Evidence against dark natural numbers

[Jim Burns]

There are two formal systems we (WM and others in sci.logic)
are concerned with here. The first formal system has rules
for manipulating numerals and the second formal system describes
the first, its numerals and its rules. It's the second that
makes statements that look like forall k, exists j: j = k+1

For the sake of clarity, I'm going to label these different
formal systems "literal arithmetic" and "symbolic arithmetic".

The "literal" formal system might be seen as inseparable from
the "symbolic", because how can we have a formal system if we
can't say what that formal system is? I'm offering another
way to describe manipulation of literals: a list of
find-and-replace operations, with instructions to repeat
until no more operations can be performed.

For example, we can _increment_ from 99 to 100 this way:
[99/] ==> [9/0] ==> [/00] ==> [100]

The rules can be written out as
0/ ==> 1
1/ ==> 2
2/ ==> 3
3/ ==> 4
4/ ==> 5
5/ ==> 6
6/ ==> 7
7/ ==> 8
8/ ==> 9
9/ ==> /0
[/ ==> [1

There is no infinity in this description of "increment. Just
strings leading to more strings leading to (we can only hope)
a final string. We can't even say that all numerals have a
successor -- we can't even *say* that, much less prove it.

The expressiveness of lists of find-and-replace operations
is greater than this simple count-up operation. Elsewhere,
I describe a method for finding the next prime following.
However, I wouldn't know how to prove the NextPrime literal
procedure terminates without introducing the symbolic formal
system, which is the one we often call "arithmetic".
https://groups.google.com/d/msg/sci.logic/vrr0ltxCi2U/_wUfu4yICwAJ
Message-ID: <f58f4c4a-63f3-2def...@att.net>
Date: Sat, 7 Jul 2018 19:24:30 -0400

One way out (the only one I know, at least) is to make a
theory of _collections_ of numbers in addition to _numbers_
We treat a claim about a variable x ranging over a
collection C as however many parallel claims, one for
each member of C. This allows us to express things we
would not be able to express, given only statements about
particular literals like 99 or 100.

Once we have variables that range over numbers and variables
that range over collections of numbers, we are able to
express _which collection_ is the collection of natural
numbers. Because _what we want to represent_ are these
literals 0,1,2,...,99,100,... we define the natural numbers
so that there are no "other" natural numbers: no dark
natural numbers. We can prove that from our definitions,
but, at its root, the reason for such definitions is that
this is what we are talking about. The same "reason" that
squares have four corners.

Your "dark numbers" are the numbers left in the collection
of naturals -- as described by symbolic arithmetic -- after
the literal numbers -- as described by literal arithmetic --
are removed. These are very different formal systems, even
if we call them both "arithmetic", with wildly different
capabilities. Your "dark numbers" are incoherent. I strongly
suspect that this is why they're incoherent: you aren't
describing any single thing but a kind of goat-headed lizard
chimera.

[George Greene]

On Saturday, January 11, 2020 at 4:15:49 PM UTC-5, Jim Burns wrote:
> I'm offering another
> way to describe manipulation of literals: a list of
> find-and-replace operations, with instructions to repeat
> until no more operations can be performed.

Please google term-rewriting.
These are generally called rewrite rules.
It's a whole field unto itself.

[Jim Burns]

Thank you.
I'm sure it's worth looking into more.
https://en.wikipedia.org/wiki/Rewriting#Arithmetic

[Ganzhinterseher]

Am Samstag, 11. Januar 2020 22:15:49 UTC+1 schrieb Jim Burns:
> There are two formal systems we (WM and others in sci.logic)
> are concerned with here. The first formal system has rules
> for manipulating numerals and the second formal system describes
> the first, its numerals and its rules. It's the second that
> makes statements that look like forall k, exists j: j = k+1
>
> For the sake of clarity, I'm going to label these different
> formal systems "literal arithmetic" and "symbolic arithmetic".

The first one distinguishes definable and undefinable numbers. But also the symbolic system cannot comprise all numbers.

For instance, in set theory it is claimed that a sequence of infinite sets can have the limit empty set. "The cardinality is not a continuous function." Set theorists are accustomed to this statement although it is counter-logical because cardinalities are based upon elements. If we have the condition that ony one element per step can be ejected, then there is no chance to get directly from aleph_0 to zero.

It is hard to understand how the first set theorists encountering this problem could be so naive. Certainly it is connected to the religious attitude towards set theory.

My personal opinion is that this is a kind of "religious debate". One can state one's belief but, with rare exceptions, there are few cases of conversion. [P.J. Cohen: "The discovery of forcing", Rocky Mountain Journal of Mathematics 32,4 (2002) pp. 1078 & 1080 & 1099]

both set theory and religious faith can claim to be in a "strong" position vis a vis skeptics, by avoiding reliance on facts which can be questioned. I reject such claims on grounds of lack of objectivity. [S.G. Simpson: "Potential versus actual infinity: insights from reverse mathematics" (2015)]

> There is no infinity in this description of "increment.

But set theory needs it.

If they ... imagine the mathematical infinite only as a magnitude which is variable and only has no limit in its growth (like some mathematicians, as we will see soon, have assumed to explain their notion) so I agree in their reproach of this notion of a magnitude only growing into the infinite but never reaching it. [Bernard Bolzano: "Paradoxien des Unendlichen", Reclam, Leipzig (1851) p. 6ff]

"Reaching the infinite" is the request and concern of set theory. But that is not possible with definable numbers.

> Because _what we want to represent_ are these
> literals 0,1,2,...,99,100,... we define the natural numbers
> so that there are no "other" natural numbers: no dark
> natural numbers.

But then there is no actual infinity.

>
> Your "dark numbers" are the numbers left in the collection
> of naturals -- as described by symbolic arithmetic -- after
> the literal numbers -- as described by literal arithmetic --
> are removed.

Yes, that is the fact! Further the literal numbers are no fixed to any finite bound.

> Your "dark numbers" are incoherent.

Infinity seems to be somehow incoherent. Anyhow without dark numbers set theory is incoherent, simply by f(x+1) = f(x) - 1.

I see that the not fixed set of definable numbers is not easy to swallow. But there is a comparatively simple way to image a not fixed set: Think of all natnumbers that have been defined already somewhere by somebody. This not fixed set does exist.

Regards, WM

[Dan Christensen]

On Sunday, January 12, 2020 at 6:40:29 AM UTC-5, Ganzhinterseher wrote:
> Am Samstag, 11. Januar 2020 22:15:49 UTC+1 schrieb Jim Burns:
> > There are two formal systems we (WM and others in sci.logic)
> > are concerned with here. The first formal system has rules
> > for manipulating numerals and the second formal system describes
> > the first, its numerals and its rules. It's the second that
> > makes statements that look like forall k, exists j: j = k+1
> >
> > For the sake of clarity, I'm going to label these different
> > formal systems "literal arithmetic" and "symbolic arithmetic".
>
> The first one distinguishes definable and undefinable numbers. But also the symbolic system cannot comprise all numbers.
>

Again, the ZFC axioms do not include an axiom that requires every object in the universe to have associated with it a finite and unique string of characters (i.e. a name or definition).

You are, of course, free to develop your own set theory which does have such an axiom, but I think you will find it be an intellectual dead end. (You probably won't even be able to prove that 1=/=2. Hee, hee!)


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Ross A. Finlayson]

What infinity?

[Ross A. Finlayson]

Come now, if you used the word "infinity" in its
mathematical context, it must have at least a partial
definition, and otherwise make what any and all the
properties cataloged under "infinity" must have a
way to replace the utterance, in all modes of inference
that used the term, with definitions in terms of the
other objects already in the context, or not.

(Constructivism and questions about the existence of
the constructible universe and it's the universe
have been well-explored for example by Feferman,
with co-consistency of ZFC and V = L.)

Burns, that the literal arithmetic _has_ an infinity
doesn't mean you can write it, but in the literals.

Introducing constants then into these "literal" and
"symbolic" languages, has quite different places where
the terms are irreplaceable, i.e. where they go, where
in the numbers infinity's after all the literals but
in the predication it's above the numbers with finity,
the property (finitudinity).

Burns', latest, "kind of goat-headed lizard chimera",
as, some, malevolent "flying rainbow sparkle pony",
has that all flying rainbow sparkle ponies are a
kind of goat-headed lizard chimera, there, for
where if you need an elephant it will do.


Here the point about rewrite rules is this:
in the symbolic it's the rule,
in the literal it's also the rewriting.

[Ganzhinterseher]

Am Sonntag, 12. Januar 2020 19:35:25 UTC+1 schrieb Dan Christensen:


> > The first one distinguishes definable and undefinable numbers. But also the symbolic system cannot comprise all numbers.
> >
>
> Again, the ZFC axioms do not include an axiom that requires every object in the universe to have associated with it a finite and unique string of characters (i.e. a name or definition).

Every thought object must be thinkable, i.e., it kust be identifyable by a finite word, picture, sound. That is a fact that all axioms of logic and mathematics are based upon. It cannot be circumvented. If an axiom contradicts this fact, then the axiom is wrong and the conclusions derived by using it are as uninteresting as games of mentally challenged chess players or "results" about inaccessible cardinals.

Regards, WM

[Dan Christensen]

On Monday, January 13, 2020 at 4:26:39 AM UTC-5, Ganzhinterseher wrote:
> Am Sonntag, 12. Januar 2020 19:35:25 UTC+1 schrieb Dan Christensen:
>
>
> > > The first one distinguishes definable and undefinable numbers. But also the symbolic system cannot comprise all numbers.
> > >
> >
> > Again, the ZFC axioms do not include an axiom that requires every object in the universe to have associated with it a finite and unique string of characters (i.e. a name or definition).
>
> Every thought object must be thinkable, i.e., it kust be identifyable by a finite word, picture, sound.

Let me guess... That sounds like an "axiom" of MuckeMath.

> That is a fact that all axioms of logic and mathematics are based upon.

It's all just "common sense," right, Mucke? Perhaps you didn't know, but axiomatic systems are supposed to be standalone systems. When you are stuck in a formal proof, you cannot wiggle out of it by simply citing "common sense." Yes, I know, this is your favourite method of "proof," but nevertheless.

[Jim Burns]

On 1/12/2020 6:40 AM, Ganzhinterseher wrote:
> Am Samstag, 11. Januar 2020 22:15:49 UTC+1
> schrieb Jim Burns:

|> For example, we can _increment_ from 99 to 100 this way:
|> [99/] ==> [9/0] ==> [/00] ==> [100]
|>
|> The rules can be written out as
|> 0/ ==> 1
|> 1/ ==> 2
|> 2/ ==> 3
|> 3/ ==> 4
|> 4/ ==> 5
|> 5/ ==> 6
|> 6/ ==> 7
|> 7/ ==> 8
|> 8/ ==> 9
|> 9/ ==> /0
|> [/ ==> [1

>> There is no infinity in this description of "increment.
>
> But set theory needs it.

It's more accurate to say first order logic (what you are
calling "set theory") *allows* infinity. It takes no side
itself in whether a domain is finite or infinite. It *allows*
us, but does not *require* us to describe an infinite domain.

I suppose we could say that the rewrite rules above *require*
"increment". They're a formal description of how to increment.

The _literals_ which the rewrite rules _operate on
0, 1, 2, ... , 99, 100, ... are more closely analogous to
the _statements_ in first-order symbolic arithmetic --
such as Ak,Ej: j = k+1, Ak: ~(k+1 = 0), etc. The rewrite rules
themselves correspond to inference rules and introducing
axioms, and they _operate on_ statements, not literals.

For literals, this is allowed when incrementing
0/ ==> 1

For statements, this is allowed when proving
Ax:P(x) ==> P(b)

For literals, this is allowed when incrementing
1/ ==> 2

For statements, this is allowed when proving
Ex:~P(x) ==> ~Ax:P(x)

For literals, this is allowed when incrementing
2/ ==> 3

For statements, this is allowed when proving
Ax:( Q \/ P(x) ) ==> ( Q \/ Ax:P(x) )

For literals, this is allowed when incrementing
3/ ==> 4

For statements, this is allowed when proving
P(x) ==> Ax:P(x)

The justification for the rules on incrementing literals
is... I don't know. It's hard to come up with something
more basic to use as justification. Luckily, you
aren't disputing how we count(?)

The justification for the rules on valid proof steps
involving "all" aren't quite that basic. They're still
pretty basic, though.

-- Ax:P(x) ==> P(b)
If all the things are P, and b is a thing,
then b is P.

-- Ex:~P(x) ==> ~Ax:P(x)
If there is a counter-example to being P,
then not all the things are P.

-- Ax:( Q \/ P(x) ) ==> ( Q \/ Ax:P(x) )
Consider De Morgan's laws.
~( Q \/ Ax:P(x) ) contradicts Ax:( Q \/ P(x) )

-- P(x) ==> Ax:P(x)
This is more notation than anything else, but it allows
us to distinguish "All children do not like chocolate"
Ax:~P(x) from "Not all children like chocolate" ~Ax:P(x)

I've been waiting for you (WM) to claim something like
| YES, we know that all the things are P and that b is
| a thing, BUT, until we know how many things there are,
| we can't tell if b is P.

Who knows? The decade is young. I may still see that.

Generally speaking, though, your (WM's) objections have been
to the _conclusions only_ while you refuse to grapple with
the arguments or the assumptions. This has a lot in common
with denying that your checking account can be overdrawn
because you still have lots of checks. This raises doubts
(with me, at least) that you have the slightest idea what
you're doing.

[Ganzhinterseher]

Am Montag, 13. Januar 2020 18:53:25 UTC+1 schrieb Dan Christensen:

> axiomatic systems are supposed to be standalone systems.

That is impossible. You could not even understand a single letter or symbol without an external explanation. In last consequence it can be put down to learning language by showing and speaking. The fact that only thinkable ideas can be thought is basic. If you cannot define the axiom or number you are choosing then you cannot apply axioms or numbers.

Regards, WM

[Ganzhinterseher]

Am Montag, 13. Januar 2020 18:58:59 UTC+1 schrieb Jim Burns:

> > But set theory needs it.
>
> It's more accurate to say first order logic (what you are
> calling "set theory") *allows* infinity. It takes no side
> itself in whether a domain is finite or infinite. It *allows*
> us, but does not *require* us to describe an infinite domain.

ZFC set theory needs an actually infinite domain and would be boring without uncountable sets.

>
> I suppose we could say that the rewrite rules above *require*
> "increment". They're a formal description of how to increment.

Your rules yield definable results and therefore potential infinity only.

> The justification for the rules on incrementing literals
> is... I don't know. It's hard to come up with something
> more basic to use as justification. Luckily, you
> aren't disputing how we count(?)

No.

> -- P(x) ==> Ax:P(x)
> This is more notation than anything else, but it allows
> us to distinguish "All children do not like chocolate"
> Ax:~P(x) from "Not all children like chocolate" ~Ax:P(x)

All natural numbers belong to a set of cardinality less than aleph_0, for instance.

>
> I've been waiting for you (WM) to claim something like
> | YES, we know that all the things are P and that b is
> | a thing, BUT, until we know how many things there are,
> | we can't tell if b is P.

On the contrary. We can tell that b is P. Every element that has the properties of a natnumber is a natnumber. But that does not yield any cardinality.

>
> Generally speaking, though, your (WM's) objections have been
> to the _conclusions only_ while you refuse to grapple with
> the arguments or the assumptions.

The assumptions are right. But try to apply the set of all natural numbers that have been defined as yet.

> This has a lot in common
> with denying that your checking account can be overdrawn
> because you still have lots of checks. This raises doubts
> (with me, at least) that you have the slightest idea what
> you're doing.

Your doubts are irrelevant. Fact is that every definable natural number can be summed. But the set of all natural numbers cannot be summed. Further, every definable endsegment is infinite. But the function f(n+1) = f(n) - 1 cannot reach 0 without passing 1. That are the facts. And that's what counts.

Regards, WM

[Jim Burns]

On 1/12/2020 6:40 AM, Ganzhinterseher wrote:

> Am Samstag, 11. Januar 2020 22:15:49 UTC+1
> schrieb Jim Burns:

|> The rules can be written out as
|> 0/ ==> 1
|> 1/ ==> 2
|> 2/ ==> 3
|> 3/ ==> 4
|> 4/ ==> 5
|> 5/ ==> 6
|> 6/ ==> 7
|> 7/ ==> 8
|> 8/ ==> 9
|> 9/ ==> /0
|> [/ ==> [1

>> There is no infinity in this description of "increment.
>
> But set theory needs it.

First order logic ("set theory") is able to describe infinite
things through the use of a formal theory of "all" (variables
and quantifiers). This theory of "all" is the same theory used
in (WM-good) "All numbers have successors" as in (WM-bad)
"The intersection of all end segments is empty".

Essentially, skipping technical details, this is
_the whole theory_
-- If all the things are P, and b is a thing, then b is P.
-- If there is a counter-example to being P, then not all
the things are P.
-- ~( Q \/ Ax:P(x) ) contradicts Ax:( Q \/ P(x) )
-- "Not all children like chocolate" does not mean
"All children do not like chocolate".

This is where you can find your "completed infinity" in our
work, in the formalization of uses of "all" that do not
*need* to distinguish between finite and infinite.

If you're going to object, object to _this_
But, if you do, bring an argument, not more blather.

[...]

> "Reaching the infinite" is the request and concern of
> set theory. But that is not possible with definable
> numbers.

How do we describe natural numbers in a way that includes
all the natural numbers and only the natural numbers?

This is a different question from "Are we there yet?"
which seems to be the question you _want_ us to ask. Or,
maybe, it's the question you insist that we _are asking_
despite being told otherwise almost infinitely many times.

Given a description of all and only the natural numbers,
we can draw some conclusions from that description about
all and only the natural numbers. Saying clearly what we
mean by "all" is key to an argument like this. "Reaching
the infinite" is NOT key, is NOT even in the argument.

----
How do we describe a sequence such as

0, 1, 2, ... , 99, 100

in such a way that it is NOT just that one sequence,
but, instead, that it would generalize to
0, 1, 2, ... , 100, 101
and
0, 1, 2, ... , 101, 102
and
0, 1, 2, ... , 98, 99
and
0, 1, 2, ... , 97, 98
and
_all_ such sequences.

What is "such a sequence"? This would get us most of
the way to an answer to "What is a natural number?"

Consider the collection { 3,5,4,7,1,0,2,6 }.

(i)
We want everything in the collection to have
a unique successor and a unique predecessor,
except for the first, 0, with only a successor,
and the last, 7, with only a predecessor.

(ii)
We want no extraneous loops, like a --> b --> c --> a.

So, the _collection_ as a whole should not have
any smaller collection satisfying our requirements.
The difference, that is, the numbers in "our" collection but
not in the (forbidden) smaller collection would be
extraneous. No smaller such collection: no extraneous "numbers".

(iii)
We do not want such a sequence to either wander off
"to infinity" or wander in "from infinity". We don't want
..., -5, -4, -3, 3, 4, 5, ...
This satisfies our other requirements, but we don't
want to include that as a sequence "from" 3 "to" -3.
(?) 3, 4, 5, ... -5, -4, -3

So, the collection should not be infinite.

And what is "infinite"? For a sequence, each having
another after -- the opposite of having a last one.
(Or, from another point of view, each having another before
-- the opposite of having a first one.) Here, being able
to say "all", as in "all items have a successor", is essential
to saying what we mean.

These, (i), (ii), (iii), are at least _necessary_ properties
of sequences like 0, 1, 2, ... , 99, 100
I think that we can show that they are also sufficient
to define natural numbers, but that's for another day.

----
I can imagine someone asking why I'm going to all of this
trouble, making something complicated out of something so
simple as 0, 1, 2, ... , 99, 100

In my opinion, it's because we have a theory of of "all" --
quantifiers and variables. We paint a portrait of the
natural numbers with whatever we have at hand, and this
is what we have at hand. Some oddness and clunkiness is
to expected, I think. I would compare it to (literal)
portraits made with whatever the artist has at hand, like
https://www.buzzfeed.com/alanwhite/stunning-portaits-made-from-found-objects

>> Because _what we want to represent_ are these
>> literals 0,1,2,...,99,100,... we define the natural
>> numbers so that there are no "other" natural numbers:
>> no dark natural numbers.
>
> But then there is no actual infinity.

If anyone knows what you mean by "actual infinity",
it would be you. Thus, I'll take your word for that.
Fine. There is no "actual infinity".

Note that we can reason about _all_ the natural numbers
by reasoning from properties that we define _all_ the
natural numbers to have, by using our theory of "all"
(variables and quantifiers). It would follow from there
being no "actual infinity" that we can do this without
"actual infinity".

Does it matter?

[Ganzhinterseher]

Am Mittwoch, 15. Januar 2020 00:14:09 UTC+1 schrieb Jim Burns:


> First order logic ("set theory") is able to describe infinite
> things through the use of a formal theory of "all" (variables
> and quantifiers). This theory of "all" is the same theory used
> in (WM-good) "All numbers have successors"

The typical natnumber has a successor. But there need not be all. In fact there cannot be all. See "Three reservoirs" in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

> as in (WM-bad)
> "The intersection of all end segments is empty".

There we need all. "Cantor's belief in the actual existence of the infinite of Set Theory still predominates in the mathematical world today." [A. Robinson: "The metaphysics of the calculus", in I. Lakatos (ed.): "Problems in the philosophy of mathematics", North Holland, Amsterdam (1967) p. 39]

"Every single finite cardinal number (1 or 2 or 3 etc.) is contained in the divine intellect (from St Augustin: De civitate Dei, lib. XII, ch. 19)." [G. Cantor, letter to I. Jeiler (20 May 1888, Whitsun)] "Dominus regnabit in infinitum (aeternum) et ultra." {{The Lord rules in eternity and beyond, from Exodus 15,18.}} [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

>
> Essentially, skipping technical details, this is
> _the whole theory_
> -- If all the things are P, and b is a thing, then b is P.
> -- If there is a counter-example to being P, then not all
> the things are P.
> -- ~( Q \/ Ax:P(x) ) contradicts Ax:( Q \/ P(x) )
> -- "Not all children like chocolate" does not mean
> "All children do not like chocolate".

There is no accout for undecidable cases, so-called dark cases.

"classical logic was abstracted from the mathematics of finite sets and their subsets. (The word finite is here to be taken in the precise sense that the members of such set are explicitly exhibited one by one.) Forgetful of this limited origin, one afterwards mistook that logic for something above and prior to all mathematics, and finally applied it, without justification, to the mathematics of infinite sets. This is the Fall and original sin of set-theory, for which it is justly punished by the antinomies." [H. Weyl: "Levels of infinity: Selected writings on mathematics and philosophy", Peter Pesic (ed.), Dover Publications (2012) p. 140f]

>
> This is where you can find your "completed infinity" in our
> work, in the formalization of uses of "all" that do not
> *need* to distinguish between finite and infinite.

"it has been shown that neither in the elementary chapters of mathematics nor in those denoted by 'infinitesimal calculus' a really infinite 'magnitude' occurs, but that rather the word 'infinite' is merely used as an abbreviating description of important facts of the finite." [G. Hessenberg: "Grundbegriffe der Mengenlehre", offprint from Abhandlungen der Fries'schen Schule, Vol. I. no. 4, Vandenhoeck & Ruprecht, Göttingen (1906) Preface and § 1]

>
> If you're going to object, object to _this_
> But, if you do, bring an argument,

F(x+1) = f(x) - 1 is the ultimate argument.

> > "Reaching the infinite" is the request and concern of
> > set theory. But that is not possible with definable
> > numbers.
>
> How do we describe natural numbers in a way that includes
> all the natural numbers and only the natural numbers?

Potential infinity.

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4,}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
...

>
> This is a different question from "Are we there yet?"

The empty intersection is just this "fact".

> which seems to be the question you _want_ us to ask. Or,
> maybe, it's the question you insist that we _are asking_
> despite being told otherwise almost infinitely many times.

We must have arrived there. Otherwise there is no empty intersection but infinitely many elements to remove.

>
> Given a description of all and only the natural numbers,
> we can draw some conclusions from that description about
> all and only the natural numbers. Saying clearly what we
> mean by "all" is key to an argument like this. "Reaching
> the infinite" is NOT key, is NOT even in the argument.

The empty intersection proves reaching the infinite. All finite numbers have been overcome.

>

> In my opinion, it's because we have a theory of of "all" --
> quantifiers and variables.

It fails without dark numbers. Simplest proof: For all natural numbers we know there are infinitely many successors. Therefore "for all" does not concern all.

> If anyone knows what you mean by "actual infinity",
> it would be you.

I use the knowledge of Cantor and many, many others reported in chapter I of my https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

> Thus, I'll take your word for that.
> Fine. There is no "actual infinity".

Then there is no empty intersection and no set theory.

"In spite of significant difference between the notions of the potential and actual infinite, where the former is a variable finite magnitude, growing above all limits, the latter a constant quantity fixed in itself but beyond all finite magnitudes, it happens deplorably often that the one is confused with the other." [Cantor, p. 374]

>
> Note that we can reason about _all_ the natural numbers
> by reasoning from properties that we define _all_ the
> natural numbers to have, by using our theory of "all"
> (variables and quantifiers). It would follow from there
> being no "actual infinity" that we can do this without
> "actual infinity".
>
> Does it matter?

It matters when we want to count beyond omega because first we must reach it.

Regards, WM

[Jim Burns]

On 1/15/2020 6:35 AM, Ganzhinterseher wrote:
> Am Mittwoch, 15. Januar 2020 00:14:09 UTC+1
> schrieb Jim Burns:

>> First order logic ("set theory") is able to describe infinite
>> things through the use of a formal theory of "all" (variables
>> and quantifiers). This theory of "all" is the same theory used
>> in (WM-good) "All numbers have successors"
>
> The typical natnumber has a successor.

If part of what defines a natnumber is that it have a
successor, then all natnumbers have a successor,
the same way that all green frogs are green.

We are trying to characterize _all_ of the entries in the
sequence 0, 1, 2, ... , 99, 100, ... , not only the typical
entries. We have rewrite rules for producing a specific
successor from a specific entry in that sequence, 0/ ==> 1,
etc. It is completely reasonable, as part of characterizing
_all_ of them, to _define_ them _all_ to have successors.

> But there need not be all.

Here is what we need for _all_ the natnumbers:

-- If _all_ the natnumbers are P, and b is a natnumber,
then b is P.
-- If there _exists_ a natnumber counter-example to being P,
then not _all_ natnumbers are P.
-- Let variable x range over natnumbers.

~( Q \/ Ax:P(x) ) contradicts Ax:( Q \/ P(x) )

-- "Not all natnumbers are P" does not mean
"All natnumbers are not P".

You know what we _don't_ need for _all_ ?
We _don't_ need to write them all one-by-one, or think
about them all one-by-one, or anything-else them all
one-by-one.

> In fact there cannot be all. See "Three reservoirs"
> in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

| Now add an intermediate reservoir M. By addition and
| subtraction of sets transfer the natural numbers, one
| after the other, from A to M and from M to Z but obeying
| the condition that number n is not transferred from M
| to Z before number n+1 has arrived in from A.

See above. Your one-by-one requirement is appropriate
for manipulating literals ( 0/ ==> 1, etc ) one-by-one.

We're doing something other than manipulating literals here.
We're characterizing _all_ natnumbers in order to draw valid
conclusions about _all_ natnumbers. The formal properties
of "all" from which we reason do not include (for example)
checking each green frog to make sure it's green.

[Ross A. Finlayson]

Frogs are usually green -

The dark numbers, "dark", numbers, these are just WM's
huge infinite numbers that fill in the end of the numeric
range before infinity, bounding all the finite models.

Obviously they are too big for space, "dark matter's numbers",
in the theory, that are that besides, all the luminous matter,
represents the free density in space. They're only "dark" the theory.

Then they are supernumbers to WM.

1, 2, 3, infinity

It's simple it just goes

1, 2, 3, 4, infinity
1, 2, 3, 4, 5, ..., infinity

1, 2, 3, 4, 5 infinity.

This is the same for WM as

1, 12, 123, 1234, 12345, ...,
also
1
2
3
4
5.

(To 10.)

(..., infinity)

After 1, 2, 3, counting to infinity
is counting all the numbers after
4, 5, 6, all the way to infinity.

So, just saying, 1, 2, 3, infinity,
means counting from 1, through 2,
all the way to infinity.

All the frogs of which are green.

[Jim Burns]

On 1/15/2020 9:31 PM, Ross A. Finlayson wrote:

> Frogs are usually green -

_Green_ frogs are always green.

> The dark numbers, "dark", numbers, these are just WM's
> huge infinite numbers that fill in the end of the numeric
> range before infinity, bounding all the finite models.

You can see how quickly that falls apart. Infinite numbers
that come before infinity? They're written down by a man who
is taller than himself, I bet.

I think that you are just guessing what WM means, and you've
made a fair attempt at an interpretation. WM doesn't do any
better than you at making sense. Maybe not as well as you.

----
How should we describe all and only the whatever-they-are
which are in this sequence? 0, 1, 2, ... , 99, 100, ...

We want that description because we can use it to reason from
it and know the conclusion applies to _all_ in that sequence
-- the way that we know that all green frogs are green.

_Naming_ the whatever-they-are "natural numbers" doesn't help.
It might even get in the way. Everyone will bring their
personal idea of what "natural number" means to the table,
and no one will listen to anyone else.

----
_All_ the things in 0, 1, 2, ... , 99, 100, ...
have a _successor_

_All_ the things in 0, 1, 2, ... , 99, 100, ...
end a _counting sequence_

We can formally describe a successor and a counting sequence.

We can provide a formalization of "all", basically:

-- If all the things are P, and b is a thing, then b is P.
-- If there is a counter-example to being P, then not all
the things are P.
-- ~( Q \/ Ax:P(x) ) contradicts Ax:( Q \/ P(x) )
-- "Not all children like chocolate" does not mean
"All children do not like chocolate".

Suppose that this serves as a description of _all_ the
"natural numbers". (How well it matches our "natural number"
intuitions is not a question of logic.) If we draw a
conclusion _from this description_ how is that NOT a
conclusion about _all_ the natural numbers? Aren't _all_
green frogs green?

[Ganzhinterseher]

Am Mittwoch, 15. Januar 2020 19:04:20 UTC+1 schrieb Jim Burns:

> See above. Your one-by-one requirement is appropriate
> for manipulating literals ( 0/ ==> 1, etc ) one-by-one.

It is not possible to have all collected in the final reservoir.

>
> We're doing something other than manipulating literals here.
> We're characterizing _all_ natnumbers in order to draw valid
> conclusions about _all_ natnumbers.

You can characterize every one. But you cannot get an empty intersection wuthout having passed all finite ontersections.

> The formal properties
> of "all" from which we reason do not include (for example)
> checking each green frog to make sure it's green.

If we get 0, then we have a fixed point that proves the existence of

...3, 2, 1, 0.

Regards, WM

[Jim Burns]

On 1/16/2020 4:12 PM, Ganzhinterseher wrote:
> Am Mittwoch, 15. Januar 2020 19:04:20 UTC+1
> schrieb Jim Burns:

>> We're doing something other than manipulating literals here.
>> We're characterizing _all_ natnumbers in order to draw valid
>> conclusions about _all_ natnumbers.
>
> You can characterize every one. But you cannot get an
> empty intersection wuthout having passed all finite
> ontersections.

Some context:
You're talking about the sequence of
The Intersection Of All The Remaining End Segments.

You do not take into account the Law of Monotonic
Intersection Sequences: I(x) = I(x+1)

which is to say
Intrsct{ E(x),E(x+1),... } = Intrsct{ E(x+1),E(x+2),... }

If you got an empty intersection, it was always empty.

If it has 137 dark numbers, it always had 137 dark numbers.

So, your claim of the sequence *passing* finite intersections
is wrong. Finite, infinite, whatever, the sequence stays
the same.

>> The formal properties
>> of "all" from which we reason do not include (for example)
>> checking each green frog to make sure it's green.
>
> If we get 0, then we have a fixed point that proves the
> existence of
> ...3, 2, 1, 0.

_The Law of Montonic Intersection Sequences_

I(x) = I(x+1)

Proof:

1. Draw a circle. Label it E(x+1).

2. Draw a circle around E(x+1). Label it E(x).
E(x+1) is inside E(x) because E(x+1) sub E(x).

3. Draw an oval that runs inside E(x+1), inside E(x),
and outside E(x). Label it I(x+2), where
I(x+2) = Intrsct{I(x+2),I(x+3),... }
Because this is the most general case for I(x+2).

4. Draw vertical hatch marks on the intersection of
E(x+1) and I(x+2). Label it I(x+1).
I(x+1) = Intrsct{ E(x+1),E(x+2),... }
Because I(x+1) = E(x+1) intrsct I(x+2)

5. Draw horizontal hatch marks on the intersection of
E(x) and I(x+1). Label it I(x).
I(x) = Intrsct{ E(x),E(x+1),... }
Because I(x) = E(x) intrsct I(x+1)

Note that the vertical hatch marks for I(x+1) and the
horizontal hatch marks for I(x) are on the same area.

Therefore, I(x) = I(x+1)

[Jim Burns]

On 1/16/2020 4:12 PM, Ganzhinterseher wrote:

> Am Mittwoch, 15. Januar 2020 19:04:20 UTC+1
> schrieb Jim Burns:

>> See above. Your one-by-one requirement is appropriate
>> for manipulating literals ( 0/ ==> 1, etc ) one-by-one.
>
> It is not possible to have all collected in the final
> reservoir.

It's not possible to move them all one-by-one. That's right.

Consider this sequence: 0, 1, 2, ... , 99, 100, ...
It's not possible to pass them all one-by-one.

Since that sequence is _what we want to describe_
so that we can draw conclusions in the style of
"All green frogs are green", it's a good thing that
it ALSO is not possible to deal with them _all_
one-by-one.

However, you disagree that it's a good thing.
Why do you disagree?

[Ross A. Finlayson]

... all the green frogs of which are green.

I like how you describe counts and bounds separately,
that's largely the idea this 1, 2, 3, infinity.

One might imagine a symmetry, or parallel, between
counting up from zero and down from infinity.

It's of a notion of a modular or clock arithmetic.

Would you please further detail your opinion of this,
it seems illustrative and on the right track, for
simple heuristics about what "infinity" is (or is not).

I seem to remember "infinity" in the vocabulary and
that its only sense was numerical, since childhood,
at least.

[Jim Burns]

On 1/16/2020 8:25 PM, Ross A. Finlayson wrote:
> On Thursday, January 16, 2020 at 12:54:09 PM UTC-8,
> Jim Burns wrote:

[...]

>> Suppose that this serves as a description of _all_ the
>> "natural numbers". (How well it matches our "natural number"
>> intuitions is not a question of logic.) If we draw a
>> conclusion _from this description_ how is that NOT a
>> conclusion about _all_ the natural numbers? Aren't _all_
>> green frogs green?
>
> ... all the green frogs of which are green.
>
> I like how you describe counts and bounds separately,
> that's largely the idea this 1, 2, 3, infinity.
>
> One might imagine a symmetry, or parallel, between
> counting up from zero and down from infinity.
>
> It's of a notion of a modular or clock arithmetic.
>
> Would you please further detail your opinion of this,
> it seems illustrative and on the right track, for
> simple heuristics about what "infinity" is (or is not).

Someone has a list (finite) of a couple dozen different
meanings of "infinity" in mathematics. For most of them,
I don't know anywhere near enough to have an opinion.

For the particular infinity of a sequence that does not
end, I have a favorite image -- which counts as an opinion,
in my opinion.

----
Consider a one-ended chain. Each link is one entry in
an infinite sequence. It starts at the one and ...
it just doesn't end. Not "it doesn't end for a really
long time". Not "it ends after passing all the finite
links". It _has no_ second end.

I've mostly been talking about the one-ended-chain
sort of infinity, lately.

Your modular-clock-arithmetic notion of infinity does
not strike me as being on the right track for the
one-ended-chain image of infinity.

There's also the Riemann sphere, very different. One maps
the complex plane onto a unit sphere set on top of (0,0).
_One_ point is added to the sphere: the "point at infinity".

I suppose that might be sort of like modular-clock-
-arithmetic.

(I don't know what to make of 1, 2, 3, infinity.)

[Ross A. Finlayson]

Frogs are usually green -

all the frogs "of which are green"

"All the frogs of which are green."

Frogs are usually green -

all the frogs of which are green.

1, 2,3, infinity is when you give up counting
and start doing.

For example "I can't count this high
as long as I'm doing this."

... all the green frogs of which are green.

Elephants: now in green.

Frog green? Not the reptile green.

(Plant frogs.)

Burns, all, frogs are usually green -
all the frogs of which are green.

1, 2, 3 is because as soon as you start
counting over, 1, 2, 3, it might as well
be 1, 2, 3, infinity.

Just, for all the time.

For example if you're not going to be starting over,
might as well be going forward, or either "it's another
one" or "I've totally lost track of how many of these there are".

It seems though whether to be restarting the count,
in intervals, for example making count or keeping time
in tens. Then, you know that it doesn't matter how far you
count, the numbers never stop growing so even though you can
count, for hours or some long time counting, it makes no sense
to count them any more.

Then the exercise the "counting in powers of two", helps,
because that you can give up after just needing a dozen,
the powers of two in decimal.

Counting by 2's? Automatically doubles the speed of the count.

Also a factor of 10....

Counting in powers of 2? Doubles the count of the count.

I'm very familiar with 1, 2, 3, and infinity.

It's all numbers in-between.

They're numbers.

"... all the green frogs of which are green."

It's, starting again going forward, and giving
up counting 1, 2, 3, and just knowing that it
goes to infinity.



I don't know what not to make with 1, 2, 3, ...,
and infinity.

It seems easier to count infinity anyways -
not keeping track of the result (for that
the infinities establish the triangle result
downward).

Or "keep count" of infinity.

... of infinity.



Infinity: biggest number, also has all the numbers.

[Ganzhinterseher]

Am Donnerstag, 16. Januar 2020 23:45:06 UTC+1 schrieb Jim Burns:
> On 1/16/2020 4:12 PM, Ganzhinterseher wrote:
> > Am Mittwoch, 15. Januar 2020 19:04:20 UTC+1
> > schrieb Jim Burns:
>
> >> We're doing something other than manipulating literals here.
> >> We're characterizing _all_ natnumbers in order to draw valid
> >> conclusions about _all_ natnumbers.
> >
> > You can characterize every one. But you cannot get an

> > empty intersection without having passed all finite
> > intersections.

>
> Some context:
> You're talking about the sequence of
> The Intersection Of All The Remaining End Segments.

The intersection of all remaining endsegments is always empty.

I consider the intersection of all finite initial sequences of endsegments in the natural well-order:

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo . (*)

This is so for all endsegments. Since however the intersection of the remaining endsegments is empty, there must be remaining endsegments beyond every endsegment defined by k.

This is very easy. Why didn't you find it independently?

>

> Intrsct{ E(x),E(x+1),... } = Intrsct{ E(x+1),E(x+2),... }

That is true for all definable endsegments. Others cannot appear at the first place.

All endsegments which can appear at first places are infinite and leave an infinite intersection (*).

> _The Law of Montonic Intersection Sequences_

> I(x) = I(x+1)

> Proof:

We need no proof but an explanation. Why are the endsegments causing an empty intersection always missing in (*). That is the question.
>
Regards, WM

[Jim Burns]

On 1/17/2020 9:04 AM, Ganzhinterseher wrote:
> Am Donnerstag, 16. Januar 2020 23:45:06 UTC+1
> schrieb Jim Burns:
>> On 1/16/2020 4:12 PM, Ganzhinterseher wrote:
>>> Am Mittwoch, 15. Januar 2020 19:04:20 UTC+1
>>> schrieb Jim Burns:

>>>> We're doing something other than manipulating literals
>>>> here. We're characterizing _all_ natnumbers in order to
>>>> draw valid conclusions about _all_ natnumbers.
>>>
>>> You can characterize every one. But you cannot get an
>>> empty intersection without having passed all finite
>>> intersections.
>>
>> Some context:
>> You're talking about the sequence of
>> The Intersection Of All The Remaining End Segments.
>
> The intersection of all remaining endsegments is always
> empty.

Because, for every natural number, there is a remaining
end segment which that natural number is not in.

Note that this is not the same as
"There is a remaining end segment which, for every
natural number, that natural number is not in."

> I consider the intersection of all finite initial
> sequences of endsegments in the natural well-order:
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k)

Because, by the Law of Monotonic Intersection Sequences,
Intrsct{ E(1),E(2),E(3),...,E(k) } =
Intrsct{ E(2),E(3),...,E(k) } =
Intrsct{ E(3),...,E(k) } =
...
Intrsct{ E(k) } =
and
Intrsct{ E(k) } = E(k)

> /\ |E(k)| = ℵo .

Because, Ak e N, Ej e N: j > k

> (*)
>
> This is so for all endsegments.
> Since however the intersection of the remaining
> endsegments is empty, there must be remaining
> endsegments beyond every endsegment defined by k.

Your quantifier dyslexia strikes again.

In order for the intersection to be empty,
for each k, there must be some E(j) such that j > k.

This is not the same as
There must be some E(j) such that, for each k, j > k.

This is the foundation of your assertion of "dark numbers"
and it is invalid.

It's NOT invalid because of some spooOOOooky stuff
happening "out at infinity" that you are clever enough
to explain. We can see it's invalid with two things
in our domain. (Note: 2 < aleph_0)

x e {1,2}, y e {1,2}

Ax,Ey: x = y TRUE

Ey,Ax: x = y FALSE

> This is very easy. Why didn't you find it independently?

A better question is why do you still think
Ax,Ey:P(x,y) -> Ey,Ax:P(x,y)
is valid?

[Exfalso Quodlibet]

On Friday, January 17, 2020 at 9:04:59 AM UTC-5, Ganzhinterseher wrote:
> The intersection of all remaining endsegments is always empty.

More precisely, the intersection over ***the entire set containing every*** endsegment is empty.

> I consider the intersection of all finite initial sequences of endsegments in the natural well-order:
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo . (*)
>
> This is so for all endsegments.

More precisely, it is so for ***each individual*** endsegment.

You use "all" ambiguously, sometimes meaning "the entire set containing every" and sometimes mean "each individual", but then falsely claim that conclusions from one meaning apply to the other.

Since you persist in this confusion despite having it pointed out to you repeatedly, we can only conclude either that you are too stupid to understand the difference and/or you are intentionally utilizing this ambiguity precisely because you understand that lying is the only recourse you have.

If you were intelligent and intellectually honest, you'd find an unambiguous way to make your claim.

EFQ

[Ganzhinterseher]

Am Freitag, 17. Januar 2020 18:18:23 UTC+1 schrieb Exfalso Quodlibet:
> On Friday, January 17, 2020 at 9:04:59 AM UTC-5, Ganzhinterseher wrote:
> > The intersection of all remaining endsegments is always empty.
>
> More precisely, the intersection over ***the entire set containing every*** endsegment is empty.

Name one endsegment that is not a last endsegment satisfying

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)

All those cannot empty the set, or can they?

>
> > I consider the intersection of all finite initial sequences of endsegments in the natural well-order:
> >
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo . (*)
> >
> > This is so for all endsegments.
>
> More precisely, it is so for ***each individual*** endsegment.

None of them can contrinute to emptying tehe set.

>
> You use "all" ambiguously, sometimes meaning "the entire set containing every" and sometimes mean "each individual", but then falsely claim that conclusions from one meaning apply to the other.

Show the difference. What definable endsegments can empty the intersection.

>
> Since you persist in this confusion despite having it pointed out to you repeatedly, we can only conclude either that you are too stupid to understand the difference and/or you are intentionally utilizing this ambiguity precisely because you understand that lying is the only recourse you have.
>
> If you were intelligent and intellectually honest, you'd find an unambiguous way to make your claim.

My claim is that none of the endsegments of (*) can empty the intersection. Either there are others or there is no empty intersection.

Regards, WM

[Ganzhinterseher]

Am Freitag, 17. Januar 2020 17:17:41 UTC+1 schrieb Jim Burns:

> We can state for all definable natural numbers:

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)

So they do not contribute to emptying the intersection.

>
> Note that this is not the same as
> "There is a remaining end segment which, for every
> natural number, that natural number is not in."

That is irrelevant. We have two points:

1) Every definable number fails to reduce the intersection below card aleph_0.

Do you understand that? If not, do you contradict that?

2) Therefore other numbers must do it.

Do you understand that none of the definable numbers of (*) can be necessary or useful in

∩{E(x), E(x+1), E(x+2), ...} = { } ?

If not you are incapable of understanding basic logic and it is useless to continue.

Regards, WM

[Exfalso Quodlibet]

On Friday, January 17, 2020 at 5:03:12 PM UTC-5, Ganzhinterseher wrote:
> Am Freitag, 17. Januar 2020 18:18:23 UTC+1 schrieb Exfalso Quodlibet:
> > On Friday, January 17, 2020 at 9:04:59 AM UTC-5, Ganzhinterseher wrote:
> > > The intersection of all remaining endsegments is always empty.
> >
> > More precisely, the intersection over ***the entire set containing every*** endsegment is empty.
>
> Name one endsegment that is not a last endsegment satisfying
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
>
> All those cannot empty the set, or can they?

They all do! The intersection over ***the entire set containing every*** endsegment is empty.

> > > I consider the intersection of all finite initial sequences of endsegments in the natural well-order:
> > >
> > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo . (*)
> > >
> > > This is so for all endsegments.
> >
> > More precisely, it is so for ***each individual*** endsegment.
>
> None of them can contrinute to emptying tehe set.

In fact, all of them do, because the intersection over ***the entire set containing every*** endsegment is empty.

> > You use "all" ambiguously, sometimes meaning "the entire set containing every" and sometimes mean "each individual", but then falsely claim that conclusions from one meaning apply to the other.
>
> Show the difference.

"All" natural numbers result in infinite cardinality, because ***the entire set containing every*** natural number is infinite.

"All" natural numbers also result in finite cardinality, because ***each individual*** natural number is finite.

Your "all" is ambiguous, whereas ***the entire set containing every*** and ***each individual*** are not ambiguous.

> What definable endsegments can empty the intersection.

***The entire set containing every*** one of them.

> > Since you persist in this confusion despite having it pointed out to you repeatedly, we can only conclude either that you are too stupid to understand the difference and/or you are intentionally utilizing this ambiguity precisely because you understand that lying is the only recourse you have.
> >
> > If you were intelligent and intellectually honest, you'd find an unambiguous way to make your claim.
>
> My claim is that none of the endsegments of (*) can empty the intersection.

And yet, the intersection over ***the entire set containing every*** endsegment is indeed empty.

> Either there are others or there is no empty intersection.

Or more obviously, you just don't understand the difference between ***the entire set containing every*** and ***each individual***.

EFQ

[Ganzhinterseher]

Am Freitag, 17. Januar 2020 23:18:48 UTC+1 schrieb Exfalso Quodlibet:


> > Name one endsegment that is not a last endsegment satisfying
> >
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> >
> > All those cannot empty the set, or can they?
>
> They all do!

None of those appearing in (*) does. They all are already in finite sets {E(1), E(2), ..., E(k)} with infinite intersection. Otherwise name definable endsegment which are not together in such a finite set with infinite intersection.

Your "they all" can only refer to undefinable endsegments.

> The intersection over ***the entire set containing every*** endsegment is empty.

But all endsegments appearing in (*) can be removed without effect.

Regards, WM

[Exfalso Quodlibet]

On Friday, January 17, 2020 at 5:32:32 PM UTC-5, Ganzhinterseher wrote:
> Am Freitag, 17. Januar 2020 23:18:48 UTC+1 schrieb Exfalso Quodlibet:
>
> > > Name one endsegment that is not a last endsegment satisfying
> > >
> > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> > >
> > > All those cannot empty the set, or can they?
> >
> > They all do!
>
> None of those appearing in (*) does.

Wrong interpretation of "all", which I why I highlighted the correct interpretation, which you disingenuously snipped: The intersection over ***the entire set containing every*** endsegment is empty.

> They all are already in finite sets {E(1), E(2), ..., E(k)} with infinite intersection.

That's the other "all". A less ambiguous phrasing would be that ***each individual*** endsegment is in a finite set with infinite intersection.

Then it would be obvious that you saying nothing at all about the intersection over ***the entire set containing every*** endsegment, because ***each individual*** is not at all the same thing as ***the entire set containing every***.

So once again, you demonstrate either that you are too stupid to comprehend the difference between ***the entire set containing every*** and ***each individual*** and/or that you are just a pathetic crank who knows there's no way to support his claims without resorting to disingenuous obfuscation by relying on the ambiguous use of "all" to cover two completely different meanings.

EFQ

[Jim Burns]

On 1/17/2020 9:04 AM, Ganzhinterseher wrote:

> The intersection of all remaining endsegments is always empty.
>
> I consider the intersection of all finite initial sequences
> of endsegments in the natural well-order:
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k)
> /\ |E(k)| = ℵo .
> (*)
>
> This is so for all endsegments.

This part is true.

> Since however the intersection of the remaining
> endsegments is empty,

This part is true.

> there must be remaining endsegments beyond every
> endsegment defined by k.

This part is hilarious.

If your argument were valid, it would also be valid to
argue that,
since, for every two distinct even naturals i and k,
there is an odd natural j between them, there must be
an odd natural j between every two distinct even naturals
i and k.

Do you also accept this conclusion from your argument?

> This is very easy. Why didn't you find it independently?

Part of the reason is that my parents did not drop me
on my head enough, when I was a babe in arms.

What is the reason that you DID find it?

[Ross A. Finlayson]

Yeah, fix it for him.

The point for inspecting and taking what he
is validating, here this "if your arguments
were valid, it would also be valid to argue ...",
that's certainly reasonable when you have
the both models each other and together,
and besides that they contradict in properties,
each other, still each is a complete little
un-broken model while you're still validating
each others' arguments. (Compared to, say,
reversing rule.) What numbers there are that
are inconsistent multiplicities to him, and,
you, and me too, everybody, still they are,
what then you're building off them as neatly
to show where the property of result comes
from, "in the infinite", that WM never had
another argument after whatever is WM's only
argument he ever had which is either "no infinite",
or "trans-finite doesn't make sense", to make
it so that the real infinite is there anyway
because besides collecting a bunch of eventually
mistaken attack against uncountability, then the
usual course of the terms in the text I can take
or leave.

So clearly you could extend his definition to
make sure to show it absurd, which then he must
instantly leave, but instead then he has the same
thing, "real infinite is absolute", or whatever
retro-finitist crankety troll, which many people are,
for example "there is no such thing as infinity",
what property is WM claiming to establish that
makes a nice mathematics for itself?

Just make up whatever mathematics it is and leave it neat.

Here it's "infinite numbers, "right after finite numbers".

Infinite numbers: none. (From here.)

The infinity is just as numbers
and the numbers go 1 - infinity in one second.

The numbers go 1-infinity in one second,
in each second.

(Of time.)

Of a drawing of a line as in time and constant velocity
mathematically....

The numbers goes 1-infinity,
in one second,
each second.

(Forever.)

Logic demands it's not any smaller than infinity,
infinity.

I.e., it's not any number.



Excuse me line-drawing is remedial and
1-infinity constant line drawing,
and with empty/full, must have some
classical notion of with the edge,
drawing the edge.

For example, Burns here might note,
..., "1-infinity" is just two labels
of arithmetic and here has a range operator
that would be easy to miss, which I draw
with "one-infinity" each measure 1.0 line segment.

Then of course it's "sure, ..., haul out the
collected your attack on the establishment of
foundations". Which I oblige.

That's just free time with 1.0 line-drawing - ....


The numbers go 1-infinity
each second
and start over
in one second.

One exactly to and through infinity.


It's a usual "continuum principle": time.


I wrote a huge mathematical attack
and seems I might've enjoyed it.


Sure, that's a modern defense.

[Ganzhinterseher]

Am Freitag, 17. Januar 2020 23:43:44 UTC+1 schrieb Exfalso Quodlibet:


> > > > Name one endsegment that is not a last endsegment satisfying
> > > >
> > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)

> A less ambiguous phrasing would be that ***each individual*** endsegment is in a finite set with infinite intersection.

Of course we can say each individual endsegment or better each endsegment individually definable. Nice to see that you begin to understand the difference.

>
> Then it would be obvious that you saying nothing at all about the intersection over ***the entire set containing every*** endsegment,

My statement concerns only the individually definable endsegments of (*). They all are of cardinality aleph_0 and their intersection has cardinality aleph_0.

> because ***each individual*** is not at all the same thing as ***the entire set containing every***.

My statement concerns all these individually definable endsegments because they all are in the sets in (*). Since the "entire set" yields the empty intersection, it is obvious that the entire set is larger than the subset of endsegments in (*).

>
> So once again, you demonstrate either that you are too stupid to comprehend the difference between ***the entire set containing every*** and ***each individual***

No, that is my essential point! The set of all endsegments that have an individual definition is different from the set of all endsegments of the entire set. Yes, there is a big difference. You have very nicely understood that.

> by relying on the ambiguous use of "all" to cover two completely different meanings.

Completely different meanings. Right. The set E_def of all endsegments of (*) is not the set E that you call the entire set, i.e., including the endsegments not appearing in (*).

Regards, WM

[Ganzhinterseher]

Am Freitag, 17. Januar 2020 23:46:38 UTC+1 schrieb Jim Burns:
> On 1/17/2020 9:04 AM, Ganzhinterseher wrote:
>
> > The intersection of all remaining endsegments is always empty.
> >
> > I consider the intersection of all finite initial sequences
> > of endsegments in the natural well-order:

> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) (*)

> > /\ |E(k)| = ℵo .
> > (*)
> >
> > This is so for all endsegments.
>
> This part is true.

Therefore they all can be removed as useless from

E(1) ∩ E(2) ∩ E(3) ∩ ... = { } (**)

Proof: You cannot define any endegment that is missing in (*) and useful in (**).

>
> > Since however the intersection of the remaining
> > endsegments is empty,
>
> This part is true.
>
> > there must be remaining endsegments beyond every
> > endsegment defined by k.
>
> This part is hilarious.

There is no alternative. If all indexed by definable natnumbers appear in (*) - and that is proven by the fact that you cannot define any endsegment not appearing in (*) -, then those are useless in (**).

>
> If your argument were valid,

If it were invalid, you could find definable endsegments not appearing in the sets of the sequence

E(1) = E(1)
E(1) ∩ E(2) = E(2)
E(1) ∩ E(2) ∩ E(3) = E(3)
... (***)

But you cannot.

> it would also be valid to
> argue that,
> since, for every two distinct even naturals i and k,
> there is an odd natural j between them, there must be
> an odd natural j between every two distinct even naturals
> i and k.

That is nonsense. Find a definable endsegment that is not in the sequence (***) or accept that no definable endsegment is missing there.

At least try to imagine why you shy away from considering this task.

Regards, WM

[Exfalso Quodlibet]

On Saturday, January 18, 2020 at 5:08:16 AM UTC-5, Ganzhinterseher wrote:
> Am Freitag, 17. Januar 2020 23:43:44 UTC+1 schrieb Exfalso Quodlibet:
>
>
> > > > > Name one endsegment that is not a last endsegment satisfying
> > > > >
> > > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
>
> > A less ambiguous phrasing would be that ***each individual*** endsegment is in a finite set with infinite intersection.
>
> Of course we can say each individual endsegment

I've been saying it all along.

> > Then it would be obvious that you saying nothing at all about the intersection over ***the entire set containing every*** endsegment,
>
> My statement concerns only the individually definable endsegments of (*).

If you are only talking about ***each individual*** endsegment, then what you say has no necessary bearing on the intersection over ***the entire set containing every*** endsegment.

> > So once again, you demonstrate either that you are too stupid to comprehend the difference between ***the entire set containing every*** and ***each individual***
>
> No,

Yes. You still don't understand.

EFQ

[Jim Burns]

On 1/18/2020 5:26 AM, Ganzhinterseher wrote:
> Am Freitag, 17. Januar 2020 23:46:38 UTC+1
> schrieb Jim Burns:
>> On 1/17/2020 9:04 AM, Ganzhinterseher wrote:

>>> Since however the intersection of the remaining
>>> endsegments is empty,
>>
>> This part is true.
>>

>>> *there must be remaining endsegments beyond every*
>>> *endsegment defined by k*
*emphasis added*

>> This part is hilarious.
>
> There is no alternative.
> If all indexed by definable natnumbers appear in (*)
> - and that is proven by the fact that you cannot define
> any endsegment not appearing in (*) -, then those are
> useless in (**).
>
>> If your argument were valid,
>
> If it were invalid, you could find definable endsegments
> not appearing in the sets of the sequence

I challenge you to find an odd natural NOT appearing
between 0 and 2,
between 2 and 4,
between 4 and 6,
...

> E(1) = E(1)
> E(1) ∩ E(2) = E(2)
> E(1) ∩ E(2) ∩ E(3) = E(3)
> ...
> (***)
>
> But you cannot.

You can't meet my challenge either, can you?

>> it would also be valid to
>> argue that,
>> since, for every two distinct even naturals i and k,
>> there is an odd natural j between them, there must be
>> an odd natural j between every two distinct even naturals
>> i and k.
>
> That is nonsense.

It's your nonsense. I drew a mustache on it, thus you didn't
recognize it as yours at first.

This:

| there must be remaining endsegments beyond every
| endsegment defined by k.

becomes this:
| there must be remaining odd numbers which are
| between every pair of even numbers defined by i and k.

I suppose that you would call them _undefinable odd numbers_
since they obviously aren't one of 1, 3, 5, 7, ...

We can _define_ an undefinable odd number == an odd number
between all pairs of even numbers. Having a definition is
not enough for one to exist. This is just as true for
"undefinable end segments".

> Find a definable endsegment that is not in the
> sequence (***) or accept that no definable endsegment
> is missing there.

No end segment is without following end segments.
No pair of even numbers is without an odd number between them.

No end segment follows all end segments.
No odd number is between all pairs of even numbers.

> At least try to imagine why you shy away from considering
> this task.

This
| Ax,Ey: P(x,y)
does not imply this:
| Ey,Ay: P(x,y)

[Ganzhinterseher]

> If you are only talking about ***each individual*** endsegment, then what you say has no necessary bearing on the intersection over ***the entire set containing every*** endsegment.

Every definable endsegment belongs to a finite set. If there is an actually infinite set, then there must be more endsegments. Those in

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

obviously fail.

>
> Yes. You still don't understand.

I cannot understand how anybody is too stupid to see that all definable endsegments are useless for the infinite set with empty intersection

∩{E(1), E(2), E(3), ...} = { } .

Apparently you don't even understand the theorem that every ne subset of an infinite set has a first element and that therefore the subset of necessary endsegments in

∩{E(x), E(x+1), E(x+2), ...} = { }

has to have a first elemnt.

Regards, WM

[Python]

Crank Wolfgang Mueckenheim sockpuppet, Ganzhinterseher wrote:
>
>
>> If you are only talking about ***each individual*** endsegment, then what you say has no necessary bearing on the intersection over ***the entire set containing every*** endsegment.
>
> Every definable endsegment belongs to a finite set.

Everything belongs to a finite set, idiot.

You cannot draw any kind of conclusion upon a property
that is true of anything, crank adjunkt lekturer Wolfgang
Mueckenheim from Hochschule Augsburg.

One day some of your victims (i.e. students) will bring
you in a dark alley in Augsburg and you'll get the punch
in the face you deserve as a charlatan, criminally teaching
fallacies.

[Ganzhinterseher]

Am Samstag, 18. Januar 2020 17:27:41 UTC+1 schrieb Jim Burns:


> > If it were invalid, you could find definable endsegments
> > not appearing in the sets of the sequence

> I challenge you to find an odd natural NOT appearing
> between 0 and 2,
> between 2 and 4,
> between 4 and 6,
> ...

Why? There is no definable one because all definable natural numbers are in the sequence of definable numbers and belong to finite initial segments
{1}, {1, 2}, {1, 2, 3}, ...

> You can't meet my challenge either, can you?

Why should I? Of course we cannot define undefinable numbers, neither me nor you.

>
> This:
> | there must be remaining endsegments beyond every
> | endsegment defined by k.
> becomes this:
> | there must be remaining odd numbers which are
> | between every pair of even numbers defined by i and k.

You have not understood. The question is whether definable endsegments are required to yield an empty intersection. This question can be denied by the theorem that every ne subset of endsegment has a first element, but the subset required for

∩{E(x), E(x+1), E(x+2), ...} = { }

has no first element. In addition we can exclude every definable endsegment by induction. E(x) is not E(1) and is not E(2). And if it is not E(n) then it is not E(n+1). Therefore it is not a definable endsegment.

Failed analogies deleted.

Regards, WM

[Ganzhinterseher]

Am Sonntag, 19. Januar 2020 16:20:09 UTC+1 schrieb Python:

> > Every definable endsegment belongs to a finite set {E(1), E(2), ..., E(k)}

>
> Everything belongs to a finite set,

Therefore actual infinity is nonsense.

> You cannot draw any kind of conclusion upon a property
> that is true of anything,

I can draw this conclusion: If every definable endsegment belongs to a finite set as given above, then it will not be useful for the set of endsegments yielding the empty intersection. Only very stupid matheologians can claim that these endsegments E(k) could be useful as E(x) in the infinite collection

∩{E(x), E(x+1), E(x+2), ...} = { } .

> One day some of your victims (i.e. students) will bring
> you in a dark alley in Augsburg and you'll get the punch
> in the face you deserve as a charlatan, criminally teaching
> fallacies.

Why should they? Even is transfinite set theory was not utter nonsense it would be completely useless for any scientific application. The simplest proof of this fact, as I always tell my students, is Scrooge McDuck
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 249.

Whether he gets bankrupt depends on the indexing of his dollars (whether he issues always the oldest dollar (bankrupt) or the newest dollar (infinite wealth) - can you follow?). Every student understands when I argue that indexing cannot be relevant for the result in any scientific application. Should all set theorists be too stupid to share this simple wisdom??

Regards, WM

[Python]

Crank Wolfgang Mueckenheim sockpuppet, Ganzhinterseher wrote:

> Am Sonntag, 19. Januar 2020 16:20:09 UTC+1 schrieb Python:
>
>>> Every definable endsegment belongs to a finite set {E(1), E(2), ..., E(k)}
>>
>> Everything belongs to a finite set,
>
> Therefore actual infinity is nonsense.

oh dear...

>> You cannot draw any kind of conclusion upon a property
>> that is true of anything,
>
> I can draw this conclusion: If every definable endsegment belongs to a finite set as given above, then it will not be useful for the set of endsegments yielding the empty intersection. Only very stupid matheologians can claim that these endsegments E(k) could be useful as E(x) in the infinite collection
>
> ∩{E(x), E(x+1), E(x+2), ...} = { } .

Only a fool and a fraud like YOU, Wofgang Mueckenheim, from Hochschule
Augsburg, may seriously pretend that this is a mathematical sound proof.

>> One day some of your victims (i.e. students) will bring
>> you in a dark alley in Augsburg and you'll get the punch
>> in the face you deserve as a charlatan, criminally teaching
>> fallacies.
>
> Why should they?

Because you are "teaching" lies, fallacies, sophistries and you
are ABUSING your position of authority as a teacher to FORCE
them to write down the same lies, fallacies and sophistries in
order to get good grades. I cannot imagine a worse act, except
rape or violence, a teacher may commit.

> Even is transfinite set theory was not utter nonsense it would be completely useless for any scientific application. The simplest proof of this fact, as I always tell my students, is Scrooge McDuck
> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 249.
>
> Whether he gets bankrupt depends on the indexing of his dollars (whether he issues always the oldest dollar (bankrupt) or the newest dollar (infinite wealth) - can you follow?). Every student understands when I argue that indexing cannot be relevant for the result in any scientific application. Should all set theorists be too stupid to share this simple wisdom??

This is complete baloney, you are not pointing out to your victims that
Ben Bacarisse wrote a complete and meaningful analysis and refutation of
your sophistic argument, are you?

http://bsb.me.uk/dd-wealth.pdf

This should be printed on A1 paper poster and put on the walls around
your infamous classroom at Hochschule Augsburg, Crank Adjunct Lekturer
Wolgang Mueckenheim, from Hochschule Augsburg.

[Ganzhinterseher]

Am Sonntag, 19. Januar 2020 17:14:48 UTC+1 schrieb Python:

> > Even is transfinite set theory was not utter nonsense it would be completely useless for any scientific application. The simplest proof of this fact, as I always tell my students, is Scrooge McDuck
> > https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 249.
> >
> > Whether he gets bankrupt depends on the indexing of his dollars (whether he issues always the oldest dollar (bankrupt) or the newest dollar (infinite wealth) - can you follow?). Every student understands when I argue that indexing cannot be relevant for the result in any scientific application. Should all set theorists be too stupid to share this simple wisdom??
>

> This is complete baloney, you are not pointing out that

> Ben Bacarisse wrote a complete and meaningful analysis and refutation of
> your sophistic argument, are you?

Ben Bacarisse has not even come close to understand my argument at all, neither have you. Otherwise you could not shout so nonsensically. Of course the dependence on indexing forbids any scientific application.

The enumeration of the rationals can be done in a way that only a small set between 0 and 1 are enumerated, like 0.1, 0.11, 0.111, ... or "all" rationals between -oo and oo. This shows to every not completely stultified person that there is no science in it. It is simply rubbish, and as I have proved, it is even self-contradictory rubbish.

Regards, WM

[Exfalso Quodlibet]

On Sunday, January 19, 2020 at 10:16:20 AM UTC-5, Ganzhinterseher wrote:
> > If you are only talking about ***each individual*** endsegment, then what you say has no necessary bearing on the intersection over ***the entire set containing every*** endsegment.
>
> Every definable endsegment belongs to a finite set.

So does everything. For all objects x, {x} is a finite set containing x.

Everything also belongs to an infinite set. For all objects x, N U {x} is an infinite set containing x.

In fact, everything belongs to an infinite number of finite sets and and infinite number of infinite sets!

> I cannot understand how anybody is too stupid to see that all definable endsegments are useless for the infinite set with empty intersection
>
> ∩{E(1), E(2), E(3), ...} = { } .

Because they aren't. All of them are there. You could see this better if you write it as:

∩{E(k) for all k in N} = { }

"all" means "all".

> Apparently you don't even understand the theorem that every ne subset of an infinite set has a first element

False. R is a non-empty subset of an infinite set, and there is no "first" element of R.

EFQ

[Jim Burns]

On 1/19/2020 10:52 AM, Ganzhinterseher wrote:
> Am Samstag, 18. Januar 2020 17:27:41 UTC+1
> schrieb Jim Burns:

>>> If it were invalid, you could find definable endsegments
>>> not appearing in the sets of the sequence
>
>> I challenge you to find an odd natural NOT appearing
>> between 0 and 2,
>> between 2 and 4,
>> between 4 and 6,
>> ...
>
> Why?

Why do I ask for that?
I'm giving you an argument to parallel your own. MAYBE mine will
give you a little perspective, a little distance from your own
argument, and you will be able to see how silly yours is, which
is so much like my silly argument.

> There is no definable one

Most people would say there are _none_ , not just
"no definable".

Do you think there is at least one "undefinable" odd number --
defined as one which is between every pair of distinct even numbers?

> because all definable natural numbers are in the
> sequence of definable numbers and belong to
> finite initial segments {1}, {1, 2}, {1, 2, 3}, ...
>
>> You can't meet my challenge either, can you?
>
> Why should I?

I don't care if you _should_
_Can_ you?

This is an (alleged) argument for odd numbers which _we are not_
_able to define_ Whether you _should_ whether you're _in the mood_
to define one or not is simply not relevant.

If you _can't_ is that because an odd number between all distinct
pairs of even numbers is (i) undefinable, or (ii) non-existent?

> Of course we cannot define undefinable numbers,
> neither me nor you.

Is the *reason* that we cannot find an odd number like that,
one between all distinct pairs of even numbers, that it would be
(i) undefinable, or (ii) non-existent?

Let me refresh your memory.

You (WM):

| Since however the intersection of the remaining endsegments

| is empty, there *must be* remaining endsegments beyond ever
| endsegment defined by k.
*emphasis added*

You (WM):

| If it were invalid, you could find definable endsegments
| not appearing in the sets of the sequence
|

| E(1) = E(1)
| E(1) ∩ E(2) = E(2)
| E(1) ∩ E(2) ∩ E(3) = E(3)
| ...
|

| But you cannot.

Is the *reason* that we cannot find an end segment like that,
one _not equal_ to the intersection of itself and preceding
end segments, that it would be (i) undefinable, or
(ii) non-existent?

We have your (WM's) answer: "must be".

If you were consistent, the same argument could be applied
to whether there is an "undefinable" odd number between
all pairs of distinct even numbers, and you would say
there "must be" such an odd number.

[...]

[Python]

Crank Wolfgang Mueckenheim sockpuppet, Ganzhinterseher wrote:

> Am Sonntag, 19. Januar 2020 17:14:48 UTC+1 schrieb Python:
>
>>> Even is transfinite set theory was not utter nonsense it would be completely useless for any scientific application. The simplest proof of this fact, as I always tell my students, is Scrooge McDuck
>>> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 249.
>>>
>>> Whether he gets bankrupt depends on the indexing of his dollars (whether he issues always the oldest dollar (bankrupt) or the newest dollar (infinite wealth) - can you follow?). Every student understands when I argue that indexing cannot be relevant for the result in any scientific application. Should all set theorists be too stupid to share this simple wisdom??
>>
>> This is complete baloney, you are not pointing out that
>> Ben Bacarisse wrote a complete and meaningful analysis and refutation of
>> your sophistic argument, are you?
>
> Ben Bacarisse has not even come close to understand my argument at all

Ben Bacarisse fully addressed your flawed argument and destroyed
100% of your sophistry in this specific case.

http://bsb.me.uk/dd-wealth.pdf

This is the only reason, you disgusting piece of dirt, you shame of
academy, you fool and you FRAUD stopped showing this article (if you
ever did so) to your victims (i.e. Hochschule Augsburg students)

Herr Mueckenheim, you are "teaching" lies, fallacies, sophistries and

[Ganzhinterseher]

Am Sonntag, 19. Januar 2020 18:18:12 UTC+1 schrieb Exfalso Quodlibet:


> > I cannot understand how anybody is too stupid to see that all definable endsegments are useless for the infinite set with empty intersection
> >
> > ∩{E(1), E(2), E(3), ...} = { } .
>
> Because they aren't. All of them are there.

All of them fail:

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

Writing them there is a mater of frauds.

You could see this better if you write it as:
>
> ∩{E(k) for all k in N} = { }
>
> "all" means "all".

All means all here too:

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

and all are failing.

Regards, WM

[Exfalso Quodlibet]

On Monday, January 20, 2020 at 11:03:32 AM UTC-5, Ganzhinterseher wrote:
> Am Sonntag, 19. Januar 2020 18:18:12 UTC+1 schrieb Exfalso Quodlibet:
>
>
> > > I cannot understand how anybody is too stupid to see that all definable endsegments are useless for the infinite set with empty intersection
> > >
> > > ∩{E(1), E(2), E(3), ...} = { } .
> >
> > Because they aren't. All of them are there.
>
> All of them fail:
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

That's a completely different scenario.

An infinite set is not the same thing as a finite set.

> You could see this better if you write it as:
> >
> > ∩{E(k) for all k in N} = { }
> >
> > "all" means "all".
>
> All means all here too:
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

No, this is a different "all"! See, you STILL don't understand the difference between ***the entire set containing every*** endsegment and ***each individual*** endsegment.

∩{E(1), E(2), E(3), ...} = { } is a statement about ***the entire set containing every*** endsegment.

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) is a statement about ***each individual*** endsegment. It says absolutely nothing at all about ***the entire set containing every*** endsegment.

In one case, the universal quantifier is inside the intersection, and in the other case, it is outside the intersection. ***IT IS NOT VALID TO FREELY MOVE THE QUANTIFIER AROUND!!!*** Changing the relative scope of the quantifier and the intersection changes the meaning.

EFQ

[Ganzhinterseher]

Am Sonntag, 19. Januar 2020 22:36:13 UTC+1 schrieb Python:

> Ben Bacarisse fully addressed your flawed argument

He said: "McDuck’s wealth is not determined by which notes he hold indefinitely, but by how many he holds."

That is a sophistry in contradiction with the fact that he gives back every note and the set theory demand of completing infinite sets.

"Prof. Mückenheim is simply telling you to use the wrong limit for the job."

Ben recommends to use the limit of the cardinalities which is increasing infinitely. That would imply, in case of enumerating fractions, that the not enumerated fractions increase infinitely and never all fractions will be enumerated. But that is certainly not what set theorists wish.

However, in order to circumvent such sophistry, I have devised the arguments with endsegments. No sophistry can avoid or "interpret" the final empty set in

∩{E(x), E(x+1), E(x+2), ...} = { }

but because of

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

none of the definable endsegments can be put in positions of E(x) except by frauds. But in this case they will be demasked by every person with average intelligence.

Regards, WM

[Ganzhinterseher]

Am Montag, 20. Januar 2020 17:21:34 UTC+1 schrieb Exfalso Quodlibet:
> On Monday, January 20, 2020 at 11:03:32 AM UTC-5, Ganzhinterseher wrote:
> > Am Sonntag, 19. Januar 2020 18:18:12 UTC+1 schrieb Exfalso Quodlibet:
> >
> >
> > > > I cannot understand how anybody is too stupid to see that all definable endsegments are useless for the infinite set with empty intersection
> > > >
> > > > ∩{E(1), E(2), E(3), ...} = { } .
> > >
> > > Because they aren't. All of them are there.
> >
> > All of them fail:
> >

> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)

>
> That's a completely different scenario.

No, it shows that only a fraud can apply one of these endsegments in

∩{E(x), E(x+1), E(x+2), ...} = { } . (**)

>
> An infinite set is not the same thing as a finite set.

An infinite set needs infinitely many elelements. But no endsegment appeasrimg in (*) would be recruited for (**) by a sober person.

>
> > You could see this better if you write it as:
> > >
> > > ∩{E(k) for all k in N} = { }
> > >
> > > "all" means "all".
> >
> > All means all here too:
> >

> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (1)

>
> No, this is a different "all"!

Chuckle. There is no difference in "all". A"All" menas that there is no counter example.

> See, you STILL don't understand the difference between ***the entire set containing every*** endsegment and ***each individual*** endsegment.

Each individual endsegment with no exception is denoed by all endsegments.

>
> ∩{E(1), E(2), E(3), ...} = { } is a statement about ***the entire set containing every*** endsegment.

E(1), E(2), E(3), ... are only inserted by frauds who try to abuse their students.

>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) is a statement about ***each individual*** endsegment. It says absolutely nothing at all about ***the entire set containing every*** endsegment.

Note that I do not even want to say anything about "the set"- My proof concerns only all definable endsegments. Only frauds can try to apply any of them in

∩{E(x), E(x+1), E(x+2), ...} = { }
>

> In one case, the universal quantifier is inside the intersection, and in the other case, it is outside the intersection.

In the latter case it concerns all definable endsegments.

***IT IS NOT VALID TO FREELY MOVE THE QUANTIFIER AROUND!!!***

Nut it is allowed to quantify over all definable endsegments because each one is the last one of a finite initial set.

When you switch to "the set", then you refrain from this sober application and allow undefinable endsegments.

> Changing the relative scope of the quantifier and the intersection changes the meaning.
>

So it is. Your quantifier inside the intersection changes the meaning heavily. It accepts undefinable endsegments. Note that individual definitions always have the form of (*), as you emphasized yourself "about ***each individual*** endsegment"!

Regards, WM

[Exfalso Quodlibet]

On Monday, January 20, 2020 at 12:23:51 PM UTC-5, Ganzhinterseher wrote:
> Am Montag, 20. Januar 2020 17:21:34 UTC+1 schrieb Exfalso Quodlibet:
> > On Monday, January 20, 2020 at 11:03:32 AM UTC-5, Ganzhinterseher wrote:
> > > Am Sonntag, 19. Januar 2020 18:18:12 UTC+1 schrieb Exfalso Quodlibet:
> > >
> > >
> > > > > I cannot understand how anybody is too stupid to see that all definable endsegments are useless for the infinite set with empty intersection
> > > > >
> > > > > ∩{E(1), E(2), E(3), ...} = { } .
> > > >
> > > > Because they aren't. All of them are there.
> > >
> > > All of them fail:
> > >
> > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> >
> > That's a completely different scenario.
>
> No,

Yes. An infinite set it not the same thing as a finite set.

> > > You could see this better if you write it as:
> > > >
> > > > ∩{E(k) for all k in N} = { }
> > > >
> > > > "all" means "all".
> > >
> > > All means all here too:
> > >
> > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (1)
> >
> > No, this is a different "all"!
>

> There is no difference in "all".

Well, at least you admit to not understanding the difference between ***the entire set containing every*** and ***each individual***, which are two different meanings of "all".

This may also explain why you don't understand how an infinite set is different from a finite set.

> Note that I do not even want to say anything about "the set"-

And yet, you keep doing so! Such as here:

> ∩{E(x), E(x+1), E(x+2), ...} = { }

EFQ

[Ganzhinterseher]

Am Sonntag, 19. Januar 2020 19:50:23 UTC+1 schrieb Jim Burns:


> I'm giving you an argument to parallel your own.

No the comparison is misleading

> Do you think there is at least one "undefinable" odd number --
> defined as one which is between every pair of distinct even numbers?

Between two defined natnumbers there cannot be an undefinable natnumber. I told you already: Try to imagine the set of all natnumbers that have been defined in the universe (or on earth, or in your office). This set has a maximum and all smaller numbers and even all larger numbers which can be constructed by finite steps like multiplication with 10^100^1000 are defined and definable.

> You (WM):
> | Since however the intersection of the remaining endsegments
> | is empty, there *must be* remaining endsegments beyond ever
> | endsegment defined by k.
> *emphasis added*
>
> You (WM):
> | If it were invalid, you could find definable endsegments
> | not appearing in the sets of the sequence
> |
> | E(1) = E(1)
> | E(1) ∩ E(2) = E(2)
> | E(1) ∩ E(2) ∩ E(3) = E(3)
> | ...
> |
> | But you cannot.
>
> Is the *reason* that we cannot find an end segment like that,
> one _not equal_ to the intersection of itself and preceding
> end segments, that it would be (i) undefinable, or
> (ii) non-existent?

In case of actual infinity we must say undefinable.
In case of potential infinity we must say not existent.

There is not much difference however, because the set of definable (in actual infinity) or existent (in potential infinity) endsegments is not fixed but can be increased as far as we wish, never becoming actually infinite though (i.e. larger than every finite number).

Regards, WM

[George Greene]

On Saturday, January 18, 2020 at 5:26:07 AM UTC-5, Ganzhinterseher wrote:
> Therefore they all can be removed as useless from
>
> E(1) ∩ E(2) ∩ E(3) ∩ ... = { } (**)

You don't even believe in infinite sets.
This infinite intersection is not something YOU can define.
INDIVIDUALLY each one of them is "useless".
SO WHAT?

> Proof: You cannot define any endegment that is missing in (*) and useful in (**).

That is true. AGAIN, SO WHAT? If you remove THAT endsegment and any other finite set of endsegments along with it, they will INDIVIDUALLY be useless.
That does not make any collective useless. The usefulness isn't an individual
thing.

> >
> > > Since however the intersection of the remaining
> > > endsegments is empty,
> >
> > This part is true.
> >
> > > there must be remaining endsegments beyond every
> > > endsegment defined by k.
> >
> > This part is hilarious.
>
> There is no alternative.

Of course there is an alternative. He called you HILARIOUS -- that means yoU FAILED TO STATE anything to which an alternative could be tried -- "every endsegment defined by k" IS HILARIOUS. It is NONSENSICAL. NO endsegment is "defined by k". You have to instantiate k before it defines an endsegment.

> If all indexed by definable natnumbers appear in (*) - and
> that is proven by the fact that you cannot define any
> endsegment not appearing in (*) -, then those are useless in (**).

Your dumb ass IS STILL EQUIVOCATING on "all", this time by using "those".
INDIVIDUALLY, EACH of them is useless. THAT FAILS to imply that ANY INFNITE
SET of them would be COLLECTIVELY useless.
It simply DOES NOT LOGICALLY FOLLOW from the fact that EACH member of something
is useless that SOME INFINITE SUBCOLLECTION of it (let alone the whole collective thing) is also useless. You seem to think that because you can add up infinitely many zeros and get zero that this has to work the same way. IT DOESN'T AND YOU CAN'T PROVE that it does. All you can do is rant in natural language while intentionally abusing adverbs.

[Jim Burns]

On 1/20/2020 12:36 PM, Ganzhinterseher wrote:
> Am Sonntag, 19. Januar 2020 19:50:23 UTC+1
> schrieb Jim Burns:

>> I'm giving you an argument to parallel your own.
>
> No the comparison is misleading

Oh, the comparison is _embarrassing_ I don't doubt that.
Misleading? No.

>> Do you think there is at least one "undefinable" odd

>> number -- *defined as one which is between every pair*
> *of distinct even numbers* ?

*emphasis added*

> Between two defined natnumbers there cannot be an undefinable
> natnumber.

If there exists an undefinable odd number *where we define*
*an indefinable odd number as one which is between every pair*
*of distinct even numbers* then that is clearly, obviously false.
If you insist upon that, (by your rules) you are a liar and/or
a fool.

> I told you already:

*NO* _I_ told _you_ already what I mean by "undefinable" here.
This argument is _parallel_ to your argument.

Do _undefinable-in-my-specified-sense_ odd numbers exist?

I think you would like to say "no" to that very much.
But, then, the next question is:

Do _undefinable-in-your-specified-sense_ end segments exist?
[1]

You have said they "must be". You base this claim on your
ability say "end segment which follows all defined end segments"
and an inability to use quantifiers correctly.

I can do that, too. Presto! "Undefinable" odd numbers --
each of which exists between all pairs of distinct even numbers.

[1]
You have specified that an undefined end segment follows all
defined end segments.

> Try to imagine the set of all natnumbers that have been
> defined in the universe (or on earth, or in your office).
> This set has a maximum and all smaller numbers and even all
> larger numbers which can be constructed by finite steps like
> multiplication with 10^100^1000 are defined and definable.

"Not defined" is not the same as "not definable".

"Try to imagine" a one-ended chain.

As a condition for _being in the same chain_ there must be a
finite piece of chain between two links. Thus, if I find a
piece of chain in my basement and you find a piece of chain
in your garage on the other side of the Atlantic, we can
reasonably conclude they are not _in the same chain_

However, there is only one end. No _other_ link in the
one-ended chain is an end.

This is a perfectly serviceable model for the natural numbers
0, 1, 2, ... , 99, 100, ...
Each link corresponds to a natural number. 0 is one end.
Every link can be finitely counted to. There is no second end.

There are no dark numbers in this model.
_In order to be in the chain_ a link must have a finite piece
of chain to 0, the single end. This disqualifies it from
being dark.

There can still exist other links, other pieces of chain.
But they won't be in the one-ended chain.

"Try to imagine" a set N which contains 0 and is closed
under successor x --> x+1, and which has no proper subset B
which contains 0 and is closed under successor.

The subset B of N defined as all x in N for which there is
a finite successor-chain from 0 and for which x is not last
-- contains 0
-- is closed under successor x --> x+1

The only subset of N which B can be is N itself.
Therefore, for all x in N, there is a finite successor-chain
from 0 and x is not last.

There are no dark numbers in N.

>> You (WM):
>> | Since however the intersection of the remaining endsegments
>> | is empty, there *must be* remaining endsegments beyond ever
>> | endsegment defined by k.
>> *emphasis added*
>>
>> You (WM):
>> | If it were invalid, you could find definable endsegments
>> | not appearing in the sets of the sequence
>> |
>> | E(1) = E(1)
>> | E(1) ∩ E(2) = E(2)
>> | E(1) ∩ E(2) ∩ E(3) = E(3)
>> | ...
>> |
>> | But you cannot.
>>
>> Is the *reason* that we cannot find an end segment like that,
>> one _not equal_ to the intersection of itself and preceding
>> end segments, that it would be (i) undefinable, or
>> (ii) non-existent?
>
> In case of actual infinity we must say undefinable.
> In case of potential infinity we must say not existent.

What do we say in case of a set which contains 0 and is closed
under successor x --> x+1, and which has no proper subset B
which contains 0 and is closed under successor?
Undefinable or non-existent?

Keep in mind that this is a set for which we can prove:
-- There is an end segment for each of its elements.
-- All of its end segments are infinite.
-- The intersection of any end segment with all of its
predecessors equals that end segment.
-- The intersection of any endsegment with all of its
successors is empty.

> There is not much difference however, because the set of
> definable (in actual infinity) or existent (in potential
> infinity) endsegments is not fixed but can be increased as
> far as we wish, never becoming actually infinite though
> (i.e. larger than every finite number).

Your distinction does not seem to be applicable in the case of
a set which contains 0 and is closed under successor x --> x+1,
and which has no proper subset B which contains 0 and is
closed under successor. B contains 0 or does not contain 0.
There _is_ some x in B such that x+1 is not in B or there
_isn't_ Whatever B is (even when B is N), B is.

I imagine that this leaves unanswered questions of potentiality
and actuality for you to ponder. It does, however, close the
book on the intersection of all the end segments (empty)
and similar well-asked questions.

[Ganzhinterseher]

Am Montag, 20. Januar 2020 19:02:25 UTC+1 schrieb George Greene:
> On Saturday, January 18, 2020 at 5:26:07 AM UTC-5, Ganzhinterseher wrote:
> > Therefore they all can be removed as useless from
> >
> > E(1) ∩ E(2) ∩ E(3) ∩ ... = { } (**)
>
> You don't even believe in infinite sets.

Why is that so? Because I can prove that they do not exist.

> This infinite intersection is not something YOU can define.

You are in error. I can. See (**).

> INDIVIDUALLY each one of them is "useless".
> SO WHAT?

Every individual endsegment is useless. That shows that the not useless endsegments are not individually definable.

>
> > Proof: You cannot define any endegment that is missing in (*) and useful in (**).
>
> That is true. AGAIN, SO WHAT? If you remove THAT endsegment and any other finite set of endsegments along with it, they will INDIVIDUALLY be useless.
> That does not make any collective useless. The usefulness isn't an individual
> thing.

But the fact the every useless endsegment can be removed and nevertheless the result persist shows that there are unremovable endsegments. I call them dark endsegments. You have discovered such paths in the Binary Tree, remember?

"You could delete all and only the finite paths from the tree, and that would delete all the nodes, yet you would never have deleted any infinite path, and they would all still exist in any case, regardless of what had been deleted!" [George Greene in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 217.

>

> NO endsegment is "defined by k". You have to instantiate k before it defines an endsegment.

k had been defined to be a natural number.

∀k ∈ ℕ: E(k) = {k, k+1, k+2, ...}

>
> It simply DOES NOT LOGICALLY FOLLOW from the fact that EACH member of something
> is useless that SOME INFINITE SUBCOLLECTION

There is no infinite subcollection of those endsegments that are the last of their subsets: {E(1), E(2), ..., E(k)}. To believe so is simply lunatic.

> of it (let alone the whole collective thing) is also useless.

Here you seem to use different "logic". Every useless endsegment lets through infinitely many natnumbers.

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)

This does not change when all are together. But in fact there are not more than given in (*):

E(1) = E(1)
E(1) ∩ E(2) = E(2)
E(1) ∩ E(2) ∩ E(3) = E(3)
...

Never more than finitely many can be gathered.

> You seem to think that because you can add up infinitely many zeros and get zero that this has to work the same way.

Of course it does.

> IT DOESN'T AND YOU CAN'T PROVE that it does.

I can prove and have proved that all definable endsegments of (*) can be removed from (**) without any effect.

> All you can do is rant

Try yourself to apply logic: Show any definable endsegment that has to stay in (**). Fail. Continue to rant.

Regards, WM

[Ganzhinterseher]

Am Montag, 20. Januar 2020 20:47:22 UTC+1 schrieb Jim Burns:


> >> Do you think there is at least one "undefinable" odd
> >> number -- *defined as one which is between every pair*
> > *of distinct even numbers* ?
>
> *emphasis added*

There is no least undefinable natural number definable, whether even or odd.

>
> > Between two defined natnumbers there cannot be an undefinable
> > natnumber.
>
> If there exists an undefinable odd number *where we define*
> *an indefinable odd number as one which is between every pair*
> *of distinct even numbers* then that is clearly, obviously false.

Between undefinable odd numbers there is an undefinable even number and vice versa. But they are undfinable,

> *NO* _I_ told _you_ already what I mean by "undefinable" here.
> This argument is _parallel_ to your argument.
>
> Do _undefinable-in-my-specified-sense_ odd numbers exist?

That is a matter of taste. In actual infinity they must exist.

>
> I think you would like to say "no" to that very much.

No is the correct answer in potential infinity.

> I can do that, too. Presto! "Undefinable" odd numbers --
> each of which exists between all pairs of distinct even numbers.
>
> [1]
> You have specified that an undefined end segment follows all
> defined end segments.

Yes, but the border is no fixed. See my explanation:

> > Try to imagine the set of all natnumbers that have been
> > defined in the universe (or on earth, or in your office).
> > This set has a maximum and all smaller numbers and even all
> > larger numbers which can be constructed by finite steps like
> > multiplication with 10^100^1000 are defined and definable.
>
> "Not defined" is not the same as "not definable".

Yes but how many numbers ever can be defined: They are finitely many. That is incontestable. But if the whole set |N is actually infinite, then infinitely many numbers remain undefinable.

If only finitely man can (we cannot say which, because there is no fixed border), then infinitely many cannot.

> "Try to imagine" a one-ended chain.
>
> As a condition for _being in the same chain_ there must be a
> finite piece of chain between two links. Thus, if I find a
> piece of chain in my basement and you find a piece of chain
> in your garage on the other side of the Atlantic, we can
> reasonably conclude they are not _in the same chain_
>
> However, there is only one end. No _other_ link in the
> one-ended chain is an end.
>
> This is a perfectly serviceable model for the natural numbers
> 0, 1, 2, ... , 99, 100, ...
> Each link corresponds to a natural number. 0 is one end.
> Every link can be finitely counted to. There is no second end.

Yes, but there is always a last link visible.

>
> There are no dark numbers in this model.

Either there is another end visisble or it is not visible. But if we know that we can reach the other end, then it must exist. This knowledge has two sources: First we can count beyond it (ω + 1) and second we can remove all links by the intersection of endsegments.

> _In order to be in the chain_ a link must have a finite piece
> of chain to 0, the single end. This disqualifies it from
> being dark.
>
> There can still exist other links, other pieces of chain.
> But they won't be in the one-ended chain.

There is no endsegment unless the chain can be covered completely.
There is no bijection with Q unless it is complete.
There is no chance to define a real number by infinitely many digits without completeness.

> There are no dark numbers in N.

Then there are no completeness.

> I imagine that this leaves unanswered questions of potentiality
> and actuality for you to ponder. It does, however, close the
> book on the intersection of all the end segments (empty)

No. There is a difference between the universal quantification over all endsegments

∩{E(1), E(2), E(3), ...} = { }

and over all definable endsegments

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)

If there were no dark endsegments, then all endsegments were those of (*) and their omission from all would yield nothing remaining in

∩{E(x), E(x+1), E(x+2), ...} = { } .

Please try to think rational enough to see that all of (*) can be omitted. And if you fail, try to find the first one that cannot be omitted.

REgards, WM

[Exfalso Quodlibet]

On Tuesday, January 21, 2020 at 9:45:39 AM UTC-5, Ganzhinterseher wrote:
> Am Montag, 20. Januar 2020 19:02:25 UTC+1 schrieb George Greene:
> > On Saturday, January 18, 2020 at 5:26:07 AM UTC-5, Ganzhinterseher wrote:
> > > Therefore they all can be removed as useless from
> > >
> > > E(1) ∩ E(2) ∩ E(3) ∩ ... = { } (**)
> >
> > You don't even believe in infinite sets.
>
> Why is that so? Because I can prove that they do not exist.

Then endsegments do not exist.

EFQ

[Jim Burns]

On 1/21/2020 10:37 AM, Ganzhinterseher wrote:
> Am Montag, 20. Januar 2020 20:47:22 UTC+1
> schrieb Jim Burns:

>> "Try to imagine" a one-ended chain.
>>
>> As a condition for _being in the same chain_ there must be a
>> finite piece of chain between two links. Thus, if I find a
>> piece of chain in my basement and you find a piece of chain
>> in your garage on the other side of the Atlantic, we can
>> reasonably conclude they are not _in the same chain_
>>
>> However, there is only one end. No _other_ link in the
>> one-ended chain is an end.
>>
>> This is a perfectly serviceable model for the natural numbers
>> 0, 1, 2, ... , 99, 100, ...
>> Each link corresponds to a natural number. 0 is one end.
>> Every link can be finitely counted to. There is no second end.
>
> Yes, but there is always a last link visible.

This does not contradict "There is no second end".

What kind of chains do you have?
Do the links of your chains stop linking if you don't
look at them? Consider getting them replaced.

>> There are no dark numbers in this model.
>
> Either there is another end visisble or it is not visible.

... or another end does not exist.
In this case, another end does not exist.

> But if we know that we can reach the other end,
> then it must exist.

... the other end that doesn't exist.

We know that we CAN'T reach "the other end", what with it
not existing and all.

I said:
| "Try to imagine" a one-ended chain.

Fail.

[...]

>> I imagine that this leaves unanswered questions of potentiality
>> and actuality for you to ponder. It does, however, close the
>> book on the intersection of all the end segments (empty)
>
> No. There is a difference between the universal quantification
> over all endsegments
> ∩{E(1), E(2), E(3), ...} = { }
>
> and over all definable endsegments
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k)
> /\ |E(k)| = ℵo

Wrong. There is no difference.

Let k e N_def.
E(k) = { j e N_def | j >= k }

Ak e N_def:
Intrsct{ E(i) sub N_def | i =< k & i e N_def } = E(k)

Ak e N_def:
card(E(k)) = card(N_def)

Intrsct{ E(k) sub N_def | k e N_def } = {}

Add to that
Ak e N_def:
Intrsct{ E(i) sub N_def | i >= k & i e N_def } = {}

> If there were no dark endsegments,
> then all endsegments were those of (*) and
> their omission from all would yield nothing remaining in
> ∩{E(x), E(x+1), E(x+2), ...} = { } .

Ah.
So, you're claiming that Intrsct{} = {} is some sort of
contradiction? Do I have that right?

Explain why Intrsct{} = {} is a contradiction.

Use this definition:
Intrsct(B) = { x e Union(B) | FORALL Y e B: x e Y }

forall x:( x e Union(B) <-> EXIST Y e B: x e Y )

> Please try to think rational enough to see that all of (*)
> can be omitted.

By the Law of Monotonic Intersection Sequences,
"can be omitted" == "is not last".

It's true that all of (*) is not last.
I haven't denied it at any point.

Note that "all of (*) is not last" implies that,
for all k e N, there is some end segment k is not in.

Which is to say: the intersection of all end segments is empty.

> And if you fail, try to find the first one that cannot
> be omitted.

Let's just settle on my failing to find a last end segment.
I don't recall every claiming there _was_ a last end segment
in { E(1),E(2),... }

_You_ on the other hand, seem to be claiming there is a
contradiction in Intrsct{} = {}
Please show me this contradiction.

(Whatever your analysis is, it shouldn't be controversial.
When last I checked, {} was finite.
If you disagree about {}, let me know.)

[Ganzhinterseher]

Am Dienstag, 21. Januar 2020 19:58:23 UTC+1 schrieb Jim Burns:


> > There is a difference between the universal quantification
> > over all endsegments

> > ∩{E(1), E(2), E(3), ...} = { } (**)

> >
> > and over all definable endsegments

> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) (*)

> > /\ |E(k)| = ℵo
>
> Wrong. There is no difference.

If there is no difference, then ℵo = 0.

> > If there were no dark endsegments,
> > then all endsegments were those of (*) and
> > their omission from all would yield nothing remaining in
> > ∩{E(x), E(x+1), E(x+2), ...} = { } .
>
> Ah.
> So, you're claiming that Intrsct{} = {} is some sort of
> contradiction?

That is not the point (although in ZF it would cause problems). But the point is that only endsegments of (*) were in (**) and none of them could deviate from what it does in (*).

> Explain why Intrsct{} = {} is a contradiction.

See MathStackExchange if you are interested. Intersection { } = U.
https://math.stackexchange.com/questions/348668/intersection-of-the-empty-set-and-vacuous-truth

But as I said, that is not the problem. The problem is the contradiction raised by different effects of endsegments. It is impossible that any endsegment of (*) can act to the effect that the intersection is empty. Endsegments are individuals.

>
> > Please try to think rational enough to see that all of (*)
> > can be omitted.
>
> By the Law of Monotonic Intersection Sequences,
> "can be omitted" == "is not last".

If there is no endsegment different from all in (*), then there cannot be a result different from (*).

>
> It's true that all of (*) is not last.
> I haven't denied it at any point.

But it is true that all of (*) do not supply an actually infinit set. Therefore the set of (*) is potentially infinite and always has a last endsegment (in an actually infinite theory. Otherwise no endsegments would exist.)

>
> Note that "all of (*) is not last" implies that,
> for all k e N, there is some end segment k is not in.

But every such set is finite and has a last endsegment.

>
> Which is to say: the intersection of all end segments is empty.

Look: The intersections of all sets

E(1) = E(1)
E(1) ∩ E(2) = E(2)
E(1) ∩ E(2) ∩ E(3) = E(3)
...

E(1) ∩ E(2) ∩ E(3) ∩ ... ∩ E(k) = E(k)
...
are infinite. With quantification like (*) this cannot be changed.

If you can't understand that the inevitable existence of aleph_0 undefined endsegments or natnumbers implies the existence of aleph_0 undefinable endsegments or natnumbers then consider whether you can accept the following statement:

There is an actually infinite set Y of aleph_0 as yet undefined natural numbers, and this will never change (although an arbitrary finite subset of Y might be removed from Y and added to the set |N_def of defined natnumbers.)

Regards, WM

[Ganzhinterseher]

Yes. But you believe in their existence. And that is in contradiction with mathematics, namely with ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

Regards, WM

[Exfalso Quodlibet]

On Wednesday, January 22, 2020 at 12:36:09 PM UTC-5, Ganzhinterseher wrote:
> Am Dienstag, 21. Januar 2020 17:55:44 UTC+1 schrieb Exfalso Quodlibet:
> > On Tuesday, January 21, 2020 at 9:45:39 AM UTC-5, Ganzhinterseher wrote:
> > > Am Montag, 20. Januar 2020 19:02:25 UTC+1 schrieb George Greene:
> > > > On Saturday, January 18, 2020 at 5:26:07 AM UTC-5, Ganzhinterseher wrote:
> > > > > Therefore they all can be removed as useless from
> > > > >
> > > > > E(1) ∩ E(2) ∩ E(3) ∩ ... = { } (**)
> > > >
> > > > You don't even believe in infinite sets.
> > >
> > > Why is that so? Because I can prove that they do not exist.
> >
> > Then endsegments do not exist.
> >
> Yes.

If they do not exist in your system, then your system can't use them to prove anything about a system in which they do exist.

EFQ

[George Greene]

On Tuesday, January 14, 2020 at 12:56:51 PM UTC-5, Ganzhinterseher wrote:
> ZFC set theory needs an actually infinite
> domain and would be boring without uncountable sets.

Oh, bullshit. You can remove the axiom of infinity (not to mention the F and the C)from Z, AND STILL have an interesting theory of nothing but finite sets.
The infinite sets would still exist (darkly) in SOME models of it but the inner
model would have only finite sets -- only finite sets could be PROVED to exist.
The fact that EVEN ONE infinite set has to be stipulated-into-existence-by-an-axiom -- without any hint at how to construct it, even with a supertask -- is
the only thing that makes all your bullshit arguments seem plausible in natural
language. The finite set theory that results from this is bi-interpretable with
Peano Arithmetic and is therefore just as interesting AS THAT is, even though
It Also Needn't Have Any infinite numbers.

[Ganzhinterseher]

Am Donnerstag, 23. Januar 2020 03:16:27 UTC+1 schrieb George Greene:
> On Tuesday, January 14, 2020 at 12:56:51 PM UTC-5, Ganzhinterseher wrote:
> > ZFC set theory needs an actually infinite
> > domain and would be boring without uncountable sets.
>

> You can remove the axiom of infinity (not to mention the F and the C)from Z, AND STILL have an interesting theory of nothing but finite sets.

Without uncountability and without Cantor's 1891 sleight of hand no-one would know set theory today except some historians perhaps.

But reality is in fact less sensational. You can see it when investigating decending sequences of ordinal numbers. There you have only definable numbers, contrary to the boasting "...". There you have correct mathematics. All these sequences are finite. Therefore the set of descending sequences of ordinals is potentially infinite.

Another example ios the set of periods of decimal representations of rational numbers or the set of prime gaps.

All that is interesting, no doubt. But there is no finished infinity let alone uncountability.

Regards, WM