c} are elements of {a, b, c}. {a, b, c} is included in {a, b, c}.

Can we conclude from this that the powerset of {a, b, c} is {{a, b,
c}}. No. We can conclude only that {{a, b, c}} is included in the
powerset. It is very plausible that there are other subsets!

Indeed,

Is {a, b} included in {a, b, c}? Yes, all elements of {a, b} are
elements of {a, b, c}. This take two verifications: we have to verify
that a belongs to {a, b, c}. And that b belongs to {a, b, c}.

Can we conclude from this that the powerset of {a, b, c} is {{a, b, c}
{a, b}}. No, we could still miss other subsets.

I accelerate a little bit.
Is {a, c} a subset of {a, b, c}? Yes, by again two easy verification.

Is there another doubleton (set with two elements) having elements in
{a, b, c}? Yes. {c, b}. It is easy to miss them, so you have to be
careful. All two elements of {c, b} are elements of {a, b, c}, as can
be verified by two easy verification.

Is {b, c} a subset of {a, b, c}. Yes, but we have already consider
it. Indeed the set {b, c} is the same set as {c, b}.

Is there another doubleton? No. Why? I search and don't find it.

is there yet some subset to find?

Yes, the set with one element, notably. They are called singleton.

Here it is easy to guess that there will be as many singletons
included in (a, b, c} that there is elements in {a, b, c}. So the
singletons are {a}, {b}, and {c}. This can be verified by one
verification for each.

Are there still subset? Yes. We have seen that the empty set { } is
included in any set. This can be (re)verify by 0 verifications, given
that there is 0 element in { }..

Conclusion:
There are 8 subsets in {a, b, c}, which are { }, {a}, {b}, {c}.{a, b},
{a, c}, {c, b} and {a, b, c}. And thus,

The powerset of {a, b, c} is the set { { }, {a}, {b}, {c}.{a, b}, {a,
c}, {c, b} {a, b, c}}.


2) What is the powerset of {a}?

Answer {{ } {a}}. It has two elements.

3) What is the powerset of { }
We could think at first sight that there are no subsets, given that
{ } is empty. But we have seen that { } is included in any set. So { }
is included in { }. Again you can verify this by zero verification!
But then the powerset of { }, which is the set of sets included in { }
is not empty: It has one element, the empty set. It is {{ }}. Think
that {{ }} is a box containing that empty box.


Attempt toward a more general conclusion.

The powerset of a set with 0 element has been shown having 1 elements,
and no more.
The powerset of a set with 1 element has been shown having 2 elements,
and no more.
The powerset of a set with 2 elements has been shown having 4
elements, and no more. (preceding post)
The powerset of a set with 3 elements has been shown having 8
elements, and no more.
The powerset of a set with 4 elements has been shown having 16
elements, and no more. (older post)

In math we like to abstract things. Let us look at the preceding line
with all the words dropped! This gives

--- 0 ---- 1 ---
--- 1 ---- 2 ---
--- 2 ---- 4 ---
--- 3 ---- 8 ---
--- 4 ---- 16 ---

On the left, we see, vertically disposed, the natural numbers,
appearing with their usual order.
On the right, we see, vertically disposed, some natural numbers, which
seems to depend in some way from what numbers appears on the left. It
looks like there is a functional relation, that is a function.

The notion of function is the most important and pervading notion in
math, physics, science in general, and we will have to come back on
that very notion soon enough.

The idea that there is a function lurking there, is the idea that we
can guess a general law, capable of providing the answer to the
general line:

"The powerset of a set with n element has been shown having ?
elements, and no more."

Can we determine ? from n. Surely it depends on n.

In this case, a simple guess can be made, by meditating on the
sequence of numbers which appear on the right. Those are, when written
horzontally:

1, 2, 4, 8, 16, ...

I guess you see what happens. Each number in that sequence is the
double of the preceding one. So the next one can be obtained, by just
continuing the multiplication by two.

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384,
32768, 65536, 131072, ... that is,

1, 2x2, 2x2x2, 2x2x2x2, 2x2x2x2x2, 2x2x2x2x2x2, ...

We will write a possibly lengthy expression like 2x2x2x2x2x ... x2, as
2^n, where n is the number of occurence of "2" in the expression.

So you can guess that in the "general line":

"The powerset of a set with n elements has been shown having ?
elements, and no more."

? does indeed depend on n, and is actually equal to 2^n.

Here is the general law, that we have guessed by experience (counting
in the simple case) and generalized by intution:

The powerset of a set with n elements has been shown having 2^n
elements, and no more.

When we found a law in such a way, we could ask if we couldn't prove
it from facts we are already believing (or guessing).

Here, the theory is the intuitive basic knowledge of logic, numbers
and sets. And the question is really: can you prove, or justify, or
explain WHY the powerset of a set of n elements has 2^n elements?

What happens which could explain why, when we add an element to a set,
its powerset becomes two time bigger. is there a reason for that
special happening. Can I convince myself that it has to be like that?

I let you think.

Hints. We have already see a "doubling scenario". Indeed, I often
mention at the step 3 of the UDA, the iterated self-duplication. Some
is cut and paste in Brussel, and copy at place W and M. Then both
individuals come back, by train, in Brussels and both do the
duplication experience again, and again, and again. The number of
individuals grows like 2, 4, 8, ... that is 2^n, with n the number of
iteration of the experience. After n iteration, each has written in
its "first person diary" the sequence of places they have visit, which
look like a string of W and M of length n.

No question? If everything is clear, I continue asap.

Oh! I have a question myself. If the law above is correct, it looks
like 2^0 = 1. It is the laws applied to the first line---the case of
the powerset of the empty set. Is it true that 2x2x2x...x2 is equal to
1 in case the number of occurence of 2 is 0? How come?

Bruno


http://iridia.ulb.ac.be/~marchal/

m.a.

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Jul 16, 2009, 9:29:03 AM7/16/09
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Bruno,
            I have no idea how to even begin to answer these questions. Have you given us the definitions we need to do so?     
                                                                                                                                                                                                 marty a.
 
 
----- Original Message -----
From: "Bruno Marchal" <mar...@ulb.ac.be>

>
> I let those interested to meditate on two questions (N is {0, 1, 2, 3, 
> 4, ...}):
>
> 1) What is common between the set of all subsets of a set with n 
> elements, and the set of all finite sequences of "0" and "1" of length 
> n.
> 2) What is common between the set of all subsets of N, and the set of 
> all infinite sequences of "0" and "1".
>
>

Bruno Marchal

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Jul 16, 2009, 10:00:02 AM7/16/09
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On 16 Jul 2009, at 15:17, John Mikes wrote:

 
I would not restrict 'a set' to what WE can conceive, or define now. (Not even within the 'math'-related domain).


Nor do I. 
I never say so. On the contrary we will see how different are sets when seen by machines and gods, but to explain this it is necessary we agree on elementary properties and definitions on sets, so that we can proceed.
Here by "we" you are free to take the  "we" by any entities (machines, humans, gods, or whatever).





the relations of the set-partners comes into play. Not only those which WE choose for 'interesting' to such set, but ALL OF THEM influencing the character of that "ONE".
Just musing.

It is OK. The idea consists in simplifying the things as much as possible, and then to realize that despite such simplification we are quickly driven to the unprovable, unnameable, un-reductible, far sooner than we could have imagine. 
 
I may suggest (or: assume?) that instead of "despite" it would make more sense to write: "AS A CONSEQUENCE"
- think about it.

It could make sense. This would lead to finitism and or mechanism, which I am trying to share with you. But again, this is an anticipation, and can hardly been made precise if we don't train ourselves to think about those simple things before.

Bruno


Bruno Marchal

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Jul 16, 2009, 10:06:08 AM7/16/09
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Read my last general post. I have to go now, you can wait for the answer. 

It is good to search without finding the answer, so as to better appreciate the answer.

When I will give the answer, ask yourself question like "how is it that I didn't think about that, how is it that I have not seen this or that, how is it that I was forgetting this or that ...

My last post should help a little bit,

Think about this is as far as you have some fun by asking yourself, and then stop.

Bruno








Bruno Marchal

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Jul 16, 2009, 3:56:54 PM7/16/09
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Ok, so now you have perhaps solve the riddle! The answer is No. I did not really provide all the definitions you could need!

I feel a bit sorry. Mathematicians are kind of summarizing the rest of the book in the exercises.

What are the sequences of "0" and "1" of length n? 

That is the question.

I give some examples of sequences of "0" and "1" of length 7:


  1111111, 0000000, 1010101, and  0100111

Here are some sequence of length three:  000, 101, 111.

Here are ALL examples of the sequences with length two:

00
01
10
11

Convince yourself that there are no other binary sequence of length 2. 

Here is an example of sequence of "0" and "1" of length 24:       

                        0111101010001000000000001

Here are the only two examples of binary sequences of length 1

0
1

And here is the only empty sequence, the one which length is 0.



You can't see it, of course. Even under a microscope! But there is one!


The problem "1)" was "What is common between the set of all subsets of a set with n elements, and the set of all sequence of "0" and 1"? 

You can search by yourself if the question is more clear, or look at the explanation below. You = anyone interested, except Kim and Marty, which have the obligation to train themselves, because they are victim of summer school (that happens in hot summers).

Take your time. The problem was certainly not clear if you did not know what is a so-called binary sequences, or sequences of "0" and "1". 

.
.
.
.




------------------------------------------------------------------------------------
The answer is that they have the same number of elements. 
And that number is 2^n. 



Why?

Let us see, and, to fix the idea, consider what happens with the set {a, b, c}, which powerset has already just be computed in a preceding post:
The powerset of {a, b, c} is {{ }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}. 

What is common between being a subset of {a, b, c} and a binary sequence of length 3?

You can see the link immediately by realizing that a subset of a set with n elements is entirely determined by the answers to n yes/no questions. With {a, b, c}, there are only three possible question? We can put them in the following order:

1)Does a belong to the subset?
2) Does b belong to subset?
3) Does  c belong to the subset? 

The poor empty subset { } is the one which get the answers: no, no, no.  
The less poor singleton {a} is the one which get the answers: yes, no, no.

Let us abstract, and let us write 0 in place of "no", and 1 in place "yes", 

------------------------------ { }  ------------------    000
------------------------------ {a}  ------------------   100
------------------------------ {b}  ------------------   010
------------------------------ {c}  ------------------   001
------------------------------ {a, b}  ---------------- 110
------------------------------ {a, c}  ------------------101
------------------------------ {b, c}  ------------------011

and the winner is:

------------------------------ {a, b,c}  ------------------111

So, there are as many subsets included in a set with n elements than there are binary sequences of length n.

And how many?

Suppose I have to invent a binary sequence of length 1.

Well, it is a binary sequence, so I have not much choice in the beginning. It is either 0 or 1. I can hesitate a long time, but I have only two possibilities. 0, and 1.

Suppose I have to invent a binary sequence of length 2.

I have two possibilities for the first digit, and for each such choice it remains two choice for the next, and last.

If I choose to begin with 0, I can still end with 1 or with 0:  01,  00. The possibilities of having 0 and 1 at a place does not depend of what happened before: so they multiply.

How many sequences of length 3? =

= (2 possibilities for the first digit) times (2 possibilities for the second digit) times (2 possibilities for the third digit)

= 2 times 2 times 2 = 8.

OK?

How many subsets of a set with three elements {a, b, c}.

If I have to "invent" a subset S
= (2 possibilities for the question "is a in the subset S") times (2 possibilities for the question "is b in the subset S") times (2 possibilities for the question "is c in the subset S")

= 2 times 2 times 2 = 8.

Question?

What about problem 2).?

I recall that N is the set of natural numbers {0, 1, 2, 3, ……… }. I let you think about the relation between subset of N, and infinite binary sequences.
Hint: the poor empty subset { } is the one getting the answer no, no, no, no, no, no, no, no, no, no, no, no, no, ....

Hot summer!

Bruno



m.a.

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Jul 20, 2009, 9:34:27 AM7/20/09
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Bruno,
             I don't know about Kim, but I'm ready to push on. I'm waiting for the answer to problem 2) see below. And could you please retstate that problem as I'm not sure which one it is? Thanks,   marty a.
 
 
----- Original Message -----
Sent: Thursday, July 16, 2009 3:56 PM
Subject: Re: The seven step series


Brian Tenneson

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Jul 20, 2009, 1:44:06 PM7/20/09
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I found a paper that might be of interest to those interested in Tegmark's work.

http://arxiv.org/abs/0904.0867

Abstract
I discuss some problems related to extreme mathematical realism, focusing on a recently proposed "shut-up-and-calculate" approach to physics (arXiv:0704.0646, arXiv:0709.4024). I offer arguments for a moderate alternative, the essence of which lies in the acceptance that mathematics is (at least in part) a human construction, and discuss concrete consequences of this--at first sight purely philosophical--difference in point of view.

-Brian

Bruno Marchal

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Jul 20, 2009, 3:17:48 PM7/20/09
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On 20 Jul 2009, at 15:34, m.a. wrote:

Bruno,
             I don't know about Kim, but I'm ready to push on. I'm waiting for the answer to problem 2) see below. And could you please retstate that problem as I'm not sure which one it is? Thanks,   marty a.
 


Let us see:



> 1) What is common between the set of all subsets of a set with n  
> elements, and the set of all finite sequences of "0" and "1" of length  
> n.
> 2) What is common between the set of all subsets of N, and the set of  
> all infinite sequences of "0" and "1".



Let me restate by introducing a "definition" (made precise later). The cardinal of a set S is the number of elements in the set S.

The cardinal of { } = 0. All singletons have cardinal one. All pairs, or doubletons, have cardinal two.  

Problem 1 has been solved. They have the same cardinal, or if you prefer, they have the same number of elements. The set of all subsets of a set with n elements has the same number of elements than the set of all strings of length n.

Let us write  B_n for the sets of binary strings of length n. So,

B_0 = { }
B_1 = {0, 1}
B_2 = {00, 01, 10, 11}
B_3 = {000, 010, 100, 110, 001, 011, 101, 111}

We have seen, without counting, that the cardinal of the powerset of a set with cardinal n is the same as the cardinal of B_n.

And then we have seen that such cardinal was given by 2^n. 
You can see this directly by seeing that adding an element in a set, double the number of subset, due to the dichotomic choice in creating a subset "placing or not placing" the new element in the subset.
Likewise with the strings. If you have already all strings of length n, you get all the strings of length n+1, by doubling them and adding zero or one correspondingly.
This is also illustrated by the iterated self-duplication W, M. Mister X is cut and paste in two rooms containing each a box, in which there is a paper with zero on it, in room W, and 1 on it in room M. After the experience, the 'Mister X' coming out from room W wrote 0 in his diary, and the 'Mister X' coming out from room M wrote 1 in his diary. And then they redo each, the experiment. The Mister-X with-0-in-his-diary redoes it, and gives a Mister-X with-0-in-his-diary coming out from room W, and adding 0 in its diary and a  Mister-X with-0-in-his-diary coming out from room M, adding 1 in its diary: they have the stories 

00

01,

and then the Mister-X-coming-from room W, and with 1 written in the diary, similarly redoes the experiment, and this gives two more Misters X, having written in their diaries

10

11.

Obviously the iteration of the self-duplication, gives as result 2x2x2x2x...x2 number of Mister X. If those four Mister X duplicate again, there will be 8 of them, with each of those guys having an element of B_3 written in his diary. OK.

And we have seen that a powerset of a set with n elements can but put in a nice correspondence with B_n.

For example: The powerset of {a, b}, that is {{ }, {a}, {b}, {a, b}}, has the following nice correspondence 

00 .............. { }
01 .............. {a}
10 ...............{b}
11 ...............{a, b}

Each 0 and 1 corresponding to the answer to the yes/no questions 'is a in the subset?', is 'b in the subset?'.

Such a nice correspondence between two sets is called a BIJECTION, and will be defined later. What we have seen, thus, is that there is a bijection between the powerset of set with n elements, and the set of binary strings B_n.

And the second question?

What is common between the subsets of N, and the set of infinite binary sequences. An infinite binary sequence is a infinite sequences of "0" and "1".
For example: 00000000000..., with only zero is such a sequence. It could be the first person story of 'the mister X who comes always from room W. Or, if the zero and one represents the result of the fair coin throw experiment, it could be the result of the infinitely unlucky guy: he always get the head.
Another one quite similar is 1111111111111111111111111111..., the infinitely lucky guy.
A more 'typical' would be 1100100100001111110110101010001000... (except that this very one *is* typical, it is PI written in binary; PI = 11. 00100...). 

The link with the subsets of N? It is really the same as above, except that we extend the idea on the infinite.

A subset of N, that is, a set included in N,  is entirely determined by the answer to the following questions:

Is 0 in the subset?
Is 1 in the subset?
Is 2 in the subset?
Is 3 in the subset?
etc.

You will tell me that nobody can answer an infinity of questions. I will answer that in many situation we can.

Let us take a simple subset of N, the set {3, 4, 7}. It seems to me we can answer to the infinite set of corresponding question:

Is 0 in the subset?    NO
Is 1 in the subset?    NO
Is 2 in the subset?  NO
Is 3 in the subset?   YES
Is 4 in the subset?    YES
Is 5 in the subset?    NO
Is 6 in the subset?   NO
Is 7 in the subset?   YES
Is 8 in the subset?   NO
Is 9 in the subset?    NO
Is 10 in the subset?    NO
Is 11 in the subset?   NO
Is 12 in the subset?    NO
Is 13 in the subset?    NO
Is 14 in the subset?  NO
... (etc. and we answer NO for all remaining questions ...!)

So, in the same spirit as above we associate the infinite binary string 000110010000000000000000... to the subset {3, 4, 7}.

Even simpler: the empty set. First, is the empty set a subset of N?But we have seen that to verify if the empty set is a subset of any set, we have 0 verification to do: the empty set is included in any set. It correspond here to 

Is 0 in the subset?    NO
Is 1 in the subset?    NO
Is 2 in the subset?  NO
Is 3 in the subset?   NO
Is 4 in the subset?   NO
Is 5 in the subset?    NO
Is 6 in the subset?   NO
Is 7 in the subset?   NO
Is 8 in the subset?   NO
Is 9 in the subset?    NO
Is 10 in the subset?    NO
Is 11 in the subset?   NO
Is 12 in the subset?    NO
Is 13 in the subset?    NO
Is 14 in the subset?  NO
... (and we answer NO for all remaining questions ...!)

the empty set, seen as a subset of N, is determined by the sequence:

000000000000000000000000000000000000000000 ...

All singleton of number, like {0}, {1}, {2}, {3}, {4}, {5}, ...

will have the following "characteristic" sequences:

{0} ---- 1000000000000000...
{1} ---- 0100000000000000...
{2} ---- 001000000000000..
etc.

Doubleton will have characteristic sequences being strings having two occurences of 1, and thus an infinity of zero.

OK, for finite set, there is a time where all answers become "NO". 
But what about infinite sets? Can we still answer the infinity of question?

Of course, at least for "simple" infinite sets.

First, do we know infinite subset of N? yes, example the set of odd numbers. {1, 3, 5, ...} is included in {0, 1, 2, 3, ...}.

What is its characteristic sequences, that is the answer to the questions: is 0 in?, is 1in?, is 2 in?,  ...


Is 0 in the subset?    NO
Is 1 in the subset?    YES
Is 2 in the subset?  NO
Is 3 in the subset?   YES
Is 4 in the subset?   NO
Is 5 in the subset?    YES
Is 6 in the subset?   NO
Is 7 in the subset?   YES
Is 8 in the subset?   NO
Is 9 in the subset?    YES
Is 10 in the subset?    NO
Is 11 in the subset?   YES
Is 12 in the subset?    NO
Is 13 in the subset?    YES
Is 14 in the subset?  NO
... (and we answer YES and NO in that way for all remaining questions ...!)

So the set of odd numbers has the characteristic sequence:

{1, 3, 5, ...} ----------- 010101010101010101010101010101010101010101010...

and the even numbers:

{0, 2, 4, 6, ...} ---------- 101010101010101010101010101010101010101010...

And ... the prime numbers

{2, 3, 5, 7, 11, ...} ------- 00110101000101000101000100000101...

OK.

Do you see that for EACH subset of N we have a corresponding infinite binary sequence, and for EACH binary sequence we have a subset.

Exercise: gives the subset (= write some elements of the subset) corresponding o the sequence "PI"

1100100100001111110110101010001000...



The genius of Cantor has consisted in allowing himself the notion of "nice correspondence", bijection, to infinite sets, including infinite sets of infinite objects, like the subset of N and the set B_infinity, that is the set of infinite binary strings.

And here, I hope you have the feeling that indeed such a nice correspondence exists between the powerset of N, and B_infinity.

I let you think, and ask some questions. Bijection will occupy us for awhile. Unfortunately, to make them precise, I have to define what we have already met a lot, but yet not define, and which is the notion of function.
Functions are utterly important because they define the crucial notion of mathematical dependency, on which eventually the mathematical notion of computation will be a very peculiar particular case.


Bruno




Bruno Marchal

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Jul 20, 2009, 3:58:08 PM7/20/09
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Exercise: criticize the following papers mentioned below in the light of the discovery of the universal machine and its main consequences from incompleteness to first person indeterminacy. Think of the identity thesis. To be sure Tegmark is less "wrong" than Jannes.

Solution: search in the archive of this list where I have already explained this, or use directly UDA, or wait for what will (perhaps) follow.

I should send some of my papers on arXiv, but up to now, only logicians understand the whole "trick", so I have to better appreciated what physicians don't understand in logic, before making a version free of references to mathematical logical baggage. Logicians are not interested in mind, nor really matter, and physicians are still naïve on the link consciousness/reality, I would say.

To be sure Tegmark is closer than most physicists except perhaps Wheeler.

Also, Tegmarks' argument for mathematicalism is invalid (even with strong non-comp axioms). But I prefer to help you to understand this by yourself through the understanding of what a universal machine is, than trying a direct argument.

According of the part of UDA (or perhaps AUDA) you understand, you can already see the weakness of such direct mathematical approach. Note that comp makes physics much more fundamental, and separate it much clearly from possible geograpies. Above all comp does not eliminate the person, which Tegmark is still doing: the frog view is not yet a first person view, in the comp sense.

Interesting stuff, still. Thanks for the references.

Bruno

Brian Tenneson

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Jul 20, 2009, 6:22:54 PM7/20/09
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Comments below.


Bruno Marchal wrote:
Exercise: criticize the following papers mentioned below in the light of the discovery of the universal machine and its main consequences from incompleteness to first person indeterminacy. Think of the identity thesis. To be sure Tegmark is less "wrong" than Jannes.

Solution: search in the archive of this list where I have already explained this, or use directly UDA, or wait for what will (perhaps) follow.

I should send some of my papers on arXiv, but up to now, only logicians understand the whole "trick", so I have to better appreciated what physicians don't understand in logic, before making a version free of references to mathematical logical baggage. Logicians are not interested in mind, nor really matter, and physicians are still naïve on the link consciousness/reality, I would say.

To be sure Tegmark is closer than most physicists except perhaps Wheeler.

Also, Tegmarks' argument for mathematicalism is invalid (even with strong non-comp axioms). But I prefer to help you to understand this by yourself through the understanding of what a universal machine is, than trying a direct argument.
I need to get a better grasp on what a universal machine is, yes.  I am interested in finding out how Tegmark's argument for mathematicalism is invalid, especially since I'm using it to motivate my research.



According of the part of UDA (or perhaps AUDA) you understand, you can already see the weakness of such direct mathematical approach. Note that comp makes physics much more fundamental, and separate it much clearly from possible geograpies. Above all comp does not eliminate the person, which Tegmark is still doing: the frog view is not yet a first person view, in the comp sense.

Interesting stuff, still. Thanks for the references.
I'll have to think more on Jannes' paper.  As I basically resting the motivation of my research on the correctness of "ERH implies MUH," I'm trying to formulate a good refutation to his paper.

Bruno Marchal

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Jul 20, 2009, 7:28:49 PM7/20/09
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On 21 Jul 2009, at 00:22, Brian Tenneson wrote:

Comments below.

Bruno Marchal wrote:
Exercise: criticize the following papers mentioned below in the light of the discovery of the universal machine and its main consequences from incompleteness to first person indeterminacy. Think of the identity thesis. To be sure Tegmark is less "wrong" than Jannes.

Solution: search in the archive of this list where I have already explained this, or use directly UDA, or wait for what will (perhaps) follow.

I should send some of my papers on arXiv, but up to now, only logicians understand the whole "trick", so I have to better appreciated what physicians don't understand in logic, before making a version free of references to mathematical logical baggage. Logicians are not interested in mind, nor really matter, and physicians are still naïve on the link consciousness/reality, I would say.

To be sure Tegmark is closer than most physicists except perhaps Wheeler.

Also, Tegmarks' argument for mathematicalism is invalid (even with strong non-comp axioms). But I prefer to help you to understand this by yourself through the understanding of what a universal machine is, than trying a direct argument.
I need to get a better grasp on what a universal machine is, yes.  I am interested in finding out how Tegmark's argument for mathematicalism is invalid, especially since I'm using it to motivate my research.


At least you are aware that a mathematicalism à-la Tegmark needs a rather sophisticated universal structure, but if we assume even very weak version of comp, the universal machine provides that structure, or that structure has to be reducible as an invariant for a set of effective transformation of that machine. We can come back on this. I may be wrong also.






According of the part of UDA (or perhaps AUDA) you understand, you can already see the weakness of such direct mathematical approach. Note that comp makes physics much more fundamental, and separate it much clearly from possible geograpies. Above all comp does not eliminate the person, which Tegmark is still doing: the frog view is not yet a first person view, in the comp sense.

Interesting stuff, still. Thanks for the references.
I'll have to think more on Jannes' paper.  As I basically resting the motivation of my research on the correctness of "ERH implies MUH," I'm trying to formulate a good refutation to his paper.

OK, nice.

My main critics is that they seem not be aware of the consciousness/reality problem. They are using an identify thesis which is not allowed by comp. The UD argument shows exactly that. It is build to show that if we keep consciousness, eventually, physics is even more fundamental than physicist imagine. The physical world(s) is(are) not just a 'sufficiently rich' part of math, it is somehow the border of the ignorance of any (Löbian) universal machine which introspects itself. This connects in some way all part 'sufficiently rich' part  of math". It explains also the non communicable part of what we can be conscious of, including physical sensations (as modalities related to self-references).

Bruno









m.a.

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Jul 21, 2009, 3:36:19 PM7/21/09
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Hi Bruno,
                   I'm not clear on the sentence in bold below, especially the word "correspondingly". The example of Mister X only confuses me more. Could you please give some simple examples? Thanks,
                                                                                                                                                                                                                                                                                        marty a.
 
 
----- Original Message -----
Sent: Monday, July 20, 2009 3:17 PM
Subject: Re: The seven step series


On 20 Jul 2009, at 15:34, m.a. wrote:

And then we have seen that such cardinal was given by 2^n. 
You can see this directly by seeing that adding an element in a set, double the number of subset, due to the dichotomic choice in creating a subset "placing or not placing" the new element in the subset.
 
Likewise with the strings. If you have already all strings of length n, you get all the strings of length n+1, by doubling them and adding zero or one correspondingly.
 
This is also illustrated by the iterated self-duplication W, M. Mister X is cut and paste in two rooms containing each a box, in which there is a paper with zero on it, in room W, and 1 on it in room M. After the experience, the 'Mister X' coming out from room W wrote 0 in his diary, and the 'Mister X' coming out from room M wrote 1 in his diary. And then they redo each, the experiment. The Mister-X with-0-in-his-diary redoes it, and gives a Mister-X with-0-in-his-diary coming out from room W, and adding 0 in its diary and a  Mister-X with-0-in-his-diary coming out from room M, adding 1 in its diary: they have the stories 

 
Bruno





Brent Meeker

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Jul 21, 2009, 3:57:59 PM7/21/09
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Each binary string of length n has two possible continuations of length
n+1, one of them by appending a 0 and one of them by appending a 1. So
to get all binary strings of length n+1 take each string of length n,
make two copies, to one copy append a 0 and to the other copy append a 1.

Brent

m.a. wrote:
> Hi Bruno,
> I'm not clear on the sentence in bold below,
> especially the word "correspondingly". The example of Mister X only
> confuses me more. Could you please give some simple examples? Thanks,
>
>
>
>
> marty a.
>
>
>
> ----- Original Message -----
> *From:* Bruno Marchal <mailto:mar...@ulb.ac.be>
> *To:* everyth...@googlegroups.com
> <mailto:everyth...@googlegroups.com>
> *Sent:* Monday, July 20, 2009 3:17 PM
> *Subject:* Re: The seven step series
>
>
> On 20 Jul 2009, at 15:34, m.a. wrote:
>
>> And then we have seen that such cardinal was given by 2^n.
> You can see this directly by seeing that adding an element in a
> set, double the number of subset, due to the dichotomic choice in
> creating a subset "placing or not placing" the new element in the
> subset.
>
> *Likewise with the strings. If you have already all strings of
> length n, you get all the strings of length n+1, by doubling them
> and adding zero or one correspondingly.*
>
> This is also illustrated by the iterated self-duplication W, M.
> Mister X is cut and paste in two rooms containing each a box, in
> which there is a paper with zero on it, in room W, and 1 on it in
> room M. After the experience, the 'Mister X' coming out from room
> W wrote 0 in his diary, and the 'Mister X' coming out from room M
> wrote 1 in his diary. And then they redo each, the experiment. The
> Mister-X with-0-in-his-diary redoes it, and gives a Mister-X
> with-0-in-his-diary coming out from room W, and adding 0 in its
> diary and a Mister-X with-0-in-his-diary coming out from room M,
> adding 1 in its diary: they have the stories
>
>
> Bruno
>
>
>
> http://iridia.ulb.ac.be/~marchal/
> <http://iridia.ulb.ac.be/%7Emarchal/>
>
>
>
>
>
> >

m.a.

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Jul 21, 2009, 4:53:47 PM7/21/09
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Thanks Brent,
                       Could you supply some illustrative examples?   
                                                                                                     marty a.
 
 
----- Original Message -----
From: "Brent Meeker" <meek...@dslextreme.com>
Sent: Tuesday, July 21, 2009 3:57 PM
Subject: Re: The seven step series

Brent Meeker

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Jul 21, 2009, 5:24:36 PM7/21/09
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Take all strings of length 2
00 01 10 11
Make two copies of each
00 00 01 01 10 10 11 11
Add a 0 to the first and a 1 to the second
000 001 010 011 100 101 110 111
and you have all strings of length 3.

Brent

m.a. wrote:
> *Thanks Brent,*
> * Could you supply some illustrative examples? *
> *
> marty a.*
> **

m.a.

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Jul 22, 2009, 9:25:10 AM7/22/09
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Hi Brent,
                I really appreciate the help and I hate to impose on your patience but...(see below)
 
----- Original Message -----
From: "Brent Meeker" <meek...@dslextreme.com>
Sent: Tuesday, July 21, 2009 5:24 PM
Subject: Re: The seven step series

>
> Take all strings of length 2
> 00             01                   10               11
> Make two copies of each
> 00      00      01      01      10      10      11      11
 
> Add a 0 to the first and a 1 to the second
> 000    001      010   011      100   101   110      111
> and you have all strings of length 3.
I can see where adding 0 to the first and 1 to the second gives 000 and 001 and I think I see how you get 010 but the rest of the permutations don't seem obvious to me. P-l-e-a-s-e  explain,  Best,
 
                                                                                                                                                                                                                                m. (mathematically hopeless)  a.

Bruno Marchal

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Jul 22, 2009, 12:20:11 PM7/22/09
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Marty,

Brent wrote:

On 21 Jul 2009, at 23:24, Brent Meeker wrote:


Take all strings of length 2
00             01                   10               11
Make two copies of each
00      00      01      01      10      10      11      11
Add a 0 to the first and a 1 to the second
000    001      010   011      100   101   110      111
and you have all strings of length 3.


Then you wrote

I can see where adding 0 to the first and 1 to the second gives 000 and 001 and I think I see how you get 010 but the rest of the permutations don't seem obvious to me. P-l-e-a-s-e  explain,  Best,
 
                                                                                                                                                                                                                                m. (mathematically hopeless)  a.
 
 


Let me rewrite Brent's explanation, with a tiny tiny tiny improvement:


Take all strings of length 2
00 
01
10
11
Make two copies of each

first copy:
00 
01
10
11

second copy
00 
01
10
11

add a 0 to the end of the strings in the first copy, and then add a 1  to the end of the strings in the second copy:

first copy:
000
010
100
110

second copy
001
011
101
111

You get all 8 elements of B_3.

You can do the same reasoning with the subsets. Adding an element to a set multiplies by 2 the number of elements of the powerset:

Exemple. take a set with two elements {a, b}. Its powerset is {{ } {a} {b} {a, b}}. How to get all the subset of {a, b, c} that is the set coming from adding c to {a, b}.

Write two copies of the powerset of {a, b}

{ }
{a}
{b}
{a, b}

{ }
{a}
{b}
{a, b}

Don't add c to the set in the first copy, and add c to the sets in the second copies. This gives

{ }
{a}
{b}
{a, b}

{c}
{a, c}
{b, c}
{a, b, c}

and that gives all subsets of {a, b, c}.

This is coherent with interpreting a subset {a, b} of a set {a, b, c}, by a string like 110, which can be conceived as a shortand for

Is a in the subset?   YES, thus 1
Is b in the subset?   YES thus  1
Is c in the subset?    NO thus   0.

OK?

You say also:

The example of Mister X only confuses me more.

Once you understand well the present post, I suggest you reread the Mister X examples, because it is a key in the UDA reasoning. If you still have problem with it, I suggest you quote it, line by line, and ask question. I will answer (or perhaps someone else).

Don't be afraid to ask any question. You are not mathematically hopeless. You are just not familiarized with reasoning in math. It is normal to go slowly. As far as you can say "I don't understand", there is hope you will understand.

Indeed, concerning the UDA I suspect many in the list cannot say "I don't understand", they believe it is philosophy, so they feel like they could object on philosophical ground, when the whole point is to present a deductive argument in a theory. So it is false, or you have to accept the theorem in the theory. It is a bit complex, because it is an "applied theory". The mystery are in the axioms of the theory, as always.

So please ask *any* question. I ask this to everyone. I am intrigued by the difficulty some people can have with such reasoning (I mean the whole UDA here). (I can understand the shock when you get the point, but that is always the case with new results: I completely share Tegmark's idea that our brain have not been prepared to have any intuition when our mind try to figure out what is behind our local neighborhood).

Bruno




Brent Meeker

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Jul 22, 2009, 12:57:50 PM7/22/09
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m.a. wrote:
> Hi Brent,
> I really appreciate the help and I hate to impose on
> your patience but...(see below)
>
> ----- Original Message -----
> From: "Brent Meeker" <meek...@dslextreme.com
> <mailto:meek...@dslextreme.com>>
> To: <everyth...@googlegroups.com
> <mailto:everyth...@googlegroups.com>>
> Sent: Tuesday, July 21, 2009 5:24 PM
> Subject: Re: The seven step series
>
> >
> > Take all strings of length 2
> > 00 01 10 11
> > Make two copies of each
> > 00 00 01 01 10 10 11 11
>
> > Add a 0 to the first and a 1 to the second
> > 000 001 010 011 100 101 110 111
> > and you have all strings of length 3.
> *I can see where adding 0 to the first and 1 to the second gives 000 and
> 001 and I think I see how you get 010 but the rest of the permutations
> don't seem obvious to me. P-l-e-a-s-e explain, Best,*
> **
>
>
>
> * m. (mathematically hopeless) a.*

They aren't permutations. They're just sticking a 0 or 1 on the end. One copy
of 01 becomes 010 and the other become 011.

Brent

m.a.

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Jul 22, 2009, 11:28:57 PM7/22/09
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Going a step further... (see below)
>>
They aren't permutations.  They're just sticking a 0 or 1 on the end.  One copy
> of 01 becomes 010 and the other become 011.
 
Then I assume the next step would be making two copies of each of those:
 
000    000       001     001      010      010       011     011     100      100       101         101         110           110             111          111
 
...and sticking a 0 or 1 at the end:
 
0000   0001    0010    0011     0100    0101    0110    0111    1000     1001     1010       1011         1100         1101           1110      1111
 
and this is the binary sequence of length 4.
 
How do these translate into ordinary numerals? 1,2,3,4...

>
> Brent
>
>

Brent Meeker

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Jul 22, 2009, 11:49:24 PM7/22/09
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m.a. wrote:
> *Going a step further... (see below)*
> **
> *Then I assume the next step would be making two copies of each of those:*
> **
> *000 **000 001 001 010 010 011 011
> 100 100 101 101 110 110
> 111 111*
> **
> *...and sticking a 0 or 1 at the end:*
> **
> *0000 0001 0010 0011 0100 0101 0110 0111
> 1000 1001 1010 1011 1100 1101
> 1110 1111*
> **
> *and this is the binary sequence of length 4.*

Right, it's all the binary strings of length 4

> **
> *How do these translate into ordinary numerals? 1,2,3,4...*

Bruno's using them to represent sets and subsets. So if we have a set {a b c}
we can represent the subset {a c} by 101 and {a b} by 110, etc. That's quite
different from using a binary string to represent a number in positional
notation. I'll leave it to Bruno whether he wants to go into that.

Brent

m.a.

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Jul 23, 2009, 8:02:43 AM7/23/09
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Hi Bruno,
                I asked Brent Meeker a question which he referred back to you. Will you be covering it? (see para in bold below)

m.a.

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Jul 23, 2009, 9:09:40 AM7/23/09
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Bruno,
            Yes, yours and Brent's explanations seem very clear. I hate to ask you to spell things out step by step all the way, but I can tell you that when I'm confronted by a dense hedge or clump of math symbols, my mind refuses to even try to disentangle them and reels back in terror. So I beg you to always advance in baby steps with lots of space between statements. I want to assure you that I'm printing out all of your 7-step lessons and using them for study and reference. Thanks for your patience,   m.a.
 
 
 
 
-- Original Message -----
Sent: Wednesday, July 22, 2009 12:20 PM
Subject: Re: The seven step series

Bruno Marchal

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Jul 23, 2009, 9:12:37 AM7/23/09
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Hi Marty,

I can if you really want it, but it is out of topic and could introduced some confusion.  I suggest we could come back on this later perhaps. But if you insist, I can do it. have you get my last post?
Note that I have also already explained how binary strings can represent number in some older post. Honestly we will not need this, so it is better not to accumulate too many "new" materials, especially when I can fear some confusion. It is good to be familiar with the object "binary strings" seen as an object by itself.

Bruno

Bruno Marchal

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Jul 23, 2009, 9:28:13 AM7/23/09
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On 23 Jul 2009, at 15:09, m.a. wrote:

Bruno,
            Yes, yours and Brent's explanations seem very clear. I hate to ask you to spell things out step by step all the way, but I can tell you that when I'm confronted by a dense hedge or clump of math symbols, my mind refuses to even try to disentangle them and reels back in terror. So I beg you to always advance in baby steps with lots of space between statements. I want to assure you that I'm printing out all of your 7-step lessons and using them for study and reference. Thanks for your patience,   m.a.


Don't worry, I understand that very well. And this illustrates also that your "despair" is more psychological than anything else. I have also abandoned the study of a mathematical book until I realize that the difficulty was more my bad eyesight than any conceptual difficulties. With good spectacles I realize the subject was not too difficult, but agglomeration of little symbols can give a bad impression, even for a mathematician.

I will make some effort, tell me if my last post, on the relation

     (a^n) * (a^m) = a^(n + m)

did help you.

You are lucky to have an infinitely patient teacher. You can ask any question, like "Bruno,

is (a^n) * (a^m) the same as a^n times a^m?" 
 Answer: yes, I use often "*", "x", as shorthand for "times", and I use "(" and ")" as delimiters in case I fear some ambiguity.

Bruno

ronaldheld

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Jul 27, 2009, 10:07:10 AM7/27/09
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Bruno:
I am following, but have not commented, because there is nothing
controversal.
When you are done, can your posts be consolidated into a paper or a
document that can be read staright through?
Ronald
> http://iridia.ulb.ac.be/~marchal/- Hide quoted text -
>
> - Show quoted text -

Bruno Marchal

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Jul 27, 2009, 1:27:50 PM7/27/09
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On 27 Jul 2009, at 16:07, ronaldheld wrote:
>
> I am following, but have not commented, because there is nothing
> controversal.

Cool. Even the sixth first steps of UDA?


>
> When you are done, can your posts be consolidated into a paper or a
> document that can be read staright through?

I should do that.

Bruno
http://iridia.ulb.ac.be/~marchal/



Bruno Marchal

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Jul 27, 2009, 4:54:40 PM7/27/09
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Hi,

OK, I will come back on the square root of 2 later.

We have talked on sets.

Sets have elements, and elements of a set define completely the set,
and a set is completely defined by its elements.

Example: here is a set of numbers {1, 2, 3}
and a set of sets of numbers {{1, 2}, {3}, { }}.

We can do some operations, like their union, or their intersection.
Examples:
{1,2,3} union {3,4,5} = {1,2,3,4,5},
{1,2,3} intersection {3,4,5} = { }.

We can verify if some relation hold for them, like equality, or
inclusion.
{1, 2} = {2, 1, 1} (yes!)
{1, 2} included-in {3, 2, 1}
{1, 2} not-included-in {1, 3}

We can compute their powerset.
Powerset {1, 2} = {{ }, {1}, {2}, {1, 2}}

We have discovered SBIJECTION between powersets of a set with cardinal
n, and the set of binary strings of length n.
And we have presented reasons for the existence of a bijection between
the powerset of N = {0, 1, 2, ...} and the set of infinite binary
strings.

OK?

Today, I suggest we look at two new operations on sets. The product of
sets, and the exponentiation of sets. Well, I will probably do only
the product today.

First I have to introduce a new, well actually *very* well known, and
absolutely important, notion: the couple.

A couple is when there is two things, but with some order. It looks
like a pair, but the order counts.

Usually a couple of things a , b is designated, in math, like this:

(a, b).

It looks like a pair {a, b}, but it is not. Indeed, {a, b} = {b, a},
but the couple (a, b) is NOT equal to the couple (b, a).

When are two couples (a, b) and (c, d) equal? Only when a = c and b = d.

Examples. the couple of number (2, 3) is not equal to the couple (3,
2), but the couple (0, 666) is equal to the couple (0, 666).

OK?

APARTE: Are couples sets? No. Nor are numbers. But yes, you can easily
represent them by sets, so we could work only with sets, but we will
not do that. Much later we will work only with numbers, in fact. The
very notion of representation will be important, though.

Now we are ready to define the so called "cartesian" product of sets.
It is indeed a cousin of Descartes' discovery that you can represent a
point of the plane by a couple of (real) numbers. I read somewhere
that Descartes discovered this by trying to describe a spider walking
on a window with squared little piece of glass. But such a
localization works also for cities like Los Angeles where you address
is something like 15th avenue 61th street. The whole field of
analytical geometry is founded on this idea.

That cartesian idea generalises on sets A and B. It is written A X B,
and it is defined by
the set of couples (x, y) such that x belongs to A, and y belongs to B.

AXB = {(x, y) such-that x belongs-to A, and y belongs to B}.
(compare with the preceding definitions).

Example: what is the product {0, 1} X {a, b}? Well it is the set of
all the couples made from elements of A in company of elements of B,
and in that order, with A = {0, 1}, and B = {a, b}.
So (0, a) is in there, and there are others. The product of {0, 1)
with {a, b} is equal to

{(0,a), (0, b), (1, a), (1, b)}

The convenient usual cartesian drawing is, for AXB, with A = {0, 1},
and B = {a, b} :


a (0, a) (1, a)

b (0, b) (1, b)

0 1

A product of numbers a and b, ab, can be conceived as the area of a
rectangle of sides a and b. Here you can see that the product of sets
AXB can fit in a rectangle when you dispose horizontally the elements
of A, and vertically the elements of B. By convention, usually A is
put horizontally, and B vertically.
But note that if the number ab is equal to the number ba, it is not
the case that the set AXB is equal to the set BXA. (0, a) does not
belong to BXA, for example.

Exercise: to the cartesian drawing for BXA.

1)
Compute
{a, b, c} X {d, e} =
{d, e} X {a, b, c} =
{a, b} X {a, b} =
{a, b} X { } =

2)
Convince yourself that the cardinal of AXB is the product of the
cardinal of A and the cardinal of B.
A and B are finite sets here. Hint: meditate on their cartesian drawing.

3) Draw a piece of NXN.

Solution and sequel tomorrow.

Any question?

Bruno

http://iridia.ulb.ac.be/~marchal/

ronaldheld

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Jul 28, 2009, 6:51:54 AM7/28/09
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Bruno:
I meant the mathematical formalism you are teaching us. When we
eventually get to the UDA steps, I wil be better able to do that
assessment.
Ronald
> >>http://iridia.ulb.ac.be/~marchal/-Hide quoted text -

m.a.

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Jul 28, 2009, 11:36:52 AM7/28/09
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Bruno,
             I have searched my notes for an exposition of BIJECTION and found only one mention in an early email which promises to define it in a later lesson. Do you have a reference to that lesson or perhaps an instant explanation of it? Thanks,
                                                                              Chief Ignoramus
 
 
 
             
----- Original Message -----
From: "Bruno Marchal" <mar...@ulb.ac.be>
Sent: Monday, July 27, 2009 4:54 PM
Subject: Re: The seven step series

>
 
 
 
 
 
 
 
 
 

>
>
>

Bruno Marchal

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Jul 28, 2009, 3:47:53 PM7/28/09
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On 28 Jul 2009, at 17:36, m.a. wrote:

Bruno,
             I have searched my notes for an exposition of BIJECTION and found only one mention in an early email which promises to define it in a later lesson. Do you have a reference to that lesson or perhaps an instant explanation of it? Thanks,
                                                                              Chief Ignoramus


I hope you are not stuck by that, given that the cartesian product does not rely on the understanding of "bijection". 

An instant explanation of bijection is this:

Suppose you have two sets A and B, and you would like to know if they have the same number of elements. For example:

A = {a,b,c,d,e,f,g}

and 

B = {1,  2,  3,  4,  5,  6,  7}

Suppose that you cannot count. You forget the lessons for counting!

But you have ficelles, I mean ropes, cords, or strings.

So you can line up the two sets, and try to attach to each elements of A a piece of rope, and joint them to one element of the set B.
IF you succeed doing that, and respecting the one-one or 1-1link, and getting all the elements of B (the "onto" condition), THEN you have shown the existence of a bijection between the two sets.

Let us see if that work on the example.  The "----------" represent the pieces of rope. OK?

a  ----------  1
b  ----------  2
c  ----------  3
d  ----------  4
e  ----------  5
f  ----------  6
g  ----------  7

So there is a bijection between A and B. 

The bijection *is* that association, as we will defined much later(*).

Other bijections can exist between A and B, like

a  ----------  7
b  ----------  2
c  ----------  3
d  ----------  4
e  ----------  5
f  ----------  6
g  ----------  1

It is enough that one exist, to conclude the sets have the same "number of elements", or same cardinal.

Convince you that if two sets have different number of elements, there is no bijection in between. It has to be 1-1, and "onto" (no missing element).

Exercise (but no hurry): 
Verify if you can see some bijections existing between the powerset of a set with 2 elements, and B_2. The same for the powerset of a set with 3 elements, and B_3.

Bruno

!*) Oh! I can give you the particular mathematical bijection, existing between A and B, given that I have already define the notion of couple.
It is the set of couples representing naturally those rope association:

{(a,1), (b,2) (c,3) (d,4), (e,5), (f,6), (g,7)}.    Take it easy, and be sure you have read what I say about the couples before.





Bruno Marchal

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Jul 29, 2009, 11:29:28 AM7/29/09
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Ronald,

On 28 Jul 2009, at 12:51, ronaldheld wrote:

>
> Bruno:
> I meant the mathematical formalism you are teaching us. When we
> eventually get to the UDA steps, I wil be better able to do that
> assessment.
>

OK.
Note that the first 6 steps have already be done recently, with Kim,
and even before. But there is no problem to come back on this, later.
The key point there consists in explaining the first person
indeterminacy, and its invariance for set of transformations (adding
delays in the computation, going from "real" to "virtual", etc.).
You may prepare yourself by reading the relevant portion of the sane04
paper. Eventually the seventh step itself somehow recapitulates the 6
preceding steps, so it is OK.

Best,

Bruno


http://iridia.ulb.ac.be/~marchal/

Bruno Marchal

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Jul 29, 2009, 11:59:26 AM7/29/09
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SOLUTIONS


OK. I give the solution of the exercises of the last session, on the cartesian product of sets.

I recall the definition of the product A X B. 

A X B    =   {(x,y) such that x belongs to A and y belongs to B}

 I gave A = {0, 1}, and B = {a, b}.


In this case, A X B = {(0,a), (0, b), (1, a), (1, b)}

The  cartesian drawing is, for AXB :



a     (0, a)   (1, a)

b     (0, b)  (1, b)

        0          1


Exercise: do the cartesian drawing for BXA.

Solution:

1     (a, 1)   (b, 1)

0     (a, 0)  (b, 0)

        a          b

You see that B X A = {(a,0), (a,1), (b,0), (b, 1)}

You should see that, not only A X B is different from B X A, but AXB and BXA have an empty intersection. They have no elements in common at all. But they do have the same cardinal 2x2 = 4.


1)
Compute
{a, b, c} X {d, e} =
I show you a method (to minimize inattention errors):

I wrote first {(a, _),  (b, _), (c, _), (a, _),  (b, _), (c, _)}  two times because I have seen that {d, e} has two elements.
Then I add the second elements of the couples, which comes from {d, e}:

{(a, d),  (b, d), (c, d), (a, e),  (b, e), (c, e)}

OK?


{d, e} X {a, b, c} = {(d, a), (d, b), (d, c), (e, a), (e, b), (e, c)}

{a, b} X {a, b} = {(a, a), (a, b), (b, a), (b, b)}

{a, b} X { } = { }.

OK?


2)
Convince yourself that the cardinal of AXB is the product of the  
cardinal of A and the cardinal of B.
A and B are finite sets here. Hint: meditate on their cartesian drawing.

Question? This should be obvious. No?


3) Draw a piece of NXN.    (with, as usual, N = {0, 1, 2, 3, ...}):

.        .          .         .         .          .         .           .
.        .          .         .         .          .         .        .
.        .          .         .         .          .         .     .
5    (0,5)  (1,5)  (2,5)  (3,5)  (4,5)  (5,5)  ...
4    (0,4)  (1,4)  (2,4)  (3,4)  (4,4)  (5,4)  ...
3    (0,3)  (1,3)  (2,3)  (3,3)  (4,3)  (5,3)  ...
2    (0,2)  (1,2)  (2,2)  (3,2)  (4,2)  (5,2)  ...
1    (0,1)  (1,1)  (2,1)  (3,1)  (4,1)  (5,1)  ...
0    (0,0)  (1,0)  (2,0)  (3,0)  (4,0)  (5,0)  ...

          0       1         2       3         4       5  ...


OK? 


N is infinite, so N X N is infinite too.


 Look at the diagonal: (0,0) (1,1) (2,2) (3,3) (4,4) (5,5) ...

definition: the diagonal of AXA, a product of a set with itself,  is the set of couples (x,y) with x = y.

All right? No question? Such diagonal will have a quite important role in the sequel.

Next: I will say one or two words on the notion of relation, and then we will define the most important notion ever discovered by the humans: the notion of function. Then, the definition of the exponentiation of sets, A^B, is very simple: it is the set of functions from B to A.
What is important will be to grasp the notion of function. Indeed, we will soon be interested in the notion of computable functions, which are mainly what computers, that is universal machine, compute. But even in physics, the notion of function is present everywhere. That notion capture the notion of dependency between (measurable) quantities. To say that the temperature of a body depends on the pressure on that body, is very well described by saying that the temperature of a body is a function of the pressure.
Most phenomena are described by relation, through equations, and most solution of those equation are functions. Functions are everywhere, somehow.

I have some hesitation, though. Functions can be described as particular case of relations, and relations can be described as special case of functions. This happens many times in math, and can lead to bad pedagogical decisions, so I have to make a few thinking, before leading you to unnecessary complications.

Please ask questions if *any*thing is unclear. I suggest the "beginners" in math take some time to invent exercises, and to solve them. Invent simple little sets, and compute their union, intersection, cartesian product, powerset.
You can compose exercises: for example: compute the cartesian product of the powerset of {0, 1} with the set {a}. It is not particularly funny, but it is like music. If you want to be able to play some music instrument, sometimes you have to "faire ses gammes",we say in french; you know, playing repetitively annoying musical patterns, if only to teach your lips or fingers to do the right movement without thinking. Math needs also such a kind of practice, especially in the beginning.
Of course, as Kim said, passive understanding of music (listening) does not need such exercises. Passive understanding of math needs, alas, many "simple" exercises. Active understanding of math, needs difficult exercises up to open problems, but this is not the goal here.

Bruno




John Mikes

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Jul 30, 2009, 11:53:41 AM7/30/09
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Hi, Bruno,
let me skip the technical part and jump on the following text.
F u n c t i o n  as I believe is - for you - the y = f(x) form. For me: the activity - shown when plotting on a coordinate system the f(x) values of the Y-s to the values on the x-axle resulting in a relation (curve). And here is my problem: who does the plotting? (Do not say: YOU are, or Iam, that would add to the function concept the homunculus to make it from a written format into a F U N C T I O N ).
 
John M

Bruno Marchal

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Jul 30, 2009, 3:22:21 PM7/30/09
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Hi John, and the other.

John motivates me to explain what is a function, "for a mathematician".

On 30 Jul 2009, at 17:53, John Mikes wrote:

Hi, Bruno,
let me skip the technical part


OK. But I remind you this current thread *is* technical.



and jump on the following text.
F u n c t i o n  as I believe is - for you - the y = f(x) form. For me: the activity - shown when plotting on a coordinate system the f(x) values of the Y-s to the values on the x-axle resulting in a relation (curve). And here is my problem: who does the plotting? (Do not say: YOU are, or Iam, that would add to the function concept the homunculus to make it from a written format into a F U N C T I O N ).


But I have not yet say what is a function. I just mentioned that "they are everywhere" to open the appetite of the audience.


F u n c t i o n  as I believe is - for you - the y = f(x) form.


You take a risk believing things - for me -. 

Actually the "y = f(x)"  form will come later, with the goal of distinguishing clearly the key difference between a function and the many possible forms of a function.


Ah! but you force me to define what is a function right on (for a mathematician of course). Take it easy. You can skip to the sum-up line below.


OK, ready? I mean the others among those who pursue this mathematical shortcut toward the seventh step (the UD step, actually).


We have already seen functions. If you remember the bijection between A = {a,b,c,d,e,f,g} and B = {1,  2,  3,  4,  5,  6,  7}.

a  ----------  7
b  ----------  2
c  ----------  3
d  ----------  4
e  ----------  5
f  ----------  6
g  ----------  1


I said that the following set of couples 

{(a,1), (b,2) (c,3) (d,4), (e,5), (f,6), (g,7)}

was a nice "set theoretical" representation of the bijection, and that the bijection is an example of function. We can give it a name, F, for example.

F = {(a,1), (b,2) (c,3) (d,4), (e,5), (f,6), (g,7)}.

A function is a mathematical object, actually a set, which embodies an association between the elements of two sets. Here the two sets involved are A and B. 
A is said to be the domain of F. B is said to be the range of F. And the function itself, F,  get a nice set theoretical  "form" of a set. The set of all the couples which determine or define the association. Here it is the set {(a,1), (b,2) (c,3) (d,4), (e,5), (f,6), (g,7)}.

Arbitrary set of couples will appear as very good way to describe relation, in general. 


But for function a key condition, the functional condition, has to be applied:

      - If (a, b) belongs to F then if (a, c) belongs to F we have that b = c.   (the functional condition).


This means, that if F is a function from the set A to the set B, you cannot associate to one object of A, many objects of B. 

For example the temperature in a place can be a function of time, because at each moment of time you will not associate two temperatures.
It is the key point for seeing that a function from A to B, describe a very general notion of dependency.

We will be interested in functions from N to N. (With N = {0, 1, 2, ...}. Where examples abound.

Take the function which associates to each natural number its successor.

The function is (or is represented "fully") by the infinite set of couples

{(0, 1), (1,2), (2,3), (3,4), (4,5), (5, 6), (6, 7), ...}

We will be interested in function having two arguments. Those will be the function from NXN to N. Example: take addition. This is a function, because when you add any numbers, 3 and 6, for example, 3+6, you don't expect two results. So the functional condition is respected. OK? So the function addition can be defined or represented by the set

{((0, 0), 0), ((0, 1) 1) ...  ((4, 8) 12) ... }

With the numbers, all the operations are functions. The same with the sets. 


To sum up: a function is a set of couples, most of the time infinite, respecting the functional condition.

A good training consists in searching all functions between little sets:


Exercise: 


1) how many functions and what are they, from the set {0, 1} to himself. What are the functions from {0, 1) to {0, 1}?

Solution:

{(0,0), (1,0)}   the constant function which associates zero to any value of its argument.
{(0,1), (1,1)}   the constant function which associates one to any value of its argument.
{(0,0), (1,1)}  the identity function, which output its argument as value.
{(0,1), (1,0)}, the NOT function, which associate 0 to 1, and 1 to 0. 

There is four functions from {0, 1} to {0, 1}.


2) how many functions, and what are they, from the set cartesian product {0, 1} X {0, 1} to {0, 1}

Among them many are celebrities, you know. The AND, the OR, and many (how many?) others.

For a beginner in math, this is not at all an easy exercise. The real useful exercise is to try to understand the enunciation of the question. We will take the time needed.


3) A bit tricky perhaps: how many functions exist from { } to { } ?


----------------------------------------- SUM UP LINE ----------------------

So functions, once mathematical objects, are just set of couples, verifying a condition. We will be interested in the functions from N to N. Each such function is an infinite set of couples.

Some function have some form, or related expression. Not all though (as we will see), and we will have to study the relation between form and function. Many functions will lack a form, and this will not prevent them to play some role in the life of those who have a form.

John, we will see who plot which functions and why. I promise.  ;)


Bruno




Mirek Dobsicek

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Aug 2, 2009, 5:20:27 PM8/2/09
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>> I am in a good mood and a bit picky :-) Do you know how many entries
>> google gave me upon entering
>> Theaetetical -marchal -bruno
>
>
> Well 144?
>
> Good way to find my papers on that. The pages refer quickly to this
> list or the FOR list.

I am sorry for the delay, I've just got back from my vacation.

Hmm. The above written search should not return any references to your
papers/letters as the minus sign in front of your name asks for an
exclusion.

Given that it works as supposed google then gives only 1 hit in my
location (Sweden). That hit is a translation of the word "Theaetetical"
into some eastern characters. Thus, I end up with zero meaningful hits
and a feeling that you might be the only one using this word.

That makes me insists a little bit more (in a very polite way) that,
occasionally, your work is
"difficult to read unless one is willing to undertake long
discussions, clarifications and position adjustments."

I am writing this in a reference to your complains that sometimes you
have troubles to get enough relevant feedback to your work.


> I let those interested to meditate on two questions (N is {0, 1, 2, 3,
> 4, ...}):
>
> 1) What is common between the set of all subsets of a set with n
> elements, and the set of all finite sequences of "0" and "1" of length
> n.
> 2) What is common between the set of all subsets of N, and the set of
> all infinite sequences of "0" and "1".
>
> Just some (finite and infinite) bread for surviving the day :)

I am going to catch up with the thread ...

Cheers,
mirek

Bruno Marchal

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Aug 3, 2009, 4:55:30 AM8/3/09
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On 02 Aug 2009, at 23:20, Mirek Dobsicek wrote:



I am in a good mood and a bit picky :-) Do you know how many entries
google gave me upon entering
Theaetetical -marchal -bruno


Well 144?

Good way to find my papers on that. The pages refer quickly to this  
list or the FOR list.

I am sorry for the delay, I've just got back from my vacation.

Hmm. The above written search should not return any references to your
papers/letters as the minus sign in front of your name asks for an
exclusion.

Given that it works as supposed google then gives only 1 hit in my
location (Sweden). That hit is a translation of the word "Theaetetical"
into some eastern characters. Thus, I end up with zero meaningful hits
and a feeling that you might be the only one using this word.

That makes me insists a little bit more (in a very polite way) that,
occasionally, your work is
"difficult to read unless one is willing to undertake long
 discussions, clarifications and position adjustments."

I am writing this in a reference to your complains that sometimes you
have troubles to get enough relevant feedback to your work.



Come on Mirek: "Theaetetical" is an adjective I have forged from "Theatetus".
"Theatetus" gives 195.000 results on Google.
"Theatetus" wiki 4310.

By "theatetical notion of knowledge", I mean the "well known" attempts to define "knowledge" by Theaetetus in Plato's Theaetetus. The most known definition is "truye justified belief", that Bill taylor just mentionned on the FOR list recently as:
"This old crock should have been given a decent burial long ago."
I guess I will have to make a comment ...


My work is, without doubt, very difficult to read because it crosses three or four fields: "mathematical logic", "philosophy of mind" and "computer science";  + quantum mechanics to evaluate the plausibility of the derived computationalist physics. This does not help in an epoch of hyper-specialization.
I am also using a deductive approach in the philosophy of mind. I am apparently the first to *postulate* "mechanism".  Most philosophers of mind accept mechanism as the only rational theory, or reject it with some passion. Few, if any, use it as an hypothesis, in a deductive strategy. Then mathematical logic is virtually unknown, except by mathematical logicians, who, for historical reasons, do not want to come back to the earlier philosophical motivations: they want to be accepted as pure mathematicians. Except the philosophical logicians, who in majority criticized classical logic, and see philosphy as a mean to criticize classical philosophy. Mathematicians are so used to classical philosophy, that they consider it as science, and hate to be remind that this is still a philosophical. 

I have no feedback for purely contingent reason related to facts which have nothing to do with the startling feature of the conclusion of the reasoning. Up to now, I heard continuously about critics on an imaginary work I have never done. The price of the best PhD thesis that I got in France has eventually only spread those rumor from Brussels to elsewhere.
All real scientist who have studied my work and have accepted to meet me, or to write a real report on it, have understood it. True, some took a rather long time to understand, but that is normal: the subject matter is very complex, and still taboo, especially for the atheists, and other religious-based thinkers. But when they study it, they quickly discover that I use the scientific method, that is I am just asking a question, what is wrong with the following reasoning? ... The reasoning is decomposed in "easy" steps, so people accepting (for personal belief or for the sake of the argument) the hypotheses and wanting to reject the conclusion have a way to put their fingers on some problems.

UDA has been judged to obvious and simple in Brussels, and that is why I have augmented the thesis with the AUDA, which unfortunately is considered as ... too much simple for logicians, and too much difficult for non logicians. But AUDA is not needed at all to understand the simple and clear result: if we are digitalisable machine, the laws of physics emerge from a statistics on computations, in a verifiable way (quantitatively and qualitatively). The result is very simple and clear: the reasoning which leads to that result is much more subtle and difficult.

I am not at all pretending that reasoning is correct. Science progress when people do errors, but we have to find them, and sometimes, if we don't find them, we have to accept momentarily the conclusion, perhaps with the hope an error will be find later. But the attitude of a (tiny but influencing) part of the community consists in hiding the reasoning, or deforming it completely. This can't help. 

Some people, even here recently (see 1Z's post) and recently on the FOR list, attributes me a curious theory, where they confuse the conclusion with the postulate (which deprives the work of *any* meaning). But the theory I am studying is the old "mechanist theory", in its modern digital version, and nothing else. So, if they have a genuine interest in the subject, we would begin to learn something if they can criticize some point in the reasoning, instead of ignoring it, or attributing it statements without ever referring to a relevant piece of text. Of course they can't point on such text, given that such information exists only in their mind. They repeat rumors, and have clearly not take time to read the papers.

The fact that the result would contradict the current paradigm does not help, of course, but is not, yet, the source of the problem.




I let those interested to meditate on two questions (N is {0, 1, 2, 3,  
4, ...}):

1) What is common between the set of all subsets of a set with n  
elements, and the set of all finite sequences of "0" and "1" of length  
n.
2) What is common between the set of all subsets of N, and the set of  
all infinite sequences of "0" and "1".

Just some (finite and infinite) bread for surviving the day :)

I am going to catch up with the thread ...

Welcome back Mirek. Feel free to ask for any clarification, position adjustments, question, at any level ...Do you understand what is the comp hypothesis? 

Bruno


John Mikes

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Aug 4, 2009, 9:05:16 AM8/4/09
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Bruno and Mirek,
 concerning Theateticus vs. Theaeteticus:
 in my strange linguistic background I make a difference betwee ai and ae - the spelling in Greek and Latin of the name. As far as I know, nobody knows for sure how did the 'ancient' Greeks pronounce their ai - maybe as the flat 'e' like in German "lehr" while the 'e' pronounciation might have been clsoer to (between) 'make' and 'peck' - the reason why the Romans transcribed it by their ONE letter "ae", (lehr) and not as English would read: 'a'+'ee'. The spelling you gave points to this latter. The Latin 'ae' is not TWO separate letters (a+e), it is a twin, as marked in the Wiki article
..."Theætetus"... and not Theaetetus
which looked strange to me from the beginning  .
(I wonder if the e-mail reproduces the (ae) one sign? look up in Wiki's Theaetetus Dialogue (in the title with the wrong spelling) the 1st line brings the merged-together double 'æ'.)
*
English spelling always does a job on classical words, the Greek 'oi' has been transcribed into Latin sometimes as 'oe' and pronounced as in "girl" (oeuvre) while many think it was a sound like what the pigs say: as "oy". then comes America, with it's Phoenix (pron: feenix)....
I don't think the Romans were much better off, centuries after and a world apart from the ancient (classical for them) Greeks.
And who knows today if the great orator was Tzitzero or Kikero to turn later into Tchitchero?
*
"The Old Man" did quite a job on us at the tower of Babel.
*
[[ - I am enjoying your 'other' post where you spelled out my own vocabulary as indeed thinking functions as relations, lately not as a static description, but also the interchanging factor - ]]
 
John

Mirek Dobsicek

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Aug 4, 2009, 9:32:25 AM8/4/09
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> Come on Mirek: "Theaetetical" is an adjective I have forged from
> "Theatetus".
> "Theatetus" gives 195.000 results on Google.
> "Theatetus" wiki 4310.

Of course, after all you reference the dialogue Theaetetus in your
papers thus one can easily match the word Theaetetical agains it.
Let me quickly summarize the experience I had with "theatetical notion
of knowledge" while reading one of your papers for the first time.

Maybe I am an ignorant, then shame on me, but I have not read the
Theaetetus. So I took a look at the Wikipedia and read

"In this dialogue, Socrates and Theaetetus discuss three definitions of
knowledge: knowledge as nothing but perception, knowledge as true
judgment, and, finally, knowledge as a true judgment with an account.
Each of these definitions are shown to be unsatisfactory."

Hmm that really helps .., I told to myself and continued with reading.
With an uneasy feeling of stepping into the water I eventually settled
down to conclusion that you likely mean something as "true justified
belief".
I really wished you wrote it more straightforwardly without turning your
readers quite unnecessarily down to the Theaetetus and inventing new
words such as "Theaetetical".

Anyway, I'd like to stop discussing this issue :-) since my only point
was to give you a hint why I said that it is not easy to read your
papers/letters.

> Feel free to ask for any clarification, position
> adjustments, question, at any level ...Do you understand what is the
> comp hypothesis?

Let us see if I get it right. Your comp hypothesis is
1) I'm a machine,
2) Each possible computation is Turing-computable,
3) Natural numbers and their relations do exist.

This should not be confused with other quite common comp hypothesis that
the universe is a big computer. This hypothesis entails the existence of
a physical computer.


Ad 1) I take the position that "I" is only a convenient temporary
pointer to a part of universe. The pointer "Socrates' thoughts" is of
the same quality.

Ad 2) Breath taking. While 1) and 3) are assumptions of the kind "OK,
let's think for a while that ...", 2) has the status of a thesis. I
don't have any firm position on what could an objective reality be (and
without a justification I tend to think it is inaccessible to us), but
if there is any objective reality, 2) could be a statement about it.

Ad 3) If natural numbers and their relations are the only entities which
do exist then me, you, everything is a recipe of a Turing-computable number.

OK, that is it. This is how I understand to your starting assumptions.

Mirek


Bruno Marchal

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Aug 4, 2009, 11:06:12 AM8/4/09
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John,

Thanks for those informations. I thought that the "æ" was just a french, if not an old french, usage.
Note that when I wrote "Theatetus", it is just a mispelling. I tend to forget that second "e", but your remark will help me to remind it. Note that Miles Burnyeat, in his book " The Theaetetus of Plato, and Levett in his traduction wrote simply "Theaetetus". But in french too, more and more people forget to attach the "o" and "e" in words like oeuvre, or soeur (sister).

Bruno

On 04 Aug 2009, at 15:05, John Mikes wrote:

Bruno and Mirek,
 concerning Theateticus vs. Theaeteticus:
 in my strange linguistic background I make a difference betwee ai and ae - the spelling in Greek and Latin of the name. As far as I know, nobody knows for sure how did the 'ancient' Greeks pronounce their ai - maybe as the flat 'e' like in German "lehr" while the 'e' pronounciation might have been clsoer to (between) 'make' and 'peck' - the reason why the Romans transcribed it by their ONE letter "ae", (lehr) and not as English would read: 'a'+'ee'. The spelling you gave points to this latter. The Latin 'ae' is not TWO separate letters (a+e), it is a twin, as marked in the Wiki article
..."Theætetus"... and not Theaetetus
which looked strange to me from the beginning  .
(I wonder if the e-mail reproduces the (ae) one sign? look up in Wiki's Theaetetus Dialogue (in the title with the wrong spelling) the 1st line brings the merged-together double 'æ'.)
*
English spelling always does a job on classical words, the Greek 'oi' has been transcribed into Latin sometimes as 'oe' and pronounced as in "girl" (oeuvre) while many think it was a sound like what the pigs say: as "oy". then comes America, with it's Phoenix (pron: feenix)....
I don't think the Romans were much better off, centuries after and a world apart from the ancient (classical for them) Greeks.
And who knows today if the great orator was Tzitzero or Kikero to turn later into Tchitchero?
*
"The Old Man" did quite a job on us at the tower of Babel.
*
[[ - I am enjoying your 'other' post where you spelled out my own vocabulary as indeed thinking functions as relations, lately not as a static description, but also the interchanging factor - ]]
 
John
 

Bruno Marchal

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Aug 4, 2009, 1:54:46 PM8/4/09
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Hi Mirek,

Long and perhaps key post.

On 04 Aug 2009, at 15:32, Mirek Dobsicek wrote:



Come on Mirek: "Theaetetical" is an adjective I have forged from
"Theatetus".
"Theatetus" gives 195.000 results on Google.
"Theatetus" wiki 4310.

Of course, after all you reference the dialogue Theaetetus in your
papers thus one can easily match the word Theaetetical agains it.
Let me quickly summarize the experience I had with "theatetical notion
of knowledge" while reading one of your papers for the first time.

Maybe I am an ignorant, then shame on me, but I have not read the
Theaetetus.

There is no shame in being ignorant. Only in staying ignorant :)
I feel a bit sorry with my last post. I hate to look like patronizing, but it is a professional deformation. Apology.

Note that all the Theaetetus' stuff is really needed just to motivate the "arithmetical" definition of the knower, alias the first person, alias the "universal soul", and this concerns AUDA (the "arithmetical UDA), which should be done normally after getting straight the UDA's point, ... unless you are mathematical logician, who are the only one who can find AUDA more "easy" than UDA.


So I took a look at the Wikipedia and read

"In this dialogue, Socrates and Theaetetus discuss three definitions of
knowledge: knowledge as nothing but perception, knowledge as true
judgment, and, finally, knowledge as a true judgment with an account.
Each of these definitions are shown to be unsatisfactory."

Socrates asks Theaetetus to define "knowledge". That is a very difficult question.
Then Socrates shows that all attempts made by Theaetetus lead to difficulties. He literally concludes that the problem is open, and this is debated in the philosophical literature since then.

The remarkable thing is that if you accept to modelize "account" by "sound machine provability", which can be done for the not too complex machine, like Peano Arithmetic provers, or Zermelo Fraenkel Set Theory prover, the definitions of Theaetetus make sense, and can be use to show, at least, that many philosopher are deductively invalid in  their critics. Actually, even the critics by Socrates have to be weakened. 

All the arithmetical hypostases (in the Plotinus paper) are variant of Theaetetus definition.

The main one (corresponding to Plotinus "primary hypostases) are the following one:
 
p (the truth of p)
Bp (the provability of p, the account of p)
Bp & p (the provability of p when p is true)

The amazing thing is that the incompleteness theorem can be used to show that, about sound machine, we have

Bp <-> Bp & p.

But this equivalence is true but not provable by the machine, making the ideal knower already obeying a different logic than the ideal prover. This introduces a non trivial notion of first person for the machines.
If you remember G and G*, the equivalence between proof and knowledge belongs to G* minus G. The corona of the true but unprovable (by the machine) statements. Yet they prove (know) the same arithmetical "p", yet, from those points of view, it appears very different.



Hmm that really helps .., I told to myself and continued with reading.
With an uneasy feeling of stepping into the water I eventually settled
down to conclusion that you likely mean something as "true justified
belief".

I have found dozen of different translations, in french and in english, of the greek expressions. 



I really wished you wrote it more straightforwardly without turning your
readers quite unnecessarily down to the Theaetetus and inventing new
words such as "Theaetetical".

In french students are burned alive if they dare to create new adjective, and I thought that in English we have more freedom, but I may be wrong. Sorry.




Anyway, I'd like to stop discussing this issue :-) since my only point
was to give you a hint why I said that it is not easy to read your
papers/letters.

There are other reasons, if only the difficulty of the subject. 




Feel free to ask for any clarification, position
adjustments, question, at any level ...Do you understand what is the
comp hypothesis?

Let us see if I get it right. Your comp hypothesis is
1) I'm a machine,

OK. This of course could be interpreted in many ways, and that is why I have introduce the quasi-operational "yes doctor". It makes clear hat the "I" is the conscious first person I, not the third person body.



2) Each possible computation is Turing-computable,

OK. That is Church thesis. Very few people doubt it, but it is a refutable statement. If a human find a well defined function with an account of how human can compute it, but no machine can, then CT will be refuted. Only Kalmar did pretend to have such a function, but eventually his "function" was not well defined.



3) Natural numbers and their relations do exist.

This is arithmetical realism. Just a way to prevent infinite discussion about intutionism and ultrafinitism.  It is no more than the belief of all mathematician that the excuded middle principle is freely used on arithmetical proposition. Except for some philsopher, or mathematicians, but only at the pause cafe, or during the week-end, this is a widespread belief. But given that the result could be considered as a bit weird, I usually prefer to make it explicit. It can be retrieved from the (classical) Church thesis. Church thesis does not make sense without arithmetical realism. This is something I intend to show precisely in the UDA-7 thread.




This should not be confused with other quite common comp hypothesis that
the universe is a big computer.

Yes. It is different in principle. It is different unless I am the physical universe, which I doubt. Comp is really indexical mechanism: "I am a machine", or "I see no change after the functional substitution at the right level", or "I survive classical teleportation", etc.


This hypothesis entails the existence of
a physical computer.

Not necessarily. It could depend by what you mean by "physical".





Ad 1) I take the position that "I" is only a convenient temporary
pointer to a part of universe. The pointer "Socrates' thoughts" is of
the same quality.

This I do not understand. "I", used by Mirek, is a person, subject of consciousness. I think you confuse that first person "I", and its (current) body.
At that stage it is better to be reasonably agnostic about a universe. "I" is the person who survives teleportation, with a new body. With mechanism, "I" can be said the owner of the body.



Ad 2) Breath taking. While 1) and 3) are assumptions of the kind "OK,
let's think for a while that ...",

Hmm... Most scientist believes implicitly or explicitly in "1)" and "3)". Only Penrose happens to be explicitly arguing that "1)" is false, but note that he defends "3)" eloquently. John Searles, a philosopher, pretends that he believes that "1)" is false, in case of digital machine, but then he reasons in most place like if it was implicitly believing it, or being very fuzzy about what is non mechanical in biology. Only "fairy tales" kind of religious person have (fairy tales) theory of the soul, in which case they would say "no to the doctor" (even in principle: I add this because I would, in most cases,  say "no" to the doctor for technical reasons probably).



2) has the status of a thesis. I
don't have any firm position on what could an objective reality be (and
without a justification I tend to think it is inaccessible to us), but
if there is any objective reality, 2) could be a statement about it.

You don't need to have a firm position on what could an objective reality be.
You need only to believe that you can be objective on a part of that reality. In particular you need to be objective with sentences like:

x divides y if and only if it exists a number z such that y = x*z.

if 4 divides 8 then 2 divides 8

For any natural number x, if 4 divides x then 2 divides x.

Goldbach conjecture is false or true.

Etc.

I think it is reasonable to bet on the objectivity of those statements. And besides, they are used in the practice of all other exact sciences all the times. You don't have to believe more than that for being an arithmetical realist.




Ad 3) If natural numbers and their relations are the only entities which
do exist then me, you, everything is a recipe of a Turing-computable number.

No. Not at all. Sorry. Gosh, you will be very surprised if you follow the UDA-7. On the contrary. Arithmetical truth VASTLY extends the computable domain. Most relations between numbers are not Turing emulable.

And then, why do you introduce suddenly the idea that numbers and their relations would be the only entities existing for me?
This is not part of the assumption. The assumption are not more than;

A - There is a level of substitution such that I can survive a digital functional substitution made at that level.
B - Church thesis  (classical Church thesis, not the intuitionist one, thus I use classical realism, classical logic, mathematical logic)

What you say is in the conclusion of the UDA reasoning (although in a rough and simplified form(*)).

If only number exists, then it would be like I am proposing a new theory. I could have done that, but this is not what I have done. What I give is a constructive proof that if I am machine, and CT is correct, then the laws of physics have to be reduced and derived from the laws of number (or any recursively isomorphic structure). The movie graph argument is what makes this obligatory. The notion of fundamental matter loose its meaning. Matter becomes a Moiré effect lived by the number when they infer relations from their point of views. You can still believe in matter if you want too, but you cannot use it to explain even the physical observation, which have to emerge from special number's points of view.

That is what UDA shows.

Now AUDA is only used to illustrate that incompleteness in computer science and logic provide consistency to such a view. But as I said, Theaetetus all what the multiple explanations of what could be knowledge are directly translatable in arithmetic thanks again to incompleteness. For this you need Solovay theorem, and you need to read Boolos or Smorynski, or Smullyan's book in logic, or study sane04. I am surfing on the shoulds of giants here: like Post, Gödel, Kleene, Löb, and Solovay.







OK, that is it. This is how I understand to your starting assumptions.


You may have to revised or I may have ben unclear, or the result is too much counter-intuitive, and things take time. People are not accustomed to see a proof in philosophy or theology, but I show, or try to show, that the very common comp belief is much strangest, and far less reductionist than what most materialists believe.
The intended result, in a nutshell, is the incompatibility of very weak forms of materialism with digital mechanism.
By a weak form of materialism I mean any doctrine which posits a primitive or substantial matter and reduces the mind to it (computationally or not computationally). Materialism and mechanism, which are always thought as ally are really just epistemologically incompatible.

OK? I mean do you see that such a result is far from trivial? UDA1-7 conveys already the basic understanding of what happens. The Movie graph captures the mind-body problem in the comp frame, and is responsible for the inadequacy of not just physics, but of any particular universal machine.  What you see is not the product of a universal machine (be it physical or mathematical), but of an infinity of them. At first sight the mystery is now: why does the physical world looks so computable. That is the unavoidable white rabbit problem.  Who knows, it could one day lead to the refutation of comp. But AUDA, computer science and self-reference logics suggest that comp could provide a coherent picture after all, capable of explaining where the laws of physics come from, without eliminating the person.  AUDA lead to identify physics with one of the Theaetetus-like definition of a form of knowledge, and this is enough to extract a logic of what is observable, and this makes comp testable. I give a tool capable of measuring the degree of computability of nature. But alas, it does not really distinguish comp and many weakening of comp. Most of the "gods" still pay taxes ("gods" are defined by NON Turing emulable self-referential entities ...). Up to now the comp logic is not contradicted by quantum logic.

I have to go. This is an important post. Other people, like Jones recently, get me wrong on the starting assumption. It is really just "I am a machine" made precise. Then computer science makes it possible to reason and prove things with that assumption. I have no original theory. I propose just a proof, a reasoning or an argument.

Bruno



Mirek Dobsicek

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Aug 4, 2009, 6:52:40 PM8/4/09
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Hi Bruno,

Bruno Marchal wrote:
> Hi Mirek,
>
> Long and perhaps key post.

Thank you a lot for a prompt and long reply. I am digesting it :-)

Just some quick comments.

> There is no shame in being ignorant. Only in staying ignorant :)

I've ordered the dialogue from a second-hand book shop :-) The Stanford
encyclopedia says
"Arguably, it is his (Plato) greatest work on anything."
So I'll give it a try :-)


>> judgment, and, finally, knowledge as a true judgment with an account.

> The remarkable thing is that if you accept to modelize "account" by
> "sound machine provability",

This is probably the key problem for me. I know next to nothing about
provability, the logic of provability, PA/ZF provers.

I know that quite often you reference Boolos 1993 - The Logic of
Provability. I took a look at it at Google Books preview but ... there
is something missing in my education. From the beginning I am puzzled
with "Why?, what?". What a headache :-)

> In french students are burned alive if they dare to create new
> adjective, and I thought that in English we have more freedom, but I may
> be wrong. Sorry.

I'd grant this freedom to rational native speakers only :-)


> x divides y if and only if it exists a number z such that y = x*z.

I don't dare to correct your english but "there is/exists a number ..."
is what I would write.

>> Ad 3) If natural numbers and their relations are the only entities which
>> do exist then me, you, everything is a recipe of a Turing-computable
>> number.
>
> No. Not at all. Sorry. Gosh, you will be very surprised if you follow
> the UDA-7. On the contrary. Arithmetical truth VASTLY extends the
> computable domain. Most relations between numbers are not Turing emulable.

Aha! Then I really have a wrong mental picture of your work. I
understood to arithmetical realism along the lines of this quotation
from the Stanford article on realism:

"According to a platonist about arithmetic, the truth of the sentence '7
is prime' entails the existence of an abstract object, the number 7.
This object is abstract because it has no spatial or temporal location,
and is causally inert. A platonic realist about arithmetic will say that
the number 7 exists and instantiates the property of being prime
independently of anyone's beliefs, linguistic practices, conceptual
schemes, and so on."

So I thought that you essentially take
a) Numbers and their properties and relations exists.
b) Now, since you don't assume existence of anything else => your body,
your bike and coffee must emerge as patterns in the world of numbers.
c) Taking the Church-Turing thesis, these patterns are Turing-computable.
d) Definitely, the vast majority of all patterns is not Turing-computable.

This is how I have thought about your working framework. Notice, that I
don't talk about what you try to show, argue for, want to end up with etc.

Cheers,
Mirek

Bruno Marchal

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Aug 5, 2009, 4:42:48 AM8/5/09
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Hi Mirek,


On 05 Aug 2009, at 00:52, Mirek Dobsicek wrote:

> I've ordered the dialogue from a second-hand book shop :-) The
> Stanford
> encyclopedia says
> "Arguably, it is his (Plato) greatest work on anything."
> So I'll give it a try :-)


I love that book, and it is also my favorite piece of Plato.
To be sure, I don't think it is needed to understand neither UDA nor
AUDA, but it can help.


> This is probably the key problem for me. I know next to nothing about
> provability, the logic of provability, PA/ZF provers.
>
> I know that quite often you reference Boolos 1993 - The Logic of
> Provability. I took a look at it at Google Books preview but ... there
> is something missing in my education. From the beginning I am puzzled
> with "Why?, what?". What a headache :-)

You miss an introductory course on mathematical logic.
Have you herad about Gödel's incompletness theorem. Boolos book
explains the sequel.

I thought that, after Hofstadter best selling book on Gödel's theorem
(Gödel, Escher, Bach), it would be possible to talk on mathematical
logic to the layman, like we can talk on physics to the layman. But I
was wrong. Gödel's theorem is not yet part of the common knowledge,
and when it is used by non mathematician, in general it is abused.

>> x divides y if and only if it exists a number z such that y = x*z.
>
> I don't dare to correct your english but "there is/exists a
> number ..."
> is what I would write.

Thanks.


>
>
>>> Ad 3) If natural numbers and their relations are the only entities
>>> which
>>> do exist then me, you, everything is a recipe of a Turing-computable
>>> number.
>>
>> No. Not at all. Sorry. Gosh, you will be very surprised if you follow
>> the UDA-7. On the contrary. Arithmetical truth VASTLY extends the
>> computable domain. Most relations between numbers are not Turing
>> emulable.
>
> Aha! Then I really have a wrong mental picture of your work. I
> understood to arithmetical realism along the lines of this quotation
> from the Stanford article on realism:
>
> "According to a platonist about arithmetic, the truth of the
> sentence '7
> is prime' entails the existence of an abstract object, the number 7.
> This object is abstract because it has no spatial or temporal
> location,
> and is causally inert. A platonic realist about arithmetic will say
> that
> the number 7 exists and instantiates the property of being prime
> independently of anyone's beliefs, linguistic practices, conceptual
> schemes, and so on."

That is quite correct. All mathematicians are realist about
arithmetic, and most are realist about sets. But set realism is a much
more stronger belief than arithmetical realism.
Comp necessitates arithmetical realism if only to be able to state
Church thesis. Theoretical computer scientist are realist, because
they belief that all machine either stop or not stop.


>
>
> So I thought that you essentially take
> a) Numbers and their properties and relations exists.

Yes, but some people put to much sense in "exists". It is the
mathematical usual sense, like when you derive "there exists a prime
number" from the statement "17 is a prime number". No need to invoke
Plato Heaven, in the assumption.

>
> b) Now, since you don't assume existence of anything else => your
> body,
> your bike and coffee must emerge as patterns in the world of numbers.

I am agnostic. I assume neither that something else exists nor that it
does not exist, and then I prove from the assumption that we are
turing emulable, that physics is no more the fundamental science. I
prove that if we are machine then matter has to be an emerging
epistemological concept, and physics is a branch of machine biology/
psychology/theology, or mathematical computer science.
"b)" is obviously non valid. The fact that bike an coffee must emerge
from numbers is really the conclusion of the whole UD reasoning. It is
not because I don't assume them, it is because their independent
existence is shown contradictory.
I show that mechanism makes physicalism epistemologically
inconsistent. Even if matter really exists, it cannot be used to
justify our belief in matter. A slight application of Occam razor
eliminates matter, at that stage.

>
> c) Taking the Church-Turing thesis, these patterns are Turing-
> computable.

Not at all. The world of number is provably not Turing-computable.
Only a very tiny part of the world of number is computable. There is a
whole branch of mathematical logic devoted to the study of the degree
of non computability of the relations existing among the numbers.
Church thesis asserts only that the *computable* patterns are Turing
computable. It is just the assertion that Turing computability can be
used to define computability.

>
> d) Definitely, the vast majority of all patterns is not Turing-
> computable.

I don't understand.


>
>
> This is how I have thought about your working framework. Notice,
> that I
> don't talk about what you try to show, argue for, want to end up
> with etc.

My framework, comp, is just the hypothesis that I can survive with an
artificial digital brain (even material, if you want). That's all.

The negation of comp is "my soul/person/consciousness" is not Turing
emulable. Or I say "no" to all doctors. Or "I don't survive classical
teleportation done at any level".

I use the term "computationalism" in its standard usual traditional
sense. If you assume explicitly that computationalism needs the brain
to be a material object, then the UDA can be seen as a reductio ad
absurdo.
The conclusion of UDA is that Materialism is incompatible with
Computationalism. It is not obvious, but given that the numerous
attempts by materialist to solve the mind body problem have failed, it
is not so astonishing that a solution of the mind body problem needs
some "scientific revolution". The "revolution" is the reversal between
physics and the "theology of numbers" (the study of what numbers can
believe in, can know, can bet on, etc.).

Bruno


http://iridia.ulb.ac.be/~marchal/

John Mikes

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Aug 5, 2009, 12:20:20 PM8/5/09
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Bruno, just to take off some mal-deserved feathers:
I think Theaetetus has two different 'e' sounds one after the other (anybody can pronounce him better?) and in Hungarian we have them (' e ' like in 'have' and e' like in 'take') with a 3rd variation where the accent is not applied: a closed and an open ' e ' sound (instrumental in dialects). So I have no problem to pronounce the discussing gentleman as The'-etetus. Maybe he called himself (?) Te-aythetos? Ask Plato you are close to him.
 
(And I always proudly thought that Hungarian - vs. English - has a simple vowel-code in an unchanging uniform pronunciation...).
German proverb: "Fremdworter sind glucksache" (= foreign words are a matter of luck). A friend added: you can NEVER know what they mean.

John

Bruno Marchal

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Aug 7, 2009, 4:45:09 AM8/7/09
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Well, given that nobody dare to ask question, I will play the role of the idiot myself.


On 30 Jul 2009, at 21:22, Bruno Marchal wrote:







Exercise: 


1) how many functions and what are they, from the set {0, 1} to himself. What are the functions from {0, 1) to {0, 1}?

Solution:

{(0,0), (1,0)}   the constant function which associates zero to any value of its argument.
{(0,1), (1,1)}   the constant function which associates one to any value of its argument.
{(0,0), (1,1)}  the identity function, which output its argument as value.
{(0,1), (1,0)}, the NOT function, which associate 0 to 1, and 1 to 0. 

There is four functions from {0, 1} to {0, 1}.


OK. But a function from A to B, (with A and B being sets) is simply a set of couples (x, y) with x in A, and y in B. No?

So it seems you are forgetting the functions  {(0, 0), (0, 1)}, no? 

ANSWER: no!   Not all set of couples are function. In particular {(0, 0), (0, 1)} is not a function because the "input" 0 has two outputs: 0 and 1. The functional condition is not met, and a function is a set of couples provided it obeys to the functional condition.

Let me introduce vocabulary, to ease the talk. If the couple (a, b) belongs to the function F, I will say that a is an argument of F, and that b is a value of F. I will write F(a) = b, for (a, b) belongs to F. We also say that F send a on b.
Later, when we will arrive at the notion of computable function: argument will be called input, and value will be called output. The idea is that a function F is a sort of generalized machine: you give argument and it gives back a value. When F *is* mechanical, you give an input and the machine gives back an output.
With that vocabulary, the functional condition can be stated by saying than a function F cannot gives back two values for one argument. A function cannot send an elements on two elements.

To help your mind, think about typical "physical" function. For example the temperature in a room in function of time. For example: we look at the temperature a 1 o'clock, then 2, 3, 4, etc.

T = {(1, 24), (2, 25) (3, 24), (4, 24), (5, 23), (6, 23), (7, 23), (8, 23), (9, 22), (10, 20), (11, 19) ...}

Meaning, at time 1 there is 24 degrees celsius. At time 2, there is 24 degrees celsius, etc.
The functional condition is respected: you cannot have two values of the temperature at the same time.

Does this help?

Another example of a typical physical function. Movement of a mobile in space SPACE in function of TIME. This can be described by a function M from the set of moment in TIME in the set of position in SPACE:

M is a function from TIME to SPACE. 
M = {(t1, p1), (t2, p2), (t3, p3) ... }
where t1, t2, t3, are time coordinate, and p1, p2, p3 are space coordinate.

Again, this will be a function because the mobile cannot be in two places at the same moment.

OK?




2) how many functions, and what are they, from the set cartesian product {0, 1} X {0, 1} to {0, 1}

Among them many are celebrities, you know. The AND, the OR, and many (how many?) others.

For a beginner in math, this is not at all an easy exercise. The real useful exercise is to try to understand the enunciation of the question. We will take the time needed.


First let us compute {0, 1} X {0, 1}. This is simple, as it is the "rectangular" set of all couples (x, y) with x and y each in {0, 1}. Thus it gives 

1   (0, 1)  (1, 1)

0   (0,0)   (1, 0)

      0        1

that is {0, 1} X {0, 1} = {(0,0), (1, 0), (0, 1), (1, 1)}.

Let us build one function F1 from {(0, 0), (1, 0), (0, 1), (1, 1)}  to  {0, 1}.

Here the arguments are (0, 0), (1, 0), (0, 1), (1, 1), and we must decide on which, among 0 and 1, those couples will be sent.

I am lazy, so I will send them all on 0. I will get the constant function 0.

F1((0, 0)) = 0
F1((1, 0)) = 0
F1((0, 1)) = 0
F1((1, 1)) = 0


OK?

Thus F1 is the set of couples {((0, 0), 0), ((1, 0), 0), ((0, 1), 0), ((1, 1), 0)}. Note that here the arguments are themselves couples.

Another one: the constant F2 which sends all couples on 1:

F2((0, 0)) = 1
F2((1, 0)) = 1
F2((0, 1)) = 1
F2((1, 1)) = 1

Thus F2 is the set of couples {((0, 0), 0), ((1, 0), 0), ((0, 1), 0), ((1, 1), 0)}.

If you remember how many binary strings of length 4 can exist, you can guess that we have 16 (2x2x2x2) such functions.

Exercise: find them all. Beginners have to train themselves, if only to develop (slowly but surely) the familiarization with the definitions and notations.



3) A bit tricky perhaps: how many functions exist from { } to { } ?

Solution: A function F from { } to { } has to be a SET of couples (x, y) with x in { }, and y in { }. But { } is empty, so there are no such couples, so F is an empty set of couples, so F is the empty set. F = { }. So there is ONE function from { } to { }. That function is the empty set itself, and is sometimes called the empty function.

OK?

Now, a few new material. It is just vocabulary.  
-------------------------------------------------------------------------------------

SET EXPONENTIATION

Definition. If A and B are sets, we define A to the power of B = A^B, by the set of all functions from A to B.

Thus the exercise above could have been written:

1) compute {0, 1} ^ {0, 1}
2) compute {0, 1} ^ ({0, 1} X {0, 1}) and card({0, 1} ^ ({0, 1} X {0, 1}))   where  "card" = "cardinal of" = "number of elements of"
3) compute { } ^ { } and card({ } ^ { })

Little subject research. If card(A) = n, and card(B) = m. What is card(A^B)? In english: how many functions exist from a set of n elements in a set of m elements? Hint: ask yourself how many choices you have at each step of the construction of a function from A to B.

For the sequel, i suggest you reread everything I have said about "bijection". All right?

Any trouble?

Courage. We are no so far from introducing the computable functions, which is an obvious prerequisite to get the mathematical notion of computation, which is needed to understand the notion of computational supervenience, and get both UDA-7 and UDA-8.

Bruno



Mirek Dobsicek

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Aug 11, 2009, 9:32:02 AM8/11/09
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> 3) compute { } ^ { } and card({ } ^ { })

> If card(A) = n, and card(B) = m. What is
> card(A^B)?

I find it neat to write | {} ^ {} | = | { {} } | = 1 :-)
It's almost like ASCII art. Just wanted to signal that I'm following.

mirek

Bruno Marchal

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Aug 11, 2009, 3:36:05 PM8/11/09
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On 11 Aug 2009, at 15:32, Mirek Dobsicek wrote:

>
>
>> 3) compute { } ^ { } and card({ } ^ { })
>
>> If card(A) = n, and card(B) = m. What is
>> card(A^B)?
>
> I find it neat to write | {} ^ {} | = | { {} } | = 1 :-)

You will make panic those who are not familiar with symbols!


>
> It's almost like ASCII art. Just wanted to signal that I'm following.

Thank for telling me.


OK, people, good time to solve the problems. Please don't read this
post, unless you find it is the good time for you to do some math. If
not, postpone until a good time. Technical posts have to be studied,
not read. Take the time needed. Tell me if I am too quick.

The solution of "3)" has been given.

Let us look at:


>> If card(A) = n, and card(B) = m. What is
>> card(A^B)?


card(A) = n
This means A is a finite set with n elements.

card(B) = m
This means B is a finite set with m elements.

Let us simplify by supposing that m = 3, and n = 2. Hoping that the
reasoning done for finding the solution on the particular case will
inspire the reasoning for finding the solution in the general case.

Let us imagine that A is the set {a, b, c}, with its three elements,
and that B is the set {1, 2}, with its two elements.

And let us try now to remember what is the question.

The question is: what is card(A^B)?

Well, card(A^B) is the number of elements of A^B. By definition of
the cardinal.

What is A^B?

Well, A^B is the set of functions from B to A. By definition of set
exponentiation.

Well, if the question was just "what is card(A^B)?", this would
provide the best solution, or the best note if you want. But the
teacher provided the information that A has n elements, and that B has
m elements, and intuitively we can bet that the number of functions
from a set to another can depend on the number of elements of each
sets involved, so that "what is card( A^B)?" meant probably how to
compute card(A^B) in function of card(A) and card(B).

Ok, we decided to look on the particular case with A = {a, b, c}, and
B = {1, 2}.

A^B = the set of all functions from B to A.

That is the set of functions from {1, 2} to {a, b, c}.

Well, let us try to find, or to build, one function from B to A.

But , here a moment of panic can occur, (empirical observation). For
the unnameable sake, what *is* a function? What is a function from B
to A.
Well, if it is an open manual home work, such panic can be eased by
looking in the math notes. You may remember the motivation or the
informal sense of what a function represents, which is a relation of
dependency, and this is in the most general sense, so that all
possible dependency are tolerated. For a function from B to A, it
means the element of A depends in function of the elements of B. Such
a dependency is well described by a couple (x, y) with x in B and y in
A.

we have (x,y) belongs-to F representing the meaning that y depends "in
the function F" of x.

Think about x as time and y as temperature.

So, a function from B to A is just a set of couples (x, y) with x in B
and y in A, with the functional restriction that x is not send to two
different values y. At each time x, you can have only one temperature y.

That is, here: a set of couples (x, y) with x in {1, 2} and y in {a,
b, c}, and such that if (1, x) belongs-to F, no other (1, y) belongs
to F.

Let us build one function from {1, 2} to {a, b, c}.

OK, 1, from B, can determine what in A ? Well, we have three
possibilities a, b and c. OK, i will use my free will to decide that
for this function I want now, 1 will determine a. So I put the couple
(1, a) in the function.

At this stage, the "function" looks like {(1, a)}.

Finished?

No, a function from a set to another one gives a values, outcomes,
outputs for all elements of its domain. I have to say what is
determine by 2, in B. OK, I will use my free will again, and decide to
add the couples (2, a).

At this stage, the function looks like {(1, a) (2, a)}.

Finished?

Yes.

We do have a function from B to A. The set {(1, a) (2, a)} describes
completely a function from B to A, a so-called "constant function".
think of 1 and 2 as moment of times, and think of a, b, c, as possible
temperature. The function {(1, a) (2, a)} describe a case here the
temperature is constant and equal to a.

Finished? No, we have to find all functions from B to A. All functions
from {1, 2} to {a, b, c}.

Well actually, we need to find only the number of such functions. For
1 I have three choices, then for 2, I have still three choices, and
the choices are independent, so that for each choice the remaining
three choice will lead to distinct functions, this make 3 X 3
functions = 9 functions:

{(1, a) (2, a)}
{(1, a) (2, b)}
{(1, a) (2, c)}

{(1, b) (2, a)}
{(1, b) (2, b)}
{(1, b) (2, c)}

{(1, c) (2, a)}
{(1, c) (2, b)}
{(1, c) (2, c)}

so A^B = {{(1, a) (2, a)}, {(1, a) (2, b)}, {(1, a) (2, c)}, ... ,
{(1, c) (2, c)}}, and card(A^B) = 9. In this case. This give all the
way the a, b, c can depend on 1 and 2.

I stop here. I let you train on the following question:

How many functions from {a, b, c} to {1, 2}?

How many functions from {1, 2, 3, 4, 5} to {a, b, c, d, e}?

What is the general solution, in term of cardinal n and m of the sets
involved ? (the original question).

Take your time, and ask any question. This is the type of stuff rather
easy for exact scientists, and rather new for those who buried math in
their unconscious in high school, so take each your own time.

I hope I am not too long. We will see many many examples of functions.

Bruno


http://iridia.ulb.ac.be/~marchal/

Mirek Dobsicek

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Aug 11, 2009, 4:24:18 PM8/11/09
to everyth...@googlegroups.com

> Well, A^B is the set of functions from B to A. By definition of set
> exponentiation.

I'd just like to point out that Bruno in his previous post in the seven
step serii made a small typo

"A^B - the set of all functions from A to B."

It should have been from B to A. The latest post is correct in this respect.

mirek

Bruno Marchal

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Aug 11, 2009, 6:02:28 PM8/11/09
to everyth...@googlegroups.com

On 11 Aug 2009, at 22:24, Mirek Dobsicek wrote:

>
>
>> Well, A^B is the set of functions from B to A. By definition of set
>> exponentiation.
>
> I'd just like to point out that Bruno in his previous post in the
> seven
> step serii made a small typo
>
> "A^B - the set of all functions from A to B."


I wrote that? I was wrong. Thanks for saying.


>
>
> It should have been from B to A.


Yes!


> The latest post is correct in this respect.

Thank God!

Apologies for typos, mispelling, and believe me, I can do even bigger
mistakes. I will. Be vigilant.

Bruno

http://iridia.ulb.ac.be/~marchal/

Bruno Marchal

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Aug 12, 2009, 1:55:31 PM8/12/09
to everyth...@googlegroups.com
On 11 Aug 2009, at 22:24, Mirek Dobsicek wrote:


And now Simplicius is coming back and asks: " but why do you define the exponentiation of sets, A^B, by the set of functions from B to A?".

The answer of the sadistic teacher: this is a DEFINITION, and is part of the program. If you have complains about the program, write a letter to the minister of education.

Hmm...

A better answer is given by the solution of the preceding exercise:


If card(A) = n, and card(B) = m. What is
card(A^B)?



It happens that if A^B is defined as the set of functions from B to A, then card(A^B) is given by card(A)^card(B)

How many functions exist from a set with m elements in a set with n elements? n^m.

Hope you see that n^m is NOT equal to m^n (when n and m are different). 3^4 = 3x3x3x3 = 81, and 4^3 = 4x4x4 = 64.
2^7 = 128, 7^2 = 49.

In that way, A^B generalizes for set what n^m is for numbers.

And why card(A^B) = card(A)^card(B) ?

You can see this in the following way: let card(A) = m, and card(B) = n. We must understand why card(A^B) = n^m.

For example a function from {a, b, c, d, e, f, g} in {0, 1, 2, 3, 4}. To fix the idea. So m = 7, and n = 5. OK?

Let us build an "arbitrary" function F. Well,we begin with "F = {(a, ...", and we have to say where "a" is sent. We have five (n) choices, and then we have to choose where b is sent, and we have again n choices, and for each first choice any second choice is acceptable so we have 5 (n) choices multiplied by 5 (n) choices, itself multiplied by 5 (n) choices, as many times there are elements in the starting set, that is 7 (m). This gives 5 x 5 x 5 x 5 x 5 x 5 x 5, that is 5^7. or more generally n x n x n x n x ... x n, m times.

OK?

We will be interested in N^N. That is, the set of functions from N to N. 
The set of computable functions will be an important subset of that set.

Let me give a precise definition of bijection, as I promise.


I need two rather useful definitions. 

 - I will say that a function from A to B is ONTO, if all elements of B appears in the couples of the function. Note that card(B) has to be less or equal to card(A) to make that possible, by the functional condition.

 - I will say a function is ONE-ONE, if two different elements of A are sent to two different elements of B. Note that card(A) has to be less or equal to card(B) to make that possible. 
The condition one-one is the reverse of the functional condition. The functional conditions says that an element cannot be sent on two different elements (a time cannot give two temperature!), and the one-one condition says that two different elements cannot be sent on one element.

Exercises: build many examples with little finite sets. You may search examples for infinite sets.


OK. The definition of bijection. I will say that a function is a bijection between A and B if it is both a function ONTO from A to B, and a function ONE-ONE from A to B. we say more quicky that f is a bijection if f is both onto and one-one.

Exercises:   for "2)"  below, the real needed exercise is:  "do you understand the question?" Unless you like to count things, but such skills are not needed for the sequel. 

1) Convince yourself that if A and B are finite sets, then there exists a bijection between A and B if and only if card(A) = card(B).

2) If A has n elements (card(A) = n), how many bijections exists from A to B? 

     Again start with simple examples, and try to generalize.

     For example, how many bijections from {a, b, c} to {1, 2}. How many bijections from (a, b, c} to {a, b, c}?

3) can you find, or define a bijection between the infinite set N, and the infinite set E = {0, 2, 4, 6, 8, ...} (E for even).

4) Key questions for the sequel, on which you can meditate:

- is there a bijection between N and NxN?      (NxN = the cartesian product of N with N)
- is there a bijection between N and N^N?


Bruno


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