Re: The modern mathematical concept of infinity is unavoidable
Albrecht
Virgil schrieb:
> In article
> <ad9d7fe8-cfcc-4bdd...@b38g2000prf.googlegroups.com>,
> Albrecht <albs...@gmx.de> wrote:
>
> > FredJeffries schrieb:
> > > On Jan 25, 5:12�am, Albrecht <albst...@gmx.de> wrote:
> > >
> > >
> > > >
> > > > Consider a body which moves from point A to point B. Consider the
> > > > barycentre Z (Schwerpunkt) of this body. Now the point Z, if it moves
> > > > from A to B, at first has to move from the point A to a point which
> > > > lays next to the point A. But there is no such next point.
> > >
> > > If there is no such point then Z does not have to move to it.
> > >
> > >
> > >
> > > > How is continuity possible?
> > >
> > > I thought that's what I was asking.
> >
> > And I was asking what the idea of actual infinity changes in this
> > concern?
>
> If there are only a finite set of points in any interval, then velocity
> and acceleration are impossible concepts.
Nobody consider only finite many points in an interval. The
alternative is not finite/infinite as you seem to imply.
Your concept of infinity is wrong, that's the problem: There is no
consistent cardinality of infinity.
Mitch Harris
> Virgil schrieb:
>
> > If there are only a finite set of points in any interval, then velocity
> > and acceleration are impossible concepts.
>
> Nobody consider only finite many points in an interval. The
> alternative is not finite/infinite as you seem to imply.
>
> Your concept of infinity is wrong, that's the problem: There is no
> consistent cardinality of infinity.
We've all been using the simple term 'infinity'. Maybe the
inconsistency you see is really the varieties of things that might be
called infinity, each consistent in its own way.
The following might give you a start:
http://en.wikipedia.org/wiki/Cardinality
http://en.wikipedia.org/wiki/Ordinal_number
You're right, it seems counter-intuitive that the set of even integers
is the same size (equinumerous, has the same number/cardinality) as
all integers. But mathematical investigation has given consistent
meanings to all these terms. If you learn the meanings of those terms,
you'll see that things work out nicely. Those meanings may not match
the ones you have now though.
Mitch
David R Tribble
> Your concept of infinity is wrong, that's the problem: There is no
> consistent cardinality of infinity.
How is Aleph_0 inconsistent?
Virgil
<faeb9459-0806-40d2...@w24g2000prd.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
> Virgil schrieb:
> > In article
> > <ad9d7fe8-cfcc-4bdd...@b38g2000prf.googlegroups.com>,
> > Albrecht <albs...@gmx.de> wrote:
> >
> > > FredJeffries schrieb:
> > > > On Jan 25, 5:12?am, Albrecht <albst...@gmx.de> wrote:
> > > >
> > > >
> > > > >
> > > > > Consider a body which moves from point A to point B. Consider the
> > > > > barycentre Z (Schwerpunkt) of this body. Now the point Z, if it moves
> > > > > from A to B, at first has to move from the point A to a point which
> > > > > lays next to the point A. But there is no such next point.
> > > >
> > > > If there is no such point then Z does not have to move to it.
> > > >
> > > >
> > > >
> > > > > How is continuity possible?
> > > >
> > > > I thought that's what I was asking.
> > >
> > > And I was asking what the idea of actual infinity changes in this
> > > concern?
> >
> > If there are only a finite set of points in any interval, then velocity
> > and acceleration are impossible concepts.
>
> Nobody consider only finite many points in an interval. The
> alternative is not finite/infinite as you seem to imply.
"->NOT<- only finitely many" is a definition of "infinitely many", so
any extension beyond "only finite many" implies infinitely many.
>
> Your concept of infinity is wrong, that's the problem: There is no
> consistent cardinality of infinity.
My concept of "infinitely any" is "NOT only finitely many".
Albrecht
Mitch Harris schrieb:
> On Jan 29, 2:15 am, Albrecht <albst...@gmx.de> wrote:
> > Virgil schrieb:
> >
> > > If there are only a finite set of points in any interval, then velocity
> > > and acceleration are impossible concepts.
> >
> > Nobody consider only finite many points in an interval. The
> > alternative is not finite/infinite as you seem to imply.
> >
> > Your concept of infinity is wrong, that's the problem: There is no
> > consistent cardinality of infinity.
>
> We've all been using the simple term 'infinity'. Maybe the
> inconsistency you see is really the varieties of things that might be
> called infinity, each consistent in its own way.
>
> The following might give you a start:
>
> http://en.wikipedia.org/wiki/Cardinality
>
> http://en.wikipedia.org/wiki/Ordinal_number
>
> You're right, it seems counter-intuitive that the set of even integers
> is the same size (equinumerous, has the same number/cardinality) as
> all integers.
That's not the problem. The probem is, that there is no concept which
is able to describe coherently the quantity of infinite many objects
As the unary system
O
OO
OOO
OOOO
OOOOO
...
easily shows is there either a natural number which describes the
quantity of the natural numbers or there is no number which discribes
the quantity of the natural numbers since the natural numbers numbered
themselves. A number or a concept which describes a quantity which is
larger than any natural number is unavoidable too large to describe
the quantity of the natural numbers or any other manifold of "the same
size".
Infinity is no size and infinity is no quantity. Infinity is a mode, a
potentiality. The concept of infinity as an actual and complete entity
is wrong and indefensible.
Best regards
Albrecht S. Storz
Virgil
<216b567d-78f6-4e9f...@t39g2000prh.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
> Mitch Harris schrieb:
> > On Jan 29, 2:15 am, Albrecht <albst...@gmx.de> wrote:
> > > Virgil schrieb:
> > >
> > > > If there are only a finite set of points in any interval, then velocity
> > > > and acceleration are impossible concepts.
> > >
> > > Nobody consider only finite many points in an interval. The
> > > alternative is not finite/infinite as you seem to imply.
> > >
> > > Your concept of infinity is wrong, that's the problem: There is no
> > > consistent cardinality of infinity.
> >
> > We've all been using the simple term 'infinity'. Maybe the
> > inconsistency you see is really the varieties of things that might be
> > called infinity, each consistent in its own way.
> >
> > The following might give you a start:
> >
> > http://en.wikipedia.org/wiki/Cardinality
> >
> > http://en.wikipedia.org/wiki/Ordinal_number
> >
> > You're right, it seems counter-intuitive that the set of even integers
> > is the same size (equinumerous, has the same number/cardinality) as
> > all integers.
>
> That's not the problem. The probem is, that there is no concept which
> is able to describe coherently the quantity of infinite many objects
How about the concept "more that expressed by any natural number"?
It works for me.
Albrecht
Virgil schrieb:
You don't understand my considerations (you must not follow my
considerations in spite of understanding my considerations!; in
reverse, I understand your considerations, and I understand the way
how they lead to wrong results).
Again: Since the manifold of all natural numbers only contains natural
numbers, and since any natural number expresses exactly the number of
the foregoing natural numbers _inclusively the number itself_ the
quantity of the natural numbers can not go above the quantity which is
expressable by every/all of the natural numbers.
The natural numbers count themselves! Infinity is not larger than _all
natural number_, it is only "larger" than _any natural number_ ---
since any natural number has a successor and con not be the largest
one.
Brouwer had said: In Invinity the concept of the 'tertium non datur'
fails. That's right.
The concept of set theoretic cardinality leads to the illusion that
the natural numbers are more than they are. But the fundamental
concept of discrete quantity leads to the essential consequence that
"there is no quantity of infinity", there is no cardinality of the
manifold of the natural numbers, there is no set of all natural
numbers, there is only a proper class of all natural numbers. --- The
lucky news: in math does this fact changes almost nothing.
Han de Bruijn
Albrecht is expressing the (IMHO overly correct) idea that infinity is
not quite different from _very large_ (but let nobody ask _how_ large).
So he is looking at an initial segment of the naturals and he concludes
that "the" naturals should have NO properties different from properties
of an arbitrary (large) such initial segment. Seems reasonable enough.
> The natural numbers count themselves! Infinity is not larger than _all
> natural number_, it is only "larger" than _any natural number_ ---
> since any natural number has a successor and con not be the largest
> one.
>
> Brouwer had said: In Invinity the concept of the 'tertium non datur'
> fails. That's right.
>
> The concept of set theoretic cardinality leads to the illusion that
> the natural numbers are more than they are. But the fundamental
> concept of discrete quantity leads to the essential consequence that
> "there is no quantity of infinity", there is no cardinality of the
> manifold of the natural numbers, there is no set of all natural
> numbers, there is only a proper class of all natural numbers. --- The
> lucky news: in math does this fact changes almost nothing.
In "real" math, not "standard" mathematics, that is.
Han de Bruijn
Albrecht
Han de Bruijn schrieb:
No. Your interpretation of my arguments is not right.
If there is a "all of the natural numbers" they coud not be more than
some of them denote since the ordinal and the cardinal aspect of
numbers are intrinsic aligned in the natural numbers. The number n is
at the nth position. There is no natural number which is on a position
"outside of the positions the natural numbers are able to denote". And
there are not more numbers than there are positions.
Infinity as a fixed quantity is an antinomy.
Best regards
Albrecht S. Storz
>
Jesse F. Hughes
> Albrecht is expressing the (IMHO overly correct) idea that infinity is
> not quite different from _very large_ (but let nobody ask _how_ large).
>
> So he is looking at an initial segment of the naturals and he concludes
> that "the" naturals should have NO properties different from properties
> of an arbitrary (large) such initial segment. Seems reasonable
> enough.
Any arbitrarily large initial segment of naturals has the following
two properties.
(1) It is finite.
(2) It has a maximal element.
Thus, according to your own claim, we are met with the same two claims
for N.
(1) There are finitely many natural numbers.
(2) There is a maximal natural number.
Do you find those two claims reasonable?
--
"Do you know why I'm tall?" "Why?"
"Because I eat apples." "Do you know why I'm short?"
"Why?" "Because I'm a kid."
--Quincy P. Hughes (age almost 4) bests his father.
Mitch Harris
I think the word 'number' and 'natural number' is being used here
unfortunately in ways that make the concepts hard to follow. But that
is what the idea of infinity helps explicate.
What is a number? What is a natural number? What is infinity? What
does 'count' or 'how many' mean?
Zero is a natural number, right? (if that bothers you then start with
one.)
If you have a natural number, and you add 1 to it, that new thing is a
natural number, right?
What does it mean that two collections have the same size? It's been
agreed that the best way to do this is to put them into one-to-one
correspondence. (Notice I didn't mention number or natural number of
infinity)
'How many' means 'what is the size of' a collection.
But what is a 'size'? Natural numbers are a good possibility for size
(and they work well using elementary arithmetic).
Now what is the size of the set of natural numbers? That sentence
works syntactically, but it -is- hard to digest because we're not sure
what 'the set of natural numbers' really is.
First, back to the defintion of 'a' natural number, it is pretty
reasonable to accept that, given -a- natural number n (we don't know
what it is at all, just that it -is- some kind of natural number), we
can -always- get another natural number bigger than n (by adding one).
That is what is meant by 'unending' or 'infinite'. Given any natural
number we can always get another, so the process of getting another
will never end. (may never?). There's no stopping point.
If you start counting at 1, you can always give the size of a
(ahem...finite) set to be the natural number you stopped on when you
stopped tallying. Yes, that's circular.
But for the set of all natural numbers, since you don't stop creating
them, doesn't have a natural number to count them (since each natural
number -is- the stopping point for some set).
The whole point is that at this point, in trying to come up with a
'size' or a 'counting number' for the set of -all- natural numbers, we
punt. We give up, we do the time honored mathematical trick of saying,
I don't know what it is, it doesn't fit anything we've seen before, it
ain't a natural number, instead of burning it at the stake, let's name
it something weird and move on. The number...sorry, the -thing- that
we'll call the size of the set of all natural numbers, we'll give it
the name aleph_0.
That thing, aleph_0, has some properties in common with natural
numbers, but a lot that are not (aleph_0 + 1 = aleph_0 for example,
for appropriate understanding of addition). This aleph_0 is the
smallest -infinite- number...er...thing that acts like 'size of a
set'.
The word 'number' might be problematic because it has overloaded and,
separately, informal uses.
> A number or a concept which describes a quantity which is
> larger than any natural number is unavoidable too large to describe
> the quantity of the natural numbers or any other manifold of "the same
> size".
>
> Infinity is no size and infinity is no quantity. Infinity is a mode, a
> potentiality. The concept of infinity as an actual and complete entity
> is wrong and indefensible.
That may be so, but it is a useful and coherent concept. Frankly, the
naming of something like aleph_0 is really just the formalization of
the idea in the middle of the above explanation that, if you define
'natural number' inductively (as ''0, or a natural number plus 1),
then there's no stopping doing that. and that is an unending process,
resulting in an unending # of possibilities for natural numbers. Which
is what is called 'infinite'.
Mitch
Albrecht
Jesse F. Hughes schrieb:
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
> > Albrecht is expressing the (IMHO overly correct) idea that infinity is
> > not quite different from _very large_ (but let nobody ask _how_ large).
> >
> > So he is looking at an initial segment of the naturals and he concludes
> > that "the" naturals should have NO properties different from properties
> > of an arbitrary (large) such initial segment. Seems reasonable
> > enough.
>
> Any arbitrarily large initial segment of naturals has the following
> two properties.
>
> (1) It is finite.
>
> (2) It has a maximal element.
>
> Thus, according to your own claim, we are met with the same two claims
> for N.
>
> (1) There are finitely many natural numbers.
>
> (2) There is a maximal natural number.
>
> Do you find those two claims reasonable?
>
In my opinion: no. Correct is:
1.) there are infinitely many natural numbers what means: the natural
numbers go on and on without end
2.) there is no quantity or cardinality of all of them since the
concept "all of them" is inconsistent.
But you has asked Han, right?
Virgil
<84ac16fe-90f7-4edd...@e1g2000pra.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
So that the number of naturals is not a ->natural<- , a conclusion with
which everyone agrees.
> And
> there are not more numbers than there are positions.
There are not more naturals than positions, but there are things which
are not naturals
>
> Infinity as a fixed quantity is an antinomy.
Only those who claim exactly one such infinity are discombobulated by
that pronouncement.
For the rest of us, infiniteness is not fixed.
Albrecht
Okay. (When starting from 1).
>
> What does it mean that two collections have the same size? It's been
> agreed that the best way to do this is to put them into one-to-one
> correspondence. (Notice I didn't mention number or natural number of
> infinity)
>
> 'How many' means 'what is the size of' a collection.
Okay.
>
> But what is a 'size'? Natural numbers are a good possibility for size
> (and they work well using elementary arithmetic).
Cardinality is size in this context, I think.
>
> Now what is the size of the set of natural numbers? That sentence
> works syntactically, but it -is- hard to digest because we're not sure
> what 'the set of natural numbers' really is.
>
> First, back to the defintion of 'a' natural number, it is pretty
> reasonable to accept that, given -a- natural number n (we don't know
> what it is at all, just that it -is- some kind of natural number), we
> can -always- get another natural number bigger than n (by adding one).
Okay.
>
> That is what is meant by 'unending' or 'infinite'. Given any natural
> number we can always get another, so the process of getting another
> will never end. (may never?). There's no stopping point.
>
> If you start counting at 1, you can always give the size of a
> (ahem...finite) set to be the natural number you stopped on when you
> stopped tallying. Yes, that's circular.
Why circular? And what is the problem if so? The natural numbers are a
inevitable given property of our being. As e.g. time, space,
structure, causality, ... They are the roots of any math,
unchangeable, perfect, good to use, ubiquitous, ...
>
> But for the set of all natural numbers, since you don't stop creating
> them, doesn't have a natural number to count them (since each natural
> number -is- the stopping point for some set).
No. And you don't have any other number or any well-defined concept
for their quantity.
>
> The whole point is that at this point, in trying to come up with a
> 'size' or a 'counting number' for the set of -all- natural numbers, we
> punt. We give up, we do the time honored mathematical trick of saying,
> I don't know what it is, it doesn't fit anything we've seen before, it
> ain't a natural number, instead of burning it at the stake, let's name
> it something weird and move on. The number...sorry, the -thing- that
> we'll call the size of the set of all natural numbers, we'll give it
> the name aleph_0.
You can't do like this. Aleph_0 is defined as being larger than any
natural number. And so it is too large to denote the quantity of all
natural numbers since they count themseves. There could not be
something out of them to count them because they count themselve and
don't go beyond them at any stage.
>
> That thing, aleph_0, has some properties in common with natural
> numbers, but a lot that are not (aleph_0 + 1 = aleph_0 for example,
> for appropriate understanding of addition). This aleph_0 is the
> smallest -infinite- number...er...thing that acts like 'size of a
> set'.
>
> The word 'number' might be problematic because it has overloaded and,
> separately, informal uses.
Is "quantity" better for you?
>
>
> > A number or a concept which describes a quantity which is
> > larger than any natural number is unavoidable too large to describe
> > the quantity of the natural numbers or any other manifold of "the same
> > size".
> >
> > Infinity is no size and infinity is no quantity. Infinity is a mode, a
> > potentiality. The concept of infinity as an actual and complete entity
> > is wrong and indefensible.
>
> That may be so, but it is a useful and coherent concept. Frankly, the
> naming of something like aleph_0 is really just the formalization of
> the idea in the middle of the above explanation that, if you define
> 'natural number' inductively (as ''0, or a natural number plus 1),
> then there's no stopping doing that. and that is an unending process,
> resulting in an unending # of possibilities for natural numbers. Which
> is what is called 'infinite'.
>
Aleph_0 is not just a name. If so, why don't use just "Infinity"? But
Infinity is no quantity and the talking about the quantity of all
natural numbers is just a talking " as if". A "facon de parler" as
someone (Gauss?) has said.
Virgil
<cbe2a148-5a0d-47be...@o40g2000prn.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
Not according to any logic with an excluded middle.
So that Albrecht must be claimng "Tertium Datur".
David R Tribble
>> The whole point is that at this point, in trying to come up with a
>> 'size' or a 'counting number' for the set of -all- natural numbers, we
>> punt. We give up, we do the time honored mathematical trick of saying,
>> I don't know what it is, it doesn't fit anything we've seen before, it
>> ain't a natural number, instead of burning it at the stake, let's name
>> it something weird and move on. The number...sorry, the -thing- that
>> we'll call the size of the set of all natural numbers, we'll give it
>> the name aleph_0.
>
Albrecht wrote:
> You can't do like this. Aleph_0 is defined as being larger than any
> natural number. And so it is too large to denote the quantity of all
> natural numbers since they count themseves. There could not be
> something out of them to count them because they count themselve and
> don't go beyond them at any stage.
Not at any /finite/ stage in the counting, of course.
And, of course, at each stage where you have counted some
number of naturals, you have a lot more naturals left that you
have /not/ counted yet. Do you have a way of describing how
many of them are left to count?
Virgil
<63f2f328-2736-416d...@a39g2000prl.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
How much too large? Let us subtract the difference and get an exact
measure.
> There could not be
> something out of them to count them because they count themselve and
> don't go beyond them at any stage.
A good reason for using the Von Neumann model which does not have that
problem.
>
> >
> > That thing, aleph_0, has some properties in common with natural
> > numbers, but a lot that are not (aleph_0 + 1 = aleph_0 for example,
> > for appropriate understanding of addition). This aleph_0 is the
> > smallest -infinite- number...er...thing that acts like 'size of a
> > set'.
> >
> > The word 'number' might be problematic because it has overloaded and,
> > separately, informal uses.
>
> Is "quantity" better for you?
Then there are natural "quantities" and non-natural "quantities", and
the quantity of all naturals quantities is not a natural quantity.
>
> >
> Aleph_0 is not just a name. If so, why don't use just "Infinity"?
Among other reasons, there are different infinities, so one must
specify which infinity. Aleph_0 is one of them.
> But
> Infinity is no quantity
But infiniteness encompasses many quantities.
> and the talking about the quantity of all
> natural numbers is just a talking " as if".
Does Albrecht mean that there is no boundary between what are naturals
and what are not?
> A "facon de parler" as
> someone (Gauss?) has said.
Gauss would probably have said it in German.
Mitch Harris
> > A "facon de parler" as someone (Gauss?) has said.
>
> Gauss would probably have said it in German.
They were so cultured back then...
"Das Unendliche ist nur eine Facon de parler"
--
Mitch
Mitch Harris
> Mitch Harris schrieb:
> > On Jan 30, 2:35 am, Albrecht <albst...@gmx.de> wrote:
> > > Mitch Harris schrieb:
>
> > > > On Jan 29, 2:15 am, Albrecht <albst...@gmx.de> wrote:
> > > > > Virgil schrieb:
> > That is what is meant by 'unending' or 'infinite'. Given any natural
> > number we can always get another, so the process of getting another
> > will never end. (may never?). There's no stopping point.
>
> > If you start counting at 1, you can always give the size of a
> > (ahem...finite) set to be the natural number you stopped on when you
> > stopped tallying. Yes, that's circular.
>
> Why circular?
Well, counting is usually defined with respect to the natural numbers,
and I was trying to explain how to count natural numbers.
> And what is the problem if so? The natural numbers are a
> inevitable given property of our being. As e.g. time, space,
> structure, causality, ... They are the roots of any math,
> unchangeable, perfect, good to use, ubiquitous, ...
Uh...neat. But that doesn't help explain anything.
> > But for the set of all natural numbers, since you don't stop creating
> > them, doesn't have a natural number to count them (since each natural
> > number -is- the stopping point for some set).
>
> No. And you don't have any other number or any well-defined concept
> for their quantity.
aleph_0 is pretty well-defined. weird, yes, but still well-defined.
> > The whole point is that at this point, in trying to come up with a
> > 'size' or a 'counting number' for the set of -all- natural numbers, we
> > punt. We give up, we do the time honored mathematical trick of saying,
> > I don't know what it is, it doesn't fit anything we've seen before, it
> > ain't a natural number, instead of burning it at the stake, let's name
> > it something weird and move on. The number...sorry, the -thing- that
> > we'll call the size of the set of all natural numbers, we'll give it
> > the name aleph_0.
>
> You can't do like this. Aleph_0 is defined as being larger than any
> natural number. And so it is too large to denote the quantity of all
> natural numbers since they count themseves.
How do the natural numbers count themselves? Which natural number is
the 'size' of the natural numbers? Remember there is no 'last' natural
number.
> There could not be
> something out of them to count them because they count themselve and
> don't go beyond them at any stage.
There's no natural number that gives you the 'size' of the natural
numbers. Right. So if you want to manipulate the concept 'the size of
the natural numbers' then you have to create a label that stands for
something that is not a natural number, yet still gives their size.
Let's call it infinity or aleph_0 (or something).
> > The word 'number' might be problematic because it has overloaded and,
> > separately, informal uses.
>
> Is "quantity" better for you?
Sure. 'size' is another.
> Aleph_0 is not just a name. If so, why don't use just "Infinity"?
There are other things that are not finite but act differently from
aleph_0. For example, the 'size' of the continuum (the 'number' of
'points' between 0 and 1).
Or the ordinal for the univariate polyomials with integer
coefficients.
> But
> Infinity is no quantity and the talking about the quantity of all
> natural numbers is just a talking " as if". A "facon de parler" as
> someone (Gauss?) has said.
Talking about 2 is a manner of speaking also.
--
Mitch
Albrecht
David R Tribble schrieb:
> Mitch Harris wrote:
> >> The whole point is that at this point, in trying to come up with a
> >> 'size' or a 'counting number' for the set of -all- natural numbers, we
> >> punt. We give up, we do the time honored mathematical trick of saying,
> >> I don't know what it is, it doesn't fit anything we've seen before, it
> >> ain't a natural number, instead of burning it at the stake, let's name
> >> it something weird and move on. The number...sorry, the -thing- that
> >> we'll call the size of the set of all natural numbers, we'll give it
> >> the name aleph_0.
> >
>
> Albrecht wrote:
> > You can't do like this. Aleph_0 is defined as being larger than any
> > natural number. And so it is too large to denote the quantity of all
> > natural numbers since they count themseves. There could not be
> > something out of them to count them because they count themselve and
> > don't go beyond them at any stage.
>
> Not at any /finite/ stage in the counting, of course.
There is no infinite stage anywhere.
>
> And, of course, at each stage where you have counted some
> number of naturals, you have a lot more naturals left that you
> have /not/ counted yet. Do you have a way of describing how
> many of them are left to count?
We have a word for that, but no quantity. Infinite is just a mode, not
a number, not a quantity. Infinity is a concept at the boundary of our
understanding. It is a helpless attempt to understand the not-
understandable.
Albrecht
Virgil schrieb:
There is no infinite quantity. But if there would be one, this
infinite quantity would be too large to denote the quantity of the
naturals.
>
>
> > There could not be
> > something out of them to count them because they count themselve and
> > don't go beyond them at any stage.
>
> A good reason for using the Von Neumann model which does not have that
> problem.
They have exactly that problem. But with them the problem is much more
veiled.
> >
> > >
> > > That thing, aleph_0, has some properties in common with natural
> > > numbers, but a lot that are not (aleph_0 + 1 = aleph_0 for example,
> > > for appropriate understanding of addition). This aleph_0 is the
> > > smallest -infinite- number...er...thing that acts like 'size of a
> > > set'.
> > >
> > > The word 'number' might be problematic because it has overloaded and,
> > > separately, informal uses.
> >
> > Is "quantity" better for you?
>
> Then there are natural "quantities" and non-natural "quantities", and
> the quantity of all naturals quantities is not a natural quantity.
> >
> > >
>
> > Aleph_0 is not just a name. If so, why don't use just "Infinity"?
>
> Among other reasons, there are different infinities, so one must
> specify which infinity. Aleph_0 is one of them.
>
> > But
> > Infinity is no quantity
>
> But infiniteness encompasses many quantities.
Senseless wording?
Albrecht
Mitch Harris schrieb:
> On Jan 30, 4:34 pm, Albrecht <albst...@gmx.de> wrote:
> > Mitch Harris schrieb:
> > > On Jan 30, 2:35 am, Albrecht <albst...@gmx.de> wrote:
> > > > Mitch Harris schrieb:
> >
> > > > > On Jan 29, 2:15 am, Albrecht <albst...@gmx.de> wrote:
> > > > > > Virgil schrieb:
>
>
> > > That is what is meant by 'unending' or 'infinite'. Given any natural
> > > number we can always get another, so the process of getting another
> > > will never end. (may never?). There's no stopping point.
> >
> > > If you start counting at 1, you can always give the size of a
> > > (ahem...finite) set to be the natural number you stopped on when you
> > > stopped tallying. Yes, that's circular.
> >
> > Why circular?
>
> Well, counting is usually defined with respect to the natural numbers,
> and I was trying to explain how to count natural numbers.
Natural numbers will be counted exactly as any other manifolds:
1.: 1, 2.: 2, 3.: 3, 4.: 4, ...
>
> > And what is the problem if so? The natural numbers are a
> > inevitable given property of our being. As e.g. time, space,
> > structure, causality, ... They are the roots of any math,
> > unchangeable, perfect, good to use, ubiquitous, ...
>
> Uh...neat. But that doesn't help explain anything.
Neat? No. Rigorous!
>
>
> > > But for the set of all natural numbers, since you don't stop creating
> > > them, doesn't have a natural number to count them (since each natural
> > > number -is- the stopping point for some set).
> >
> > No. And you don't have any other number or any well-defined concept
> > for their quantity.
>
> aleph_0 is pretty well-defined. weird, yes, but still well-defined.
Well-definde in spite of the fact that the concept is antinome?
>
>
> > > The whole point is that at this point, in trying to come up with a
> > > 'size' or a 'counting number' for the set of -all- natural numbers, we
> > > punt. We give up, we do the time honored mathematical trick of saying,
> > > I don't know what it is, it doesn't fit anything we've seen before, it
> > > ain't a natural number, instead of burning it at the stake, let's name
> > > it something weird and move on. The number...sorry, the -thing- that
> > > we'll call the size of the set of all natural numbers, we'll give it
> > > the name aleph_0.
> >
> > You can't do like this. Aleph_0 is defined as being larger than any
> > natural number. And so it is too large to denote the quantity of all
> > natural numbers since they count themseves.
>
> How do the natural numbers count themselves? Which natural number is
> the 'size' of the natural numbers? Remember there is no 'last' natural
> number.
There is no coherent concept of the quantity of the naturals. Yet
forgotten?
>
> > There could not be
> > something out of them to count them because they count themselve and
> > don't go beyond them at any stage.
>
> There's no natural number that gives you the 'size' of the natural
> numbers. Right. So if you want to manipulate the concept 'the size of
> the natural numbers' then you have to create a label that stands for
> something that is not a natural number, yet still gives their size.
> Let's call it infinity or aleph_0 (or something).
Any concept of a quantity of the naturals is wrong since either it is
too small (a natural number) or it is too large (something which is
larger than any natural number). So the concept of a quantity of the
natural numbers is antinome.
>
>
> > > The word 'number' might be problematic because it has overloaded and,
> > > separately, informal uses.
> >
> > Is "quantity" better for you?
>
> Sure. 'size' is another.
>
>
> > Aleph_0 is not just a name. If so, why don't use just "Infinity"?
>
> There are other things that are not finite but act differently from
> aleph_0. For example, the 'size' of the continuum (the 'number' of
> 'points' between 0 and 1).
> Or the ordinal for the univariate polyomials with integer
> coefficients.
This concepts are all senseless since there is no infinite set.
>
> > But
> > Infinity is no quantity and the talking about the quantity of all
> > natural numbers is just a talking " as if". A "facon de parler" as
> > someone (Gauss?) has said.
>
> Talking about 2 is a manner of speaking also.
>
> --
Arguing with someone who claims that the concept of 2 and the concept
of infinity is the same seems senseless to me.
Virgil
<7d89c10d-3981-4935...@w24g2000prd.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
> David R Tribble schrieb:
> > Mitch Harris wrote:
> > >> The whole point is that at this point, in trying to come up with a
> > >> 'size' or a 'counting number' for the set of -all- natural numbers, we
> > >> punt. We give up, we do the time honored mathematical trick of saying,
> > >> I don't know what it is, it doesn't fit anything we've seen before, it
> > >> ain't a natural number, instead of burning it at the stake, let's name
> > >> it something weird and move on. The number...sorry, the -thing- that
> > >> we'll call the size of the set of all natural numbers, we'll give it
> > >> the name aleph_0.
> > >
> >
> > Albrecht wrote:
> > > You can't do like this. Aleph_0 is defined as being larger than any
> > > natural number. And so it is too large to denote the quantity of all
> > > natural numbers since they count themseves.
Which natural does Aleph_0 count twice or what does it count that is
not a natural? Unless you can answer this, you are dead wrong.
> > > There could not be
> > > something out of them to count them because they count themselve and
> > > don't go beyond them at any stage.
Then do you claim that there is some natural that counts all naturals?
Unless you do, you cannot say that they count themselves, but only that
one of them counts one of them.
> >
> > Not at any /finite/ stage in the counting, of course.
>
> There is no infinite stage anywhere.
Perhaps not in Albrecht's philosophy, but we can all see how little that
is worth.
>
> >
> > And, of course, at each stage where you have counted some
> > number of naturals, you have a lot more naturals left that you
> > have /not/ counted yet. Do you have a way of describing how
> > many of them are left to count?
>
> We have a word for that, but no quantity. Infinite is just a mode, not
> a number, not a quantity. Infinity is a concept at the boundary of our
> understanding. It is a helpless attempt to understand the not-
> understandable.
That something is not understandable by Albrecht only makes it not
understandable by Albrecht, but in no way makes it necessarily not
understandable in general.
Virgil
> David R Tribble schrieb:
> > Mitch Harris wrote:
> > >> The whole point is that at this point, in trying to come up with a
> > >> 'size' or a 'counting number' for the set of -all- natural numbers, we
> > >> punt. We give up, we do the time honored mathematical trick of saying,
> > >> I don't know what it is, it doesn't fit anything we've seen before, it
> > >> ain't a natural number, instead of burning it at the stake, let's name
> > >> it something weird and move on. The number...sorry, the -thing- that
> > >> we'll call the size of the set of all natural numbers, we'll give it
> > >> the name aleph_0.
> > >
> >
> > Albrecht wrote:
> > > You can't do like this. Aleph_0 is defined as being larger than any
> > > natural number. And so it is too large to denote the quantity of all
> > > natural numbers since they count themseves. There could not be
> > > something out of them to count them because they count themselve and
> > > don't go beyond them at any stage.
> >
> > Not at any /finite/ stage in the counting, of course.
>
> There is no infinite stage anywhere.
There is in ZFC.
>
> >
> > And, of course, at each stage where you have counted some
> > number of naturals, you have a lot more naturals left that you
> > have /not/ counted yet. Do you have a way of describing how
> > many of them are left to count?
>
> We have a word for that, but no quantity.
We, on the other hand, have both, the quantity being Aleph_0, which is
as much a quantity as any natural
> Infinite is just a mode, not
> a number, not a quantity.
So is "natural number", but particular examples of each are numbers.
> Infinity is a concept at the boundary of our
> understanding. It is a helpless attempt to understand the not-
> understandable.
You speak only for yourself, Albrecht. There are others with better
understanding.
Virgil
<7ec2d158-2f7c-44db...@e1g2000pra.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
> Virgil schrieb:
> > In article
> > <63f2f328-2736-416d...@a39g2000prl.googlegroups.com>,
> > Albrecht <albs...@gmx.de> wrote:
> >
....
> > > > But for the set of all natural numbers, since you don't stop creating
> > > > them, doesn't have a natural number to count them (since each natural
> > > > number -is- the stopping point for some set).
> > >
> > > No. And you don't have any other number or any well-defined concept
> > > for their quantity.
Storz may not but others do.
> >
> > > You can't do like this.
We HAVE done it.
> > > Aleph_0 is defined as being larger than any
> > > natural number. And so it is too large to denote the quantity of all
> > > natural numbers since they count themseves.
No natural "counts" any of its successors so that the naturals do NOT
count themselves in any reasonable sense.
> >
> > How much too large? Let us subtract the difference and get an exact
> > measure.
>
> There is no infinite quantity. But if there would be one, this
> infinite quantity would be too large to denote the quantity of the
> naturals.
Repeating a falsehood does not make it any less false.
>
> >
> >
> > > There could not be
> > > something out of them to count them because they count themselve and
> > > don't go beyond them at any stage.
> >
> > A good reason for using the Von Neumann model which does not have that
> > problem.
>
> They have exactly that problem. But with them the problem is much more
> veiled.
Neither version has any such problem as no natural in either system can
"count" any of its successors, so that the naturals cannot "count
themselves".
> > Among other reasons, there are different infinities, so one must
> > specify which infinity. Aleph_0 is one of them.
> >
> > > But
> > > Infinity is no quantity
> >
> > But infiniteness encompasses many quantities.
>
> Senseless wording?
Only to the senseless.
>
> >
> > > and the talking about the quantity of all
> > > natural numbers is just a talking " as if".
> >
> > Does Albrecht mean that there is no boundary between what are naturals
> > and what are not?
Can't find an answer?
Virgil
<cb926a08-a34d-4aa2...@y23g2000pre.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
> Mitch Harris schrieb:
> > On Jan 30, 4:34 pm, Albrecht <albst...@gmx.de> wrote:
> > > Mitch Harris schrieb:
> > > > On Jan 30, 2:35 am, Albrecht <albst...@gmx.de> wrote:
> > > > > Mitch Harris schrieb:
> > >
> > > > > > On Jan 29, 2:15 am, Albrecht <albst...@gmx.de> wrote:
> > > > > > > Virgil schrieb:
> >
> >
> > > > That is what is meant by 'unending' or 'infinite'. Given any natural
> > > > number we can always get another, so the process of getting another
> > > > will never end. (may never?). There's no stopping point.
> > >
> > > > If you start counting at 1, you can always give the size of a
> > > > (ahem...finite) set to be the natural number you stopped on when you
> > > > stopped tallying. Yes, that's circular.
> > >
> > > Why circular?
> >
> > Well, counting is usually defined with respect to the natural numbers,
> > and I was trying to explain how to count natural numbers.
>
> Natural numbers will be counted exactly as any other manifolds:
>
> 1.: 1, 2.: 2, 3.: 3, 4.: 4, ...
Cardinality necessarily starts with zero.
Ordinality starts with "first".
>
>
> >
> > > And what is the problem if so? The natural numbers are a
> > > inevitable given property of our being. As e.g. time, space,
> > > structure, causality, ... They are the roots of any math,
> > > unchangeable, perfect, good to use, ubiquitous, ...
> >
> > Uh...neat. But that doesn't help explain anything.
>
> Neat? No. Rigorous!
Not rigorous as described by Storz.
>
> >
> >
> > > > But for the set of all natural numbers, since you don't stop creating
> > > > them, doesn't have a natural number to count them (since each natural
> > > > number -is- the stopping point for some set).
> > >
> > > No. And you don't have any other number or any well-defined concept
> > > for their quantity.
> >
> > aleph_0 is pretty well-defined. weird, yes, but still well-defined.
>
> Well-definde in spite of the fact that the concept is antinome?
Since that "antinome" does not exist for most of us, aleph_0 is
well-defined for all but that unhappy antinomous few.
> >
> > How do the natural numbers count themselves? Which natural number is
> > the 'size' of the natural numbers? Remember there is no 'last' natural
> > number.
>
>
> There is no coherent concept of the quantity of the naturals. Yet
> forgotten?
Then they do not count themselves. In particular, none may "count" any
of its infinitely many successors. so that no natural may count more
than a miniscule minority of naturals.
>
> Any concept of a quantity of the naturals is wrong
Only to those who assume a priori that it is wrong.
> >
> > > Aleph_0 is not just a name. If so, why don't use just "Infinity"?
> >
> > There are other things that are not finite but act differently from
> > aleph_0. For example, the 'size' of the continuum (the 'number' of
> > 'points' between 0 and 1).
> > Or the ordinal for the univariate polyomials with integer
> > coefficients.
>
> This concepts are all senseless since there is no infinite set.
The senselessness is all in Albrecht, who rejects what he is too thick
to comprehend.
>
> Arguing with someone who claims that the concept of 2 and the concept
> of infinity is the same seems senseless to me.
Then stop doing it. only the terminally stupid kdeep doing what they
believe to be senseless.
We, on the other hand, while knowing that it is useless to persuade
sense into Storz, keep posting to prevent Storz from infecting others.
Mitch Harris
> > > Mitch Harris schrieb:
> > > > If you start counting at 1, you can always give the size of a
> > > > (ahem...finite) set to be the natural number you stopped on when you
> > > > stopped tallying. Yes, that's circular.
>
> > > Why circular?
>
> > Well, counting is usually defined with respect to the natural numbers,
> > and I was trying to explain how to count natural numbers.
>
> Natural numbers will be counted exactly as any other manifolds:
>
> 1.: 1, 2.: 2, 3.: 3, 4.: 4, ...
I think that's the circularity I'm thinking of.
> > > And what is the problem if so? The natural numbers are a
> > > inevitable given property of our being. As e.g. time, space,
> > > structure, causality, ... They are the roots of any math,
> > > unchangeable, perfect, good to use, ubiquitous, ...
>
> > Uh...neat. But that doesn't help explain anything.
>
> Neat? No. Rigorous!
Rigorous? A repetition of adjectives is considered reasoning? OK, but
it is not very convincing reasoning to me.
> > aleph_0 is pretty well-defined. weird, yes, but still well-defined.
>
> Well-definde in spite of the fact that the concept is antinome?
Nope just well-defined. You think it is whatever 'antnome' means, I
don't.
> > How do the natural numbers count themselves? Which natural number is
> > the 'size' of the natural numbers? Remember there is no 'last' natural
> > number.
>
> There is no coherent concept of the quantity of the naturals. Yet
> forgotten?
I don't think you've established that by any reasoning. I think I have
established it by explanation. Yes, I must have already forgotten
that. What explanation did you offer? (link,summary?)
> > There's no natural number that gives you the 'size' of the natural
> > numbers. Right. So if you want to manipulate the concept 'the size of
> > the natural numbers' then you have to create a label that stands for
> > something that is not a natural number, yet still gives their size.
> > Let's call it infinity or aleph_0 (or something).
>
> Any concept of a quantity of the naturals is wrong since either it is
> too small (a natural number)
Right. the quantity of the natural numbers -cannot- be a natural
number. Excellent, we've mutually established that.
> or it is too large (something which is
> larger than any natural number).
Hm. Yes, the quantity of natural numbers is larger than any natural
number. Why is that wrong?
> So the concept of a quantity of the
> natural numbers is antinome.
What is 'antinome'? Is that 'die Antinomie'? A paradox? A
contradiction?
If a contradiction, one of those possibilities works. It's not the
first one. You haven't shown how the second one is wrong.
> > > Aleph_0 is not just a name. If so, why don't use just "Infinity"?
>
> > There are other things that are not finite but act differently from
> > aleph_0. For example, the 'size' of the continuum (the 'number' of
> > 'points' between 0 and 1).
> > Or the ordinal for the univariate polyomials with integer
> > coefficients.
>
> This concepts are all senseless since there is no infinite set.
- 'is'? There is no 2 either.
- They make sense to some people. And they have some useful
consequences. Could we get those useful consequences without bothering
with the 'nonsense' of infinity? Maybe, but I'll let you do that work.
> > > But
> > > Infinity is no quantity and the talking about the quantity of all
> > > natural numbers is just a talking " as if". A "facon de parler" as
> > > someone (Gauss?) has said.
>
> > Talking about 2 is a manner of speaking also.
>
> Arguing with someone who claims that the concept of 2 and the concept
> of infinity is the same seems senseless to me.
- I don't think I called them the same. I pointed out one similarity.
- "Arguing with someone who claims" ... "seems senseless to me".
Touche
Mitch
David Formosa (aka ? the Platypus)
[...]
> 2.) there is no quantity or cardinality of all of them since the
> concept "all of them" is inconsistent.
Using the rules of standard mathmatics and logic please derive the statement
A = ~A from the concept "all of them".
LauLuna
At every finite stage in the generation of naturals, we find a natural
that tells us how many naturals have been produced SO FAR. But 'so
far' is never 'in the end', concerning naturals.
Now, there is no reason to assume that what holds for the finite also
holds for the infinite. There might be no natural counting all natural
numbers. And the fact that every finite amount of naturals is counted
by a natural is no good argument against the existence of a nonnatural
number counting all naturals.
I do think infinity is no cardinality but your way of arguing it, is
not the best.
There is no inconsistency in admitting infinite cardinalities, most
probably because it is ultimately a question of how we name things;
you can restrict or expand what you understand by a cardinality at
will.
Nevertheless, there are consistent impossibilities, impossibilities
that do not imply contradiction (in the sense that no contradiction is
a logical consequence of them). Consider the negation ~G of Gödel's
sentence G, or consider 'something pops out of absolute nothingness'.
As I see it, these are consistent impossibilities.
So perhaps, looking for an inconsistency in the theory of the
transfinite is not the best way to disprove it.
Regards
> A number or a concept which describes a quantity which is
> larger than any natural number is unavoidable too large to describe
> the quantity of the natural numbers or any other manifold of "the same
> size".
>
> Infinity is no size and infinity is no quantity. Infinity is a mode, a
> potentiality. The concept of infinity as an actual and complete entity
> is wrong and indefensible.
>
> Best regards
> Albrecht S. Storz
>
>
>
> > But mathematical investigation has given consistent
> > meanings to all these terms. If you learn the meanings of those terms,
> > you'll see that things work out nicely. Those meanings may not match
> > the ones you have now though.- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
LauLuna
<dform...@usyd.edu.au> wrote:
> On Fri, 30 Jan 2009 13:16:22 -0800 (PST), Albrecht <albst...@gmx.de> wrote:
>
> [...]
>
> > 2.) there is no quantity or cardinality of all of them since the
> > concept "all of them" is inconsistent.
>
> Using the rules of standard mathmatics and logic please derive the statement
> A = ~A from the concept "all of them".
Yes, that's a good point. Most likely, there is no inconsistency in
that concept.
Nevertheless, don't forget there are consistent impossibilities.
Jesse F. Hughes
> Nevertheless, don't forget there are consistent impossibilities.
If you are not speaking of logical impossibilities nor of physical
impossibilities, then what kind do you have in mind?
--
Jesse F. Hughes
"I mean it's just kind of like ... whatever ... I'm here for a
purpose. I know what my purpose is. I'm not a ... moron, you know
what I mean." -- "Kristen", Gov. Spitzer's philosopher/"escort"
Albrecht
LauLuna schrieb:
1.) I'm convinced that it is possible to create a stringent theory of
the discrete quantities (the natural numbers) with at least the same
strenght as there is a theory of sets like ZFC (respectively theories
of sets, e.g. ZFC, NBG, ...). Up to now Math lacks of such a theory.
Set theories are not the most fundamental theories of Math. Discrete
quantities (natural numbers) are more fundamental than sets. Natural
numbers are the essential roots of any mathematics, inclusively of any
set theories.
2.) To discuss the problem of infinite quantities we have to consider
the relation "larger than" (>).
The set theoretic definiton is an "artificial" definition which don't
satisfy our intuitive understanding of that relation.
I claim that the definition of "larger than" of discrete quantities
has to cover the condition:
If a > b than a = b+1 or a > b+1
or, for sure, in another notation:
a > b <=> a = b+1 or a > b+1
I hope the notation is clear.
Some may claim that this notation is circular, but we are satisfied
with the condition that a has to be at least one unity larger than b
to be larger than b (thus a > b holds).
With this considerations, if someone claims that the quantity of the
natural numbers (let's call it A*) is larger than any natural number,
he has to show that:
A* > n forall n in N (with N is the proper class of the natural
numbers) is true.
Thus: The quantity of the natural numbers has to be at least one unity
larger than any natural number.
The axiom of Infinity implies this claim since infinite sets has to
have a cardinality and cardinality is the quantity of the elements of
a set.
This claim leads to two contradictory sentences:
(1) If the cardinality (or quantity) of the natural numbers is at
least one unity larger than any natural number, there has to be at
least one element more in the manifold of the natural numbers as there
are natural numbers.
But (2) The manifold of natural numbers consists only of natural
numbers (by definition).
This facts are easely apparent with the unary notation of the natural
numbers:
O
OO
OOO
OOOO
OOOOO
...
or with the notation of the ordianl and cardinal oneness of the
natural numbers:
1. <-> 1
1. 2. <-> 2
1. 2. 3. <-> 3
1. 2. 3. 4. <-> 4
1. 2. 3. 4. 5. <-> 5
...
The pretence of the soundness of a quantity of infinity disappears
with putting down the set-theoretical glasses.
WM
> On Jan 30, 8:35 am, Albrecht <albst...@gmx.de> wrote:
>
> > As the unary system
>
> > O
> > OO
> > OOO
> > OOOO
> > OOOOO
> > ...
>
> > easily shows is there either a natural number which describes the
> > quantity of the natural numbers or there is no number which discribes
> > the quantity of the natural numbers since the natural numbers numbered
> > themselves
>
> At every finite stage in the generation of naturals, we find a natural
> that tells us how many naturals have been produced SO FAR. But 'so
> far' is never 'in the end', concerning naturals.
>
> Now, there is no reason to assume that what holds for the finite also
> holds for the infinite.
But every natural number is finite. There is no infinite natural
number. So there is no reason why the well proved finite laws should
not apply here.
There is a simply proof to show that they do: The sequence of all
natural numbers cannot be larger than a natural number (of course we
cannot name that number. Therefore let us call it ?). The proof is as
follows:
Theorem: For all n in N: the initial segment {1, 2, 3, ..., n} is
finite and has cardinality n.
Proof by induction. For "all" n.
So, if there should be an actually infinite set (different from a
potentially infinite set that is always finite but not fixed), then it
is forced to contain elements that are not subject of my theorem. But
that would be non-natural numbers, as my theorem covers all natural
numbers.
Can you really imagine a reason why this theorem should not hold?
Further we should analyze the concept of "all". What does it mean?
1) If you consider a number, then you can tell whether it is a
natural. That is ok, but that does not say that all naturals "are
there".
2) Actually, all naturals are there. Then the concept of "all" has to
be proved by showing all - not of course by the ridiculous three
points or by the even more ridiculous "axiom".
Regards, WM
Virgil
<7996c514-6675-4635...@e1g2000pra.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
>
> 1.) I'm convinced that it is possible to create a stringent theory of
> the discrete quantities (the natural numbers) with at least the same
> strenght as there is a theory of sets like ZFC (respectively theories
> of sets, e.g. ZFC, NBG, ...).
Then go away until you have found it
> Set theories are not the most fundamental theories of Math.
At least in Storz's opinion. There is, as yet, no conclusive evidence
that any there is any other theory more basic, which certainly leaves
set theory in the running for "most basic".
> Discrete
> quantities (natural numbers) are more fundamental than sets.
Since Storz's "discrete quantities" can be generated within set theory,
but not vice-versa, it would appear that sets are more basic.
> Natural
> numbers are the essential roots of any mathematics, inclusively of any
> set theories.
A considerable amount of geometry can be done without a single natural
number.
> 2.) To discuss the problem of infinite quantities we have to consider
> the relation "larger than" (>).
> The set theoretic definiton is an "artificial" definition which don't
> satisfy our intuitive understanding of that relation.
The set-theoretic definition defines such strict order relations as
being asymmetric, irreflexive and transitive. Which of those properties
does not satisfy Albrecht's "inuitive understanding" of that relation?
> I claim that the definition of "larger than" of discrete quantities
> has to cover the condition:
> If a > b than a = b+1 or a > b+1
> or, for sure, in another notation:
> a > b <=> a = b+1 or a > b+1
Since Albrecht has as yet defined neither "discrete quantities" nor
"a+1" where "a" is a discrete quantity, his alleged definition sucks.
> With this considerations, if someone claims that the quantity of the
> natural numbers (let's call it A*) is larger than any natural number,
> he has to show that:
> A* > n forall n in N (with N is the proper class of the natural
> numbers) is true.
>
> Thus: The quantity of the natural numbers has to be at least one unity
> larger than any natural number.
> The axiom of Infinity implies this claim since infinite sets has to
> have a cardinality and cardinality is the quantity of the elements of
> a set.
>
> This claim leads to two contradictory sentences:
>
> (1) If the cardinality (or quantity) of the natural numbers is at
> least one unity larger than any natural number, there has to be at
> least one element more in the manifold of the natural numbers as there
> are natural numbers.
Non sequitur. What is the case is that there is at least one more in the
"quantity" than up to any single natural.
> But (2) The manifold of natural numbers consists only of natural
> numbers (by definition).
But is not "numbered" by any one of its members.
[too often repeated nonsense snipped]
Virgil
<6a2d45e2-7733-4a42...@i20g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 2 Feb., 16:01, LauLuna <laureanol...@yahoo.es> wrote:
> > On Jan 30, 8:35 am, Albrecht <albst...@gmx.de> wrote:
> >
> > > As the unary system
> >
> > > O
> > > OO
> > > OOO
> > > OOOO
> > > OOOOO
> > > ...
> >
> > > easily shows is there either a natural number which describes the
> > > quantity of the natural numbers or there is no number which discribes
> > > the quantity of the natural numbers since the natural numbers numbered
> > > themselves
> >
> > At every finite stage in the generation of naturals, we find a natural
> > that tells us how many naturals have been produced SO FAR. But 'so
> > far' is never 'in the end', concerning naturals.
> >
> > Now, there is no reason to assume that what holds for the finite also
> > holds for the infinite.
>
> But every natural number is finite. There is no infinite natural
> number. So there is no reason why the well proved finite laws should
> not apply here.
One of the "well proved finite laws" says that for any finite
->ordered<- assembly of objects, there is a largest.
From which those with any sense may conclude that unless there is a
largest natural, the naturally ordered assembly of naturals is not
finite.
>
> There is a simply proof to show that they do: The sequence of all
> natural numbers cannot be larger than a natural number (of course we
> cannot name that number. Therefore let us call it ?). The proof is as
> follows:
>
> Theorem: For all n in N: the initial segment {1, 2, 3, ..., n} is
> finite and has cardinality n.
> Proof by induction. For "all" n.
That "theorem" only applies to "initial segments with a largest
element", so either produce that alleged largest natural or concede your
error.
>
> So, if there should be an actually infinite set (different from a
> potentially infinite set that is always finite but not fixed), then it
> is forced to contain elements that are not subject of my theorem.
Your theorem does not cover any segment of naturals without a largest
element, of which there s at least one.
>
> Can you really imagine a reason why this theorem should not hold?
It does not hold for any segment of naturals lacking a largest natural.
So WM is, as usual, wrong.
David Formosa (aka ? the Platypus)
> On Feb 1, 10:22 pm, "David Formosa (aka ? the Platypus)"
><dform...@usyd.edu.au> wrote:
>> On Fri, 30 Jan 2009 13:16:22 -0800 (PST), Albrecht <albst...@gmx.de> wrote:
>>
>> [...]
>>
>> > 2.) there is no quantity or cardinality of all of them since the
>> > concept "all of them" is inconsistent.
>>
>> Using the rules of standard mathmatics and logic please derive the statement
>> A = ~A from the concept "all of them".
>
> Yes, that's a good point. Most likely, there is no inconsistency in
> that concept.
Thankyou.
> Nevertheless, don't forget there are consistent impossibilities.
Can you give an example? Mathematics tends only to require consistency.
MoeBlee
> 1.) I'm convinced that it is possible to create a stringent theory of
> the discrete quantities (the natural numbers) with at least the same
> strenght as there is a theory of sets like ZFC (respectively theories
> of sets, e.g. ZFC, NBG, ...).
There is a technical definition of 'relative strength', but I doubt
you have any particular definition in mind.
> Up to now Math lacks of such a theory.
Are you at all familiar with the wide range of theories that have been
proposed in the mathematical literature?
> Set theories are not the most fundamental theories of Math. Discrete
> quantities (natural numbers) are more fundamental than sets.
Would that you had a definition of 'more fundamental'. In any case,
there are theories that take both numbers and sets as primitives (cf.
Myhill's constructive set theory), but I'm not aware of one that
defines sets by way of numbers (perhaps there is one; I don't know).
> 2.) To discuss the problem of infinite quantities we have to consider
> the relation "larger than" (>).
> The set theoretic definiton is an "artificial" definition which don't
> satisfy our intuitive understanding of that relation.
Satisfies my intuitive understanding.
> I claim that the definition of "larger than" of discrete quantities
> has to cover the condition:
> If a > b than a = b+1 or a > b+1
> or, for sure, in another notation:
> a > b <=> a = b+1 or a > b+1
For various kinds of numbers, those are theorems of set theory too. I
don't see what problem there is in this regard. (Of course, though,
those are not DEFINITIONS.)
> I hope the notation is clear.
> Some may claim that this notation is circular, but we are satisfied
> with the condition that a has to be at least one unity larger than b
> to be larger than b (thus a > b holds).
As a definition, of course it's circular. And circularity is not
avoided unless 'larger' itself is defined (without circle back to '<')
or primitive.
A non-circular definition is
x>y <-> Ez(z not=0 & y=x+z).
> With this considerations, if someone claims that the quantity of the
> natural numbers (let's call it A*) is larger than any natural number,
> he has to show that:
> A* > n forall n in N (with N is the proper class of the natural
> numbers) is true.
Of course, yes, if union with some singleton is required of every
member in the field of the relation, then cardinality doesn't work
with all infinite sets. We know that.
> Thus: The quantity of the natural numbers has to be at least one unity
> larger than any natural number.
Yes, if you stipulate that a definition of '>' for sets, not just for
integers and certain other numbers, must have the property you
mentioned earlier. But no one is obligated to accept your stipulation.
Do you understand? You are make a pure stipulation. That is fine for
whatever system you like to build. But the rest of humanity is not
obligated to accept your stipulations applied to all other kinds of
systems, unless, of course, we are mistaken in our belief that you are
not God from Whom all truth derives.
> The axiom of Infinity implies this claim since infinite sets has to
> have a cardinality and cardinality is the quantity of the elements of
> a set.
>
> This claim leads to two contradictory sentences:
>
> (1) If the cardinality (or quantity) of the natural numbers is at
> least one unity larger than any natural number, there has to be at
> least one element more in the manifold of the natural numbers as there
> are natural numbers.
> But (2) The manifold of natural numbers consists only of natural
> numbers (by definition).
Again, that contradicts your stipulation about how a 'greater than'
relation must be defined. We know that. But, again, we are not
obligated to your stipulation, unless we are mistaken in our appraisal
that you are not God from Whom all truth derives.
MoeBlee
Dik T. Winter
...
> 2.) To discuss the problem of infinite quantities we have to consider
> the relation "larger than" (>).
> The set theoretic definiton is an "artificial" definition which don't
> satisfy our intuitive understanding of that relation.
> I claim that the definition of "larger than" of discrete quantities
> has to cover the condition:
> If a > b than a = b+1 or a > b+1
> or, for sure, in another notation:
> a > b <=> a = b+1 or a > b+1
So you claim that in the floating-point numbers on a computer (actually
discrete quantities and finitely many too), the relation
1.5 > 1
is false? Bizarre.
> With this considerations, if someone claims that the quantity of the
> natural numbers (let's call it A*) is larger than any natural number,
> he has to show that:
> A* > n forall n in N (with N is the proper class of the natural
> numbers) is true.
But that is easy. For each n in N, A* > n+1 (as n+1 is also in N), so it
satisfies the condition.
> Thus: The quantity of the natural numbers has to be at least one unity
> larger than any natural number.
No. It has to be at least one unity larger than *each* natural number.
Actually it is infinitely much larger than *each* natural number.
> (1) If the cardinality (or quantity) of the natural numbers is at
> least one unity larger than any natural number, there has to be at
> least one element more in the manifold of the natural numbers as there
> are natural numbers.
Unfounded conclusion. Why has there to be at least one element more?
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
WM
> In article
> <6a2d45e2-7733-4a42-bf6a-19dbd31bb...@i20g2000prf.googlegroups.com>,
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 2 Feb., 16:01, LauLuna <laureanol...@yahoo.es> wrote:
> > > On Jan 30, 8:35 am, Albrecht <albst...@gmx.de> wrote:
>
> > > > As the unary system
>
> > > > O
> > > > OO
> > > > OOO
> > > > OOOO
> > > > OOOOO
> > > > ...
>
> > > > easily shows is there either a natural number which describes the
> > > > quantity of the natural numbers or there is no number which discribes
> > > > the quantity of the natural numbers since the natural numbers numbered
> > > > themselves
>
> > > At every finite stage in the generation of naturals, we find a natural
> > > that tells us how many naturals have been produced SO FAR. But 'so
> > > far' is never 'in the end', concerning naturals.
>
> > > Now, there is no reason to assume that what holds for the finite also
> > > holds for the infinite.
>
> > But every natural number is finite. There is no infinite natural
> > number. So there is no reason why the well proved finite laws should
> > not apply here.
>
> One of the "well proved finite laws" says that for any finite
> ->ordered<- assembly of objects, there is a largest.
That law is easily proven wrong. What is the largest photon, with
respect to energy, say, in the finite part of the universe that we can
see? What is the largest man with respect to mass of brain?
>
> From which those with any sense may conclude that unless there is a
> largest natural, the naturally ordered assembly of naturals is not
> finite.
>
Only those with no sense for real assemblies of numbers try to
conclude this obviously senseless notion.
>
>
> > There is a simply proof to show that they do: The sequence of all
> > natural numbers cannot be larger than a natural number (of course we
> > cannot name that number. Therefore let us call it ?). The proof is as
> > follows:
>
> > Theorem: For all n in N: the initial segment {1, 2, 3, ..., n} is
> > finite and has cardinality n.
> > Proof by induction. For "all" n.
>
> That "theorem" only applies to "initial segments with a largest
> element", so either produce that alleged largest natural or concede your
> error.
That theorem applies to every natural number, i.e., to all natural
numbers. Or do you know a natural number that is outside of a finite
initial segment?
>
>
>
> > So, if there should be an actually infinite set (different from a
> > potentially infinite set that is always finite but not fixed), then it
> > is forced to contain elements that are not subject of my theorem.
>
> Your theorem does not cover any segment of naturals without a largest
> element, of which there s at least one.
My theorem covers all natural numbers. That is enough for me. If you
insist that your "infinite segment" contains only natural numbers,
then you should specify a natural number that belongs to your segment
but is not subject to my theorem. Otherwise I cannot but judge your
saying as handwaving without reason.
>
>
>
> > Can you really imagine a reason why this theorem should not hold?
>
> It does not hold for any segment of naturals lacking a largest natural.
Please do not try to change the subject. I am not interested in the
fact that my theorem does not supply your desired result. I am
interested in the question whether my conclusion is lacking logic.
Logic, not wishful thinking or unreasonable belief, is the fundament
of mathematics.
Regards, WM
WM
>
> > That's not the problem. The probem is, that there is no concept which
> > is able to describe coherently the quantity of infinite many objects
>
> How about the concept "more that expressed by any natural number"?
>
> It works for me
But it does not work with natural numbers.
Regards, WM
Albrecht
Dik T. Winter schrieb:
> In article <7996c514-6675-4635...@e1g2000pra.googlegroups.com> Albrecht <albs...@gmx.de> writes:
> ...
> > 2.) To discuss the problem of infinite quantities we have to consider
> > the relation "larger than" (>).
> > The set theoretic definiton is an "artificial" definition which don't
> > satisfy our intuitive understanding of that relation.
> > I claim that the definition of "larger than" of discrete quantities
> > has to cover the condition:
> > If a > b than a = b+1 or a > b+1
> > or, for sure, in another notation:
> > a > b <=> a = b+1 or a > b+1
>
> So you claim that in the floating-point numbers on a computer (actually
> discrete quantities and finitely many too), the relation
> 1.5 > 1
> is false? Bizarre.
A first element (1.) together with five elements (5) is greater than
1? But thats true, not false! :-)
>
> > With this considerations, if someone claims that the quantity of the
> > natural numbers (let's call it A*) is larger than any natural number,
> > he has to show that:
> > A* > n forall n in N (with N is the proper class of the natural
> > numbers) is true.
>
> But that is easy. For each n in N, A* > n+1 (as n+1 is also in N), so it
> satisfies the condition.
Hahahah.
>
> > Thus: The quantity of the natural numbers has to be at least one unity
> > larger than any natural number.
>
> No. It has to be at least one unity larger than *each* natural number.
Okay. *Each* number.
> Actually it is infinitely much larger than *each* natural number.
There is no quantity "infinity". And surely not actually. Let's say:
Potentially it is infinitely larger than each natural number. The
question is only: What is that "it"? There is nothing which could be
larger than any natural number, since the natural numbers increases
infinitely. That means: without end ... unendingly ... in
eternally ... for ever ... beyond all boundaries ...
infinitely ... ... ... ....
>
> > (1) If the cardinality (or quantity) of the natural numbers is at
> > least one unity larger than any natural number, there has to be at
> > least one element more in the manifold of the natural numbers as there
> > are natural numbers.
>
> Unfounded conclusion. Why has there to be at least one element more?
By definiton of "larger than".
William Hughes
... every natural number is finite. There is no infinite natural
> number. So there is no reason why the well proved finite laws should
> not apply here.
>
> There is a simply proof to show that they do: The sequence of all
> natural numbers cannot be larger than a natural number (of course we
> cannot name that number. Therefore let us call it ?). The proof is as
> follows:
>
> Theorem: For all n in N: the initial segment {1, 2, 3, ..., n} is
> finite and has cardinality n.
> Proof by induction. For "all" n.
>
>
No sequence that has an end is the sequence of all natural
numbers. So any property that holds for each of theses sequences
may may not hold for the union of theses sequences
the initial sequence {1,2,3,...} that does not have a end.
Indeed, this initial sequence is not finite and
does not have a cardinality that is a natural number.
- William Hughes
Dik T. Winter
> Dik T. Winter schrieb:
> > In article <7996c514-6675-4635...@e1g2000pra.googlegroups.com> Albrecht <albs...@gmx.de> writes:
> > > If a > b than a = b+1 or a > b+1
> > > or, for sure, in another notation:
> > > a > b <=> a = b+1 or a > b+1
> >
> > So you claim that in the floating-point numbers on a computer (actually
> > discrete quantities and finitely many too), the relation
> > 1.5 > 1
> > is false? Bizarre.
>
> A first element (1.) together with five elements (5) is greater than
> 1? But thats true, not false! :-)
Eh? You claim:
a > b <=> a = b+1 or a > b+1
for 1.5 and 1 that means:
1.5 > 1 <=> 1.5 = 1+1 or 1.5 > 1+1
that is *your* definition. So according to *your* definition 1.5 > 1 is
false.
> > > With this considerations, if someone claims that the quantity of the
> > > natural numbers (let's call it A*) is larger than any natural number,
> > > he has to show that:
> > > A* > n forall n in N (with N is the proper class of the natural
> > > numbers) is true.
> >
> > But that is easy. For each n in N, A* > n+1 (as n+1 is also in N), so it
> > satisfies the condition.
>
> Hahahah.
Yeah, so whats is wrong with it?
> > > Thus: The quantity of the natural numbers has to be at least one unity
> > > larger than any natural number.
> >
> > No. It has to be at least one unity larger than *each* natural number.
>
> Okay. *Each* number.
Why each *number* and not natural number? You state *explicitly* has to
show A* > n forall n in N. What that statement by you wrong?
> > Actually it is infinitely much larger than *each* natural number.
>
> There is no quantity "infinity". And surely not actually.
Do I claim that? If so, where?
> Let's say:
> Potentially it is infinitely larger than each natural number. The
> question is only: What is that "it"? There is nothing which could be
> larger than any natural number, since the natural numbers increases
> infinitely. That means: without end ... unendingly ... in
> eternally ... for ever ... beyond all boundaries ...
> infinitely ... ... ... ....
Right. But why is it not possible that something is still larger, except
that it hurts your feelings?
> > > (1) If the cardinality (or quantity) of the natural numbers is at
> > > least one unity larger than any natural number, there has to be at
> > > least one element more in the manifold of the natural numbers as there
> > > are natural numbers.
> >
> > Unfounded conclusion. Why has there to be at least one element more?
>
> By definiton of "larger than".
Give a proof that there has to be at least one element more. *That* was
the unfounded conclusion.
Albrecht
Dik T. Winter schrieb:
> In article <68576828-007b-40c4...@z6g2000pre.googlegroups.com> Albrecht <albs...@gmx.de> writes:
> > Dik T. Winter schrieb:
> > > In article <7996c514-6675-4635...@e1g2000pra.googlegroups.com> Albrecht <albs...@gmx.de> writes:
> > > > If a > b than a = b+1 or a > b+1
> > > > or, for sure, in another notation:
> > > > a > b <=> a = b+1 or a > b+1
> > >
> > > So you claim that in the floating-point numbers on a computer (actually
> > > discrete quantities and finitely many too), the relation
> > > 1.5 > 1
> > > is false? Bizarre.
> >
> > A first element (1.) together with five elements (5) is greater than
> > 1? But thats true, not false! :-)
>
> Eh? You claim:
> a > b <=> a = b+1 or a > b+1
> for 1.5 and 1 that means:
> 1.5 > 1 <=> 1.5 = 1+1 or 1.5 > 1+1
> that is *your* definition. So according to *your* definition 1.5 > 1 is
> false.
Nice. Have you some other funny arithmetic challenges? (Apart from
that: I'd talked about discrete quantities/natural numbers. But I'm
sure, in fact you'd understood that.)
>
> > > > With this considerations, if someone claims that the quantity of the
> > > > natural numbers (let's call it A*) is larger than any natural number,
> > > > he has to show that:
> > > > A* > n forall n in N (with N is the proper class of the natural
> > > > numbers) is true.
> > >
> > > But that is easy. For each n in N, A* > n+1 (as n+1 is also in N), so it
> > > satisfies the condition.
> >
> > Hahahah.
>
> Yeah, so whats is wrong with it?
And what is with: For each n in N, A* > 1+n
>
> > > > Thus: The quantity of the natural numbers has to be at least one unity
> > > > larger than any natural number.
> > >
> > > No. It has to be at least one unity larger than *each* natural number.
> >
> > Okay. *Each* number.
>
> Why each *number* and not natural number? You state *explicitly* has to
> show A* > n forall n in N. What that statement by you wrong?
Okay. *Each* *natural* number.
Now you are able to ask: Why "*Each* *natural* number" and not "Each
natural number"? And then: Why "Each natural number" and not Each
natural number? And so on. Very funny.
>
> > > Actually it is infinitely much larger than *each* natural number.
> >
> > There is no quantity "infinity". And surely not actually.
>
> Do I claim that? If so, where?
hahahah.
>
> > Let's say:
> > Potentially it is infinitely larger than each natural number. The
> > question is only: What is that "it"? There is nothing which could be
> > larger than any natural number, since the natural numbers increases
> > infinitely. That means: without end ... unendingly ... in
> > eternally ... for ever ... beyond all boundaries ...
> > infinitely ... ... ... ....
>
> Right. But why is it not possible that something is still larger, except
> that it hurts your feelings?
Hahahahhaha
>
> > > > (1) If the cardinality (or quantity) of the natural numbers is at
> > > > least one unity larger than any natural number, there has to be at
> > > > least one element more in the manifold of the natural numbers as there
> > > > are natural numbers.
> > >
> > > Unfounded conclusion. Why has there to be at least one element more?
> >
> > By definiton of "larger than".
>
> Give a proof that there has to be at least one element more. *That* was
> the unfounded conclusion.
Are you part of a dutch clown company? Very good. Very funny.
Mitch Harris
> Dik T. Winter schrieb:
> > Albrecht <albst...@gmx.de> writes:
> >
> > > Thus: The quantity of the natural numbers has to be at least one unity
> > > larger than any natural number.
>
> > No. It has to be at least one unity larger than *each* natural number.
>
> Okay. *Each* number.
>
> > Actually it is infinitely much larger than *each* natural number.
>
> There is no quantity "infinity".
If by 'quantity' you mean some natural number, then yes, of course
that is the case. Nobody is disputing the fact, everyone is in
agreement that each (or any, or every) natural number is finite. Of
all the possible things that 'infinite' might mena, we certainly don't
mean to call any particular natural number infinite.
But...
The thing the rest of us are calling infinity is what we are using to
put a handle on the 'quantity' (a generalized concept that goes beyond
getting the size of sets in one-to-one correspondence with the natural
numbers).
You seem to not like calling the collection of natural numbers a set,
and you'd prefer to call it a class. That's fine...whatever you'd like
to call it (it's not following the standard technical meanings of
those terms but no matter).
But that collection of things (1, 2, 3, etc., keep going) does it have
a member of it that counts the -entire- collection? No, it can't. Does
17 count that -collection-? No, because you can add 1 to 17 to get
another item that's in the collection. Consider any natural n...can it
count all the items in the collection? No because you can add 1 to get
another item that -is- in the collection, so your number n cannot be
the 'size'.
I think you are also saying this. There is no natural number that
counts all the natural numbers.
But now, it might be useful (you might disagree) to have a thing, a
concept that somehow captures -something- about that entire collection
of natural numbers, and that concept has many similarities with
particular natural numbers when they are used for sizes of finite
sets. That's what all this infinity talk is about, allowing people to
talk coherently about size when applied to something other than finite
sets. It may sound weird, and the concept of infinity may not act
exactly like a natural number, in fact very counter to many of the
usual properties of natural numbers, but it still works.
Don't try to force the concept of infinity to be just like a natural
number and you might get past your distaste for it. Think of negative
numbers... you can't hold negative 3 apples in your hand like you can
positive 3 apples in your hand, and -1*-1 = 1 (totally weird, right?)
but you get used to manipulating negatives, and it all seems to work
out.
Mitch
WM
Please refrain from elucidating your convictions but show the point
where you think that the logic of my proof fails.
Regards, WM
Ralf Bader
You, of course. But the brain of a horse has still more mass.
Albrecht
Mitch Harris schrieb:
Negative numbers are a sound enhancement of the concept of numbers (i.
e. discrete quantities) which starts from the naturally given natural
numbers which are essentially this numbers which perform the oneness
of the ordinal and cardinal numbers.
Actual infinity, an infinite quantity, the cardinality of an infinite
set is no sound enhancement of the concept of numbers because infinity
is no quantity and is not able to carry properties of quantities.
Actual infinity is not sound and not consistent and math lose it's
ability of prediction with the use of the concept of actual infinity.
William Hughes
You have given a correct proof of the wrong thing.
You need to show that the *union* of all initial seqments
of the form {1, 2, 3, ..., n} (where n is a natural number)
has as a cardinality something that is not larger
than every natural number.
You show that all initial segments of the form
{1, 2, 3, ..., n} (where n is a natural number)
have as a cardinality something that is not larger
than every natural number.
(This theorem is correct and your proof is correct,)
The point at which the logic of your proof fails is when
you try to take a theorem that holds for every set in the
union and say that it holds for the union.
- William Hughes
Your Theorem is correct, your proof
MoeBlee
It fails exactly where people have told it fails for about, at least,
the last three years.
It is correct that any finite sequence of natural numbers has finite
cardinality. But you've shown no rule of logic by which that entails
that the cardinality of an infinite sequence of natural numbers is a
natural number. You may posit an axiom that there are no infinite
sets, thus no infinite sequences. No one disputes that with such an
axiom (while dropping the axiom of infinity), there are no infinite
cardinalities and that there is no sequence that does not have a
finite cardinality. Or, you can posit an axiom schema that what holds
for all finite set holds for all infinite sets, but that is self-
inconsistent, since the property of being finite is not one held by
any infinite set (with Tarski definition of 'finite' and 'infinite').
So, the best you could do is to posit an axiom: If all finite
sequences have finite cardinality then all infinite sequences have
finite cardinality. But that also is inconsistent with other set
theory axioms.
This has been explained to you over and over and over by many people
and in many variations. Yet you remained so fixated that you refuse to
understand the most simple things.
MoeBlee
Virgil
<bb4ca98e-9f7f-4f20...@v18g2000pro.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
Except his clowning is infinitely closer to truth than yours ever gets.
You claim that what is true for each member of the "quantity of
naturals" must be true of "the quantity of naturals", which is nor more
true than that a quantity of empty sets must be an empty set.
Virgil
<d507f755-9749-45b6...@g1g2000pra.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Your "theorem" speaks only of finite intitial segments of N, so until
you prove that N is one of its own initial segments, you have failed to
prove anything about N itself.
Do you claim that N I one of its own initial segments?
If so, you should be able to name the largest member of N, or at least
prove that there is one.
But the successor rule for naturals makes such a proof impossible, so
that WM's arguments are refuted by the very nature of naturals.
Virgil
<c50279c1-d5c8-424b...@40g2000prx.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
> Negative numbers are a sound enhancement of the concept of numbers (i.
> e. discrete quantities) which starts from the naturally given natural
> numbers which are essentially this numbers which perform the oneness
> of the ordinal and cardinal numbers.
Does one have to stand on one's head, as Storz does, to "perform the
David R Tribble
> Theorem: For all n in N: the initial segment {1, 2, 3, ..., n} is
> finite and has cardinality n.
> Proof by induction. For "all" n.
>
> So, if there should be an actually infinite set (different from a
> potentially infinite set that is always finite but not fixed), then it
> is forced to contain elements that are not subject of my theorem.
Correction: If there should be an actually infinite set, then it is
forced to be of a form other than {1,2,3,...,n}, because all the
sets of this form are finite initial segments, and thus an actually
infinite set is not subject to your theorem.
> But that would be non-natural numbers, as my theorem covers all natural
> numbers.
> Can you really imagine a reason why this theorem should not hold?
Obviously it holds for all finite initial sets of the form
{1,2,3,...,n}.
Can you really imagine a reason why your theorem should
hold for sets other than that form, such as the set {1,2,3,...}
which has no last element n and is not an initial finite segment?
Mitch Harris
> Mitch Harris schrieb:
> > Don't try to force the concept of infinity to be just like a natural
> > number and you might get past your distaste for it. Think of negative
> > numbers... you can't hold negative 3 apples in your hand like you can
> > positive 3 apples in your hand, and -1*-1 = 1 (totally weird, right?)
> > but you get used to manipulating negatives, and it all seems to work
> > out.
>
> Negative numbers are a sound enhancement of the concept of numbers (i.
> e. discrete quantities) which starts from the naturally given natural
> numbers which are essentially this numbers which perform the oneness
> of the ordinal and cardinal numbers.
Yes, negative numbers are a sound, coherent enhancement or extension
of the natural numbers.
Can you produce an -actual- negative number?
> Actual infinity, an infinite quantity, the cardinality of an infinite
> set is no sound enhancement of the concept of numbers because infinity
> is no quantity and is not able to carry properties of quantities.
Don't worry about this actual or potential stuff. It's a red herring.
I think your definition of 'quantity' is restricted to natural numbers
and so by definition -cannot- apply coherently to whatever infinity
might be because it is just not allowed to be applied at all.
So now my advice is to not not bother with 'quantity' at all. And just
worry about one-to-one correspondences.
When you do that, you'll find that it is coherent to say that a set
that can be placed in 1-1 correspondence with the first n natural
numbers has a property involving n and is identical to what you call
quantity.
And then you can also have a 1-1 correspondence between some sets and
the -entire- collection of natural numbers. That correspondence
(explained elsewhere in this thread) is sound, coherent, kosher, all
sorts of other adjectives connoting 'OK'.
And so here is the weird thing...many people call that particular kind
of correspondence aleph_0 or infinity. And despite its weirdness, it
has many properties in common (surely not all) with those things that
you call quantity.
> Actual infinity is not sound and not consistent and math lose it's
> ability of prediction with the use of the concept of actual infinity.
Math loses its ability of prediction? Really? How? Any examples?
Mitch
herbzet
What was to be be proven:
The sequence of natural numbers cannot be larger than a natural number.
What was proven:
Every natural number fails to be as large as the sequence of natural numbers.
How amusing!
--
hz
Dik T. Winter
> Dik T. Winter schrieb:
> > > > So you claim that in the floating-point numbers on a computer
> > > > (actually discrete quantities and finitely many too), the relation
> > > > 1.5 > 1
> > > > is false? Bizarre.
> > Eh? You claim:
> > a > b <=> a = b+1 or a > b+1
> > for 1.5 and 1 that means:
> > 1.5 > 1 <=> 1.5 = 1+1 or 1.5 > 1+1
> > that is *your* definition. So according to *your* definition 1.5 > 1 is
> > false.
>
> Nice. Have you some other funny arithmetic challenges? (Apart from
> that: I'd talked about discrete quantities/natural numbers. But I'm
> sure, in fact you'd understood that.)
In what way are the floating-point numbers in which I am doing this *not*
discrete quantities? I thought you had read my original post!
So I ask again. What about your definition of ">" for discrete quantities?
> > > > But that is easy. For each n in N, A* > n+1 (as n+1 is also in N),
> > > > so it satisfies the condition.
> > >
> > > Hahahah.
> >
> > Yeah, so whats is wrong with it?
>
> And what is with: For each n in N, A* > 1+n
Nothing, because that is also true. But what is the relevance?
> > > > No. It has to be at least one unity larger than *each* natural
> > > > number.
> > >
> > > Okay. *Each* number.
> >
> > Why each *number* and not natural number? You state *explicitly* has to
> > show A* > n forall n in N. What that statement by you wrong?
>
> Okay. *Each* *natural* number.
Now we are back were we where, in what way is A* not at least one unity
larger than each natural number?
> > > > Actually it is infinitely much larger than *each* natural number.
> > >
> > > There is no quantity "infinity". And surely not actually.
> >
> > Do I claim that? If so, where?
>
> hahahah.
Yeah. If you do not have an answer you weasel out.
> > > Let's say:
> > > Potentially it is infinitely larger than each natural number. The
> > > question is only: What is that "it"? There is nothing which could be
> > > larger than any natural number, since the natural numbers increases
> > > infinitely. That means: without end ... unendingly ... in
> > > eternally ... for ever ... beyond all boundaries ...
> > > infinitely ... ... ... ....
> >
> > Right. But why is it not possible that something is still larger, except
> > that it hurts your feelings?
>
> Hahahahhaha
See my remark above.
> > > > > (1) If the cardinality (or quantity) of the natural numbers is at
> > > > > least one unity larger than any natural number, there has to be at
> > > > > least one element more in the manifold of the natural numbers as there
> > > > > are natural numbers.
> > > >
> > > > Unfounded conclusion. Why has there to be at least one element more?
> > >
> > > By definiton of "larger than".
> >
> > Give a proof that there has to be at least one element more. *That* was
> > the unfounded conclusion.
>
> Are you part of a dutch clown company? Very good. Very funny.
See my remark aboven.
WM
That need not be shown because I do not consider the union at all. I
consider only every natural number. The only thing of interest is: Is
there a natural number in N that requires or causes or even allows N
to be actually infinite. This is disproven.
Your approach comes from the other side. You assume that N is larger
than every initial segment. But, as my proof shows, that approach is
wrong from the outset, because you cannot justify it by showing at
least one natural number that is not covered by my proof.
>
> You show that all initial segments of the form
> {1, 2, 3, ..., n} (where n is a natural number)
> have as a cardinality something that is not larger
> than every natural number.
>
> (This theorem is correct and your proof is correct,)
>
> The point at which the logic of your proof fails is when
> you try to take a theorem that holds for every set in the
> union and say that it holds for the union.
So you think that the union is more than every initial segment. Either
your logic is invalid or there is a natural number outside of every
initial segment.
Regards, WM
William Hughes
> That need not be shown because I do not consider the union at all. I
> consider only every natural number. The only thing of interest is: Is
> there a natural number in N that requires or causes or even allows N
> to be actually infinite. This is disproven.
There is no single natural number in N that makes
N infinite. This is correct and totally beside the point. There is
no single natural number that makes N infinite, but this does not
mean that N is not infinite. If a union of elements has property
P, this does not mean there is a single element of the union
that means that the union has property P (e.g. P is "has no
largest element").
- William Hughes
FredJeffries
>
> > You need to show that the *union* of all initial seqments
> > of the form {1, 2, 3, ..., n} (where n is a natural number)
> > has as a cardinality something that is not larger
> > than every natural number.
>
> That need not be shown because I do not consider the union at all. I
> consider only every natural number.
It seems we have someone here who does not believe in teamwork...
WM
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 3 Feb., 12:36, William Hughes <wpihug...@hotmail.com> wrote:
> > > On Feb 2, 2:25 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > There is a simply proof to show that they do: The sequence of all
> > > > natural numbers cannot be larger than a natural number (of course we
> > > > cannot name that number. Therefore let us call it ?). The proof is as
> > > > follows:
>
> > > > Theorem: For all n in N: the initial segment {1, 2, 3, ..., n} is
> > > > finite and has cardinality n.
> > > > Proof by induction. For "all" n.
>
> > > No sequence that has an end is the sequence of all natural
> > > numbers. So any property that holds for each of theses sequences
> > > may may not hold for the union of theses sequences
> > > the initial sequence {1,2,3,...} that does not have a end.
> > > Indeed, this initial sequence is not finite and
> > > does not have a cardinality that is a natural number.
>
> > Please refrain from elucidating your convictions but show the point
> > where you think that the logic of my proof fails.
>
> Your "theorem" speaks only of finite intitial segments of N, so until
> you prove that N is one of its own initial segments, you have failed to
> prove anything about N itself.
My theorem holds for every n, for every finite initial segments and
for every union
U[k=<n] {1, 2, 3, ..., k}.
Unless there is something in N that does not belong to such a union,
my proof holds for N.
>
> Do you claim that N I one of its own initial segments?
Of course. That is the most basic property of nartural numbers.
> If so, you should be able to name the largest member of N, or at least
> prove that there is one.
I cannot name it - among other reasons, because it is not fixed
But the proof has been given:
Consider all natural numbers by letting n run through 1, 2, 3, ...,
n, ....
Obviously this does not leave out any natural number.
And obviously you will never get in trouble with infinity, if you only
consider all segments that are not larger than {1, 2, 3, ..., n}.
But if this does not leave out any natural number, then it holds for
all of them, doesn't it?
>
> But the successor rule for naturals makes such a proof impossible, so
> that WM's arguments are refuted by the very nature of naturals.
But if my proof does not leave out any initial segment of natural
numbers, then it holds for all of them, doesn't it?
The contradictory requirements posed by you and by me do not show that
one of them were false but they show that it is nonsense to attribute
a cardinal number to infinity.
Otherwise you must claim that a proof that does not leave out any
finite initial segment of natural numbers, does not hold for all
natural numbers. And that is obviously nonsense, if N is nothing but
the assembly of all its finite initial segments.
Regards, WM
Virgil
<74d82273-f8f4-4c54...@i24g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
When you talk about the totality of finite initial segments, you bring
into consideration its union, whether you care to notice it or not.
> I
> consider only every natural number.
Then you must not be considering finite intial segments of naturals,
which are not themselves naturals, except in the von Neumann model.
The only thing of interest is: Is
> there a natural number in N that requires or causes or even allows N
> to be actually infinite. This is disproven.
While no one element of an infinite set makes the difference between it
and finite sets, the totality of such elements does.
>
> Your approach comes from the other side. You assume that N is larger
> than every initial segment.
If you assume otherwise, then can you display that hypothetical initial
segment that N is NOT larger than?
Only when you have done that will anyone accept your silly thesis.
> But, as my proof shows, that approach is
> wrong from the outset, because you cannot justify it by showing at
> least one natural number that is not covered by my proof.
Your "proof" proves nothing except your own incompetence.
> >
> > You show that all initial segments of the form
> > {1, 2, 3, ..., n} (where n is a natural number)
> > have as a cardinality something that is not larger
> > than every natural number.
> >
> > (This theorem is correct and your proof is correct,)
> >
> > The point at which the logic of your proof fails is when
> > you try to take a theorem that holds for every set in the
> > union and say that it holds for the union.
>
> So you think that the union is more than every initial segment.
Not quite. What we DO think is that every ->finite<- initial segment is
a proper subset of the union of all of those initial segments.
And we will keep thinking that until someone can show a ->finite<-
initial segment as large as N.
Can you show us such a finite initial segment, WM?
> Either
> your logic is invalid or there is a natural number outside of every
> initial segment.
False dichotomy, as logically there is nothing prohibiting a non-finite
initial segment containing all naturals.
It is trivial that for each finite initial segment of naturals(FISONs),
there is a natural outside it, but that does not imply that there is a
natural outside the union of all those FISONs, which is an ISON but not
a FISON.
WM
> WM wrote:
> > Theorem: For all n in N: the initial segment {1, 2, 3, ..., n} is
> > finite and has cardinality n.
> > Proof by induction. For "all" n.
>
> > So, if there should be an actually infinite set (different from a
> > potentially infinite set that is always finite but not fixed), then it
> > is forced to contain elements that are not subject of my theorem.
>
> Correction: If there should be an actually infinite set, then it is
> forced to be of a form other than {1,2,3,...,n}, because all the
> sets of this form are finite initial segments, and thus an actually
> infinite set is not subject to your theorem.
But all sets consisting of natural numbers only are of this form
{1,2,3,...,n}.
You seem to claim that N is not covered by a proof that is correct
forall n.
A n : U[k =< n] {1,2,3,...,k} is finite.
I am happy with the set that is subject to this proof.
>
> > But that would be non-natural numbers, as my theorem covers all natural
> > numbers.
> > Can you really imagine a reason why this theorem should not hold?
>
> Obviously it holds for all finite initial sets of the form
> {1,2,3,...,n}.
>
> Can you really imagine a reason why your theorem should
> hold for sets other than that form, such as the set {1,2,3,...}
> which has no last element n and is not an initial finite segment?
I cannot imagine what you understand by (a set that has) "no last
element". This "no last element" obviously is not a natural number.
Have you in omega in mind? Of course, my proof holds only for natural
numbers - indeed for all of them.
Consider all natural numbers by letting n run through 1, 2, 3, ...,
n, ....
Obviously this does not leave out any natural number.
And obviously you will never get in trouble with infinity, if you
only
consider all segments that are not larger than {1, 2, 3, ..., n}.
But if this does not leave out any natural number, then it holds for
all of them, doesn't it?
Otherwise you must claim that a proof that does not leave out any
WM
Consider all natural numbers (keep in mind: all)
by letting n run through 1, 2, 3, ..., n, ....
Obviously this does not leave out any natural number.
And obviously you will never get in trouble with infinity, if you
only
consider all segments that are not larger than {1, 2, 3, ..., n}.
But if this does not leave out any natural number, then it holds for
all of them, doesn't it?
But if my proof does not leave out any initial segment of natural
numbers, then it holds for all of them, doesn't it?
If you deny that, then you must claim that a proof that does not leave
WM
> On Feb 4, 10:17 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > That need not be shown because I do not consider the union at all. I
> > consider only every natural number. The only thing of interest is: Is
> > there a natural number in N that requires or causes or even allows N
> > to be actually infinite. This is disproven.
>
> There is no single natural number in N that makes
> N infinite. This is correct and totally beside the point. There is
> no single natural number that makes N infinite, but this does not
> mean that N is not infinite.
My proof dhows that every natural number and all smaller natural
numbers are within a finite set.
> If a union of elements has property
> P, this does not mean there is a single element of the union
> that means that the union has property P (e.g. P is "has no
> largest element").
Obviously you imagine some union N that is different from the assembly
of all natural numbers and in particular is not covered by the
assembly of all finite initial segments {1, 2, 3, ..., n}. The
assembly of all finite initial segments {1, 2, 3, ..., n} of natural
numbers is not actually infinite. To see this you cannnot assume an
actual infinity assembly of such segments, because for every segments
you csan prove that there are only n-1 segments that are smaller. So
you can do the proof by letting n start from 1 and run through all
natural numbers. At no n you have to consider infinitely many
segments, because at no n you have infinitely many such segments. This
proof has the advantage that is covers all (infinitely many, if
available) natural numbers but need not and cannot run into the
problem of starting with the assumption of an infinitude.
A n : {1, 2, 3, ..., n} is finite
and
A n : U[k =< n] {1, 2, 3, ..., k} is finite
If you feel that
for all n the union U[k =< n] {1, 2, 3, ..., k}
is different from N, then you have only two possible logical excuses
for lack of comprehension:
Either you assume that N is not covered by the union of all finite
initial segments, or you do not accept that a proof covering all
finite initial segments of natural numbers covers all finite initial
segments of natural numbers. Well the latter is not really an excuse.
Regards, WM
>
> - William Hughes
WM
Not quite. My sentence above should by understood as: That need not be
shown because I do not consider the union at all. I consider only
every natural number and all its predecessors. Of course this covers
the complete assembly of all natural numbers. But I would like to
avoid the magic action of set theoretic unions. I simply mean every
and all natural numbers.
Regards, WM
MoeBlee
> But all sets consisting of natural numbers only are of this form
> {1,2,3,...,n}.
No, they're not.
{1 3} is a set of natural numbers not of that form.
{1 3 100} is a set of natural numbers not of that form.
w (omega) is a set of natural numbers not of that form.
{n | n is even} is a set of natural numbers not of that form.
{3 5} u {n| n is even} is a set of natural numbers not of that form.
{p | p is prime} is a set of natural numbers not of that form.
{n | n is the Godel number of a sentence of PA true in the standard
model for the language of PA} is a set of natural numbers not of that
form.
Did you fall on your head this morning?
> You seem to claim that N is not covered by a proof that is correct
> forall n.
>
> A n : U[k =< n] {1,2,3,...,k} is finite.
Yes, for each n in w, we have U[k=1 to n] {1 2 ... k} is finite.
For each n in w, we have that U[k=1 to n] {1 2 ... k} is a finite
union of finite sets. And a finite union of finite sets is finite.
So what? It doesn't prove that the set of natural numbers is finite.
MoeBlee
Brian Chandler
> On 3 Feb., 22:42, David R Tribble <da...@tribble.com> wrote:
> > Can you really imagine a reason why your theorem should
> > hold for sets other than that form, such as the set {1,2,3,...}
> > which has no last element n and is not an initial finite segment?
> I cannot imagine what you understand by (a set that has) "no last
> element".
Well, consider the English expression "A German with no brain". Can
you imagine what that would mean? Can you imagine that saying "An X
with no Y" means an X for which there is no Y?
A "potato with no blemish" refers to a potato on which no blemish can
be found. A person, whether a German with no brain or not, would be
unable to find a blemish on this potato, because this potato has no
blemish. Here's an exercise to test whether you have understood so
far:
----------------
Q: Then where is the blemish on the potato?
What would the correct answer to this question be?
----------------
Now consider your next statement:
> This "no last element" obviously is not a natural number.
Does this bring the meaning of "A German with no brain" into sharper
focus? A German with no brain is a German inside whose head no brain
can be found. A set that has no last element is a set such that
person, whether a German with no brain or not, would be unable to find
a last element in this set, because there is no last element. If there
is no last element to this set, it is not possible to ask for any
properties of the last element to this set, because we have just said
there isn't one.
Are you _really_ this stupid? Is it really possible to imagine a man
who, told there is no beer to drink, insists on asking for thousands
of Usenet posts what the flavour of the beer is like?
> Have you in omega in mind?
"Omega in mind" for what?? The last element of the set that has no
last element perhaps?
Grief.
Brian Chandler
http://imaginatorium.org
MoeBlee
wrote:
> WM wrote:
> > This "no last element" obviously is not a natural number.
Yes, this is one of the most ridiculous comments I've come across in a
while. If only it were funny, it would be deserving of Groucho.
MoeBlee
Virgil
<bf33a7c4-18b2-4dea...@w1g2000prk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 3 Feb., 22:11, Virgil <Vir...@gmale.com> wrote:
> > In article
> > <d507f755-9749-45b6-a3a3-d23951c9a...@g1g2000pra.googlegroups.com>,
> > Your "theorem" speaks only of finite intitial segments of N, so until
> > you prove that N is one of its own initial segments, you have failed to
> > prove anything about N itself.
>
> My theorem holds for every n, for every finite initial segments and
> for every union
> U[k=<n] {1, 2, 3, ..., k}.
Your "theorem" does not hold for U[k in N] {1, 2, 3, ..., k}, which is
the only relevant case.
>
> Unless there is something in N that does not belong to such a union,
> my proof holds for N.
For every such union, U[k=<n] {1, 2, 3, ..., k},
n+1 belongs to N but does not belong to the union.
> >
> > Do you claim that N I one of its own initial segments?
>
> Of course. That is the most basic property of nartural numbers.
It is not a FINITE initial segment unless is contains a maximal member.
Which N does not.
>
> > If so, you should be able to name the largest member of N, or at least
> > prove that there is one.
>
> I cannot name it - among other reasons, because it is not fixed
That is because it can not exist. The existence of any member of N
requires the existence of its successor, which is larger.
>
> But the proof has been given:
> Consider all natural numbers by letting n run through 1, 2, 3, ...,
> n, ....
> Obviously this does not leave out any natural number.
> And obviously you will never get in trouble with infinity, if you only
> consider all segments that are not larger than {1, 2, 3, ..., n}.
> But if this does not leave out any natural number, then it holds for
> all of them, doesn't it?
> >
> > But the successor rule for naturals makes such a proof impossible, so
> > that WM's arguments are refuted by the very nature of naturals.
>
> But if my proof does not leave out any initial segment of natural
> numbers, then it holds for all of them, doesn't it?
There is one initial segment that it leaves out, the one which has no
largest member, which is the union of all the ones which do have largest
elements.
>
> The contradictory requirements posed by you and by me do not show that
> one of them were false but they show that it is nonsense to attribute
> a cardinal number to infinity.
It shows that it is nonsense to attribute sense to WM. In ZFC, for
example, infinite cardinalities are quite normal.
>
> Otherwise you must claim that a proof that does not leave out any
> finite initial segment of natural numbers, does not hold for all
> natural numbers.
A proof that holds for each and every FISON need not be true for the
union of all FISONS.
And that is obviously nonsense, if N is nothing but
> the assembly of all its finite initial segments.
I do not know what the mathematical properties of an "assembly" are, but
while N is the union of all infinitely many of its FISONs, that does not
make it a FISON.
At least unless WM can name a natural so large that it has no successor.
Virgil
<6c71034d-fb37-40b7...@i18g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 3 Feb., 22:42, David R Tribble <da...@tribble.com> wrote:
> > WM wrote:
> > > Theorem: For all n in N: the initial segment {1, 2, 3, ..., n} is
> > > finite and has cardinality n.
> > > Proof by induction. For "all" n.
> >
> > > So, if there should be an actually infinite set (different from a
> > > potentially infinite set that is always finite but not fixed), then it
> > > is forced to contain elements that are not subject of my theorem.
> >
> > Correction: If there should be an actually infinite set, then it is
> > forced to be of a form other than {1,2,3,...,n}, because all the
> > sets of this form are finite initial segments, and thus an actually
> > infinite set is not subject to your theorem.
>
> But all sets consisting of natural numbers only are of this form
> {1,2,3,...,n}.
Nonsense. The set {n+1}, consisting of a successor natural number only,
is not of that form.
> You seem to claim that N is not covered by a proof that is correct
> forall n.
I equally claim that a proof about all eagles need not hold for frogs.
And a proof about apples need not hold for oranges.
So what?
> I cannot imagine what you understand by (a set that has) "no last
> element". This "no last element" obviously is not a natural number.
> Have you in omega in mind? Of course, my proof holds only for natural
> numbers - indeed for all of them.
For each perhaps but not for the set of all, which is a different thing
entirely.
>
> Consider all natural numbers by letting n run through 1, 2, 3, ...,
> n, ....
> Obviously this does not leave out any natural number.
> And obviously you will never get in trouble with infinity, if you
> only
> consider all segments that are not larger than {1, 2, 3, ..., n}.
> But if this does not leave out any natural number, then it holds for
> all of them, doesn't it?
But what is true of the members of a set may need be true of the set
itself.
A set of apples is not an apple, and the set of naturals is not a
natural.
>
> Otherwise you must claim that a proof that does not leave out any
> finite initial segment of natural numbers, does not hold for all
> natural numbers.
I need not do that. I need only claim that the set of all naturals is
not a natural.
> And that is obviously nonsense, if N is nothing but
> the assembly of all its finite initial segments.
What sort of "assemblies" are required to be the same as the things
assembled?
Is a basket of apples an apple?
Virgil
<44a0409f-eff9-423a...@p2g2000prn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
What holds for each may not hold for the collection. Each is a single
number but the collection is not a single number.
>
> But if my proof does not leave out any initial segment of natural
> numbers, then it holds for all of them, doesn't it?
Each individually, but not collectively.
Each object in a basket may be an apple without the container being an
apple.
Virgil
<8499f9e4-04aa-4011...@r15g2000prh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 4 Feb., 17:04, William Hughes <wpihug...@hotmail.com> wrote:
> > On Feb 4, 10:17 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > That need not be shown because I do not consider the union at
> > > all. I consider only every natural number. The only thing of
> > > interest is: Is there a natural number in N that requires or
> > > causes or even allows N to be actually infinite. This is
> > > disproven.
> >
> > There is no single natural number in N that makes N infinite. This
> > is correct and totally beside the point. There is no single
> > natural number that makes N infinite, but this does not mean that N
> > is not infinite.
>
> My proof dhows that every natural number and all smaller natural
> numbers are within a finite set.
But does not show that all larger numbers can also be put into a single
finite set. So it proves nothing relevant to WM's delusions.
>
> > If a union of elements has property
> > P, this does not mean there is a single element of the union that
> > means that the union has property P (e.g. P is "has no largest
> > element").
>
> Obviously you imagine some union N that is different from the
> assembly of all natural numbers and in particular is not covered by
> the assembly of all finite initial segments {1, 2, 3, ..., n}.
Is an "assembly" of sets the same as a union of the sets assembled or
the same as a set of the sets assembled, or neither. If it is the
union, then it is not a member of itself, and if it is the set, then it
contains not naturals at all.
And if it is something else, it is nonsense until defined, which it has
not yet been.
> The assembly
Undefined nonsense! The rest snipped.
Virgil
<26de8643-56d2-40b0...@p23g2000prp.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 4 Feb., 18:37, FredJeffries <FredJeffr...@gmail.com> wrote:
> > On Feb 4, 7:17 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> > > > You need to show that the *union* of all initial seqments
> > > > of the form {1, 2, 3, ..., n} (where n is a natural number)
> > > > has as a cardinality something that is not larger
> > > > than every natural number.
> >
> > > That need not be shown because I do not consider the union at all. I
> > > consider only every natural number.
> >
> > It seems we have someone here who does not believe in teamwork...
>
> Not quite. My sentence above should by understood as: That need not be
> shown because I do not consider the union at all.
Then you must NOT allow yourself to consider finite initial segments of
naturals (FISONs), as they are not themselves natural numbers.
But in any case, WM's refusal to consider unions, does not mean that we
cannot consider them, and given any family of sets such as the family of
FISONs that WM DOES consider, the issue of its union is quite natural.
Virgil
<95f2c3e9-c4e0-4834...@o40g2000prn.googlegroups.com>,
MoeBlee <jazz...@hotmail.com> wrote:
Do you suppose WM can tell us which "no last element" he is referring to?
WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 3 Feb., 22:11, Virgil <Vir...@gmale.com> wrote:
> > > In article
> > > <d507f755-9749-45b6-a3a3-d23951c9a...@g1g2000pra.googlegroups.com>,
> > > Your "theorem" speaks only of finite intitial segments of N, so until
> > > you prove that N is one of its own initial segments, you have failed to
> > > prove anything about N itself.
>
> > My theorem holds for every n, for every finite initial segments and
> > for every union
> > U[k=<n] {1, 2, 3, ..., k}.
>
> Your "theorem" does not hold for U[k in N] {1, 2, 3, ..., k}, which is
> the only relevant case.
Which k do you have in mind to support your claim?
>
>
>
> > Unless there is something in N that does not belong to such a union,
> > my proof holds for N.
>
> For every such union, U[k=<n] {1, 2, 3, ..., k},
> n+1 belongs to N but does not belong to the union.
It is not necessary to consider n+1, because the proof is based on the
method t consider only k =< n. The n+1 is included by the next step. n
+ m is included by the m-th step. Which natural number do you have in
mind that is not included in my proof?
>
>
>
> > > Do you claim that N I one of its own initial segments?
>
> > Of course. That is the most basic property of nartural numbers.
>
> It is not a FINITE initial segment unless is contains a maximal member.
>
> Which N does not.
Please do not claim unjustified ideas. Show me where my proof is
lacking logic.
>
Regards, WM
MoeBlee
We've shown you a THOUSAND times. It's lacking at the step where you
don't state a rule of inference that says what holds for finite sets
also holds for infinite sets.
In a post I made to you yesterday, I gave even more detail about this.
Over and over and over, people show you EXACTLY what is incorrect in
your proof. You cover your ears during those explanations. Then you
claim no one has explained it to you. That you do that continually
suggests that you do suffer a least a little from a certain kind of
madness.
MoeBlee
David R Tribble
>> Do you claim that N I one of its own initial segments?
>> prove that there is one.
>
WM wrote:
> I cannot name it - among other reasons, because it is not fixed
Is saying "it is not fixed" the same as saying "it does not exist"?
William Hughes
> On 4 Feb., 17:04, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Feb 4, 10:17 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > That need not be shown because I do not consider the union at all. I
> > > consider only every natural number. The only thing of interest is: Is
> > > there a natural number in N that requires or causes or even allows N
> > > to be actually infinite. This is disproven.
>
> > There is no single natural number in N that makes
> > N infinite. This is correct and totally beside the point. There is
> > no single natural number that makes N infinite, but this does not
> > mean that N is not infinite.
>
> My proof dhows that every natural number and all smaller natural
> numbers are within a finite set.
However, you want to show that something is true about N. Showing
that something is true about all elements of N is not the same thing,
>
> > If a union of elements has property
> > P, this does not mean there is a single element of the union
> > that means that the union has property P (e.g. P is "has no
> > largest element").
>
> Obviously you imagine some union N that is different from the assembly
> of all natural numbers and in particular is not covered by the
> assembly of all finite initial segments {1, 2, 3, ..., n}.
No. This is exaclty the union I do imagine. Call it N.
> The assembly of all finite initial segments {1, 2, 3, ..., n} of natural
> numbers is not actually infinite.
> To see this you cannnot assume an
> actual infinity assembly of such segments, because for every segments
> you csan prove that there are only n-1 segments that are smaller. So
> you can do the proof by letting n start from 1 and run through all
> natural numbers. At no n you have to consider infinitely many
> segments, because at no n you have infinitely many such segments.
Indeed, *at no n*. You have shown that something is true for every
element
of N. However, you need to show that something is true
for N. Showing that something is true for every element of N is
not the same thing.
> This
> proof has the advantage that is covers all (infinitely many, if
> available) natural numbers but need not and cannot run into the
> problem of starting with the assumption of an infinitude.
>
> A n : {1, 2, 3, ..., n} is finite
> and
> A n : U[k =< n] {1, 2, 3, ..., k} is finite
>
> If you feel that
> for all n the union U[k =< n] {1, 2, 3, ..., k}
> is different from N,
Since there is no n for which the union U[k =< n] {1, 2, 3, ..., k}
is equal to N we have:
for all n the union U[k =< n] {1, 2, 3, ..., k} is
different from n.
> then you have only two possible logical excuses
> for lack of comprehension:
> Either you assume that N is not covered by the union of all finite
> initial segments,
No, it is obvious that N is covered by the union of all finite
initial segments. However for no n is
the union U[k =< n] {1, 2, 3, ..., k}
equal to the union of all finite segments.
> or you do not accept that a proof covering all
> finite initial segments of natural numbers covers all finite initial
> segments of natural numbers.
No, a proof covering all finite intial segments of the natural numbers
covers all finite intial segments of the natural numbers. However, it
does not cover all initial segments of the natural numbers (it
does not cover the intitial segment with no largest element). The
fact that
P is true for every element of a set which covers N does not mean that
P is true for N.
- William Hughes
cbr...@cbrownsystems.com
> But all sets consisting of natural numbers only are of this form
> {1,2,3,...,n}.
> You seem to claim that N is not covered by a proof that is correct
> forall n.
>
> A n : U[k =< n] {1,2,3,...,k} is finite.
>
If you claim to truly have a proof of:
A n : U[k =< n] {1,2,3,...,k} is finite
which holds for /all/ n, then I would like to know what you mean, for
example, by saying that "k is a natural and k =< n" when n = {f : f is
a function from R^3 to C and for all x in R^3, |f(x)| < 15.5}.
For which values of k is "k =< n" true in that case?
Cheers - Chas
herbzet
WM wrote:
Absolutely correct in every detail.
You have unquestionably proven of each and every natural number that
it fails to be as large as the sequence of natural numbers.
Well done!
--
hz
Virgil
<8ba9491e-1610-4397...@p2g2000prn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 4 Feb., 21:52, Virgil <Vir...@gmale.com> wrote:
> > In article
> > <bf33a7c4-18b2-4dea-b09e-38f7e0920...@w1g2000prk.googlegroups.com>,
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 3 Feb., 22:11, Virgil <Vir...@gmale.com> wrote:
> > > > In article
> > > > <d507f755-9749-45b6-a3a3-d23951c9a...@g1g2000pra.googlegroups.com>,
> > > > Your "theorem" speaks only of finite intitial segments of N, so until
> > > > you prove that N is one of its own initial segments, you have failed to
> > > > prove anything about N itself.
> >
> > > My theorem holds for every n, for every finite initial segments and
> > > for every union
> > > U[k=<n] {1, 2, 3, ..., k}.
> >
> > Your "theorem" does not hold for U[k in N] {1, 2, 3, ..., k}, which is
> > the only relevant case.
>
> Which k do you have in mind to support your claim?
The "k" in " U[k in N] {1, 2, 3, ..., k}" represents every one of the
infinitely many members of N, as anyone even marginally competent in
mathematics would have known.
> >
> >
> >
> > > Unless there is something in N that does not belong to such a union,
> > > my proof holds for N.
> >
> > For every such union, U[k=<n] {1, 2, 3, ..., k},
> > n+1 belongs to N but does not belong to the union.
>
> It is not necessary to consider n+1, because the proof is based on the
> method t consider only k =< n.
Then YOUR considerations are all incomplete, since each of them ignores
more naturals than it considers,, i.e., {n+1, n+2, ...,n+n+1}
> The n+1 is included by the next step. n
> + m is included by the m-th step. Which natural number do you have in
> mind that is not included in my proof?
Each of your sets of naturals ignores more naturals that it includes, so
I have in mind all infinitely many of those infinite sets of naturals
that you ignore, of form N minus a FISON.
> >
> >
> >
> > > > Do you claim that N I one of its own initial segments?
> >
> > > Of course. That is the most basic property of nartural numbers.
> >
> > It is not a FINITE initial segment unless is contains a maximal member.
> >
> > Which N does not.
>
> Please do not claim unjustified ideas. Show me where my proof is
> lacking logic.
Your claim that what is true for each member of N is necessarily true
for N is unjustified, though that is far from the only unjustified and
unjustifiable claim you make.
Virgil
<596fa0cd-2901-42af...@g1g2000pra.googlegroups.com>,
David R Tribble <da...@tribble.com> wrote:
What "it is not fixed" means to me is that WM's logic is not fixed, but
is still broken.
Virgil
> On 4 Feb., 17:04, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Feb 4, 10:17 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > That need not be shown because I do not consider the union at all. I
> > > consider only every natural number. The only thing of interest is: Is
> > > there a natural number in N that requires or causes or even allows N
> > > to be actually infinite. This is disproven.
>
> > There is no single natural number in N that makes
> > N infinite. This is correct and totally beside the point. There is
> > no single natural number that makes N infinite, but this does not
> > mean that N is not infinite.
>
> My proof dhows that every natural number and all smaller natural
> numbers are within a finite set.
But it is the natural numbers LARGER than any given natural that give
rise to the infiniteness of N that WM is too blind to see.
>
> > If a union of elements has property
> > P, this does not mean there is a single element of the union
> > that means that the union has property P (e.g. P is "has no
> > largest element").
>
> Obviously you imagine some union N that is different from the assembly
> of all natural numbers
Depends what you mean by the "assembly" of all natuals.
If it is at all like N, the SET of all naturals, then it is clearly not
finite by any reasonable definition of finiteness,
> and in particular is not covered by the
> assembly of all finite initial segments {1, 2, 3, ..., n}.
Actually, each natural is "covered by" infinitely many FISONs.
> The assembly of all finite initial segments {1, 2, 3, ..., n} of natural
> numbers is not actually infinite.
If an assembly is at all like a set, that assembly of all FISONs cannot
possibly be finite.
If an assembly is not like a set, then it is nonsense until given an
adequate mathematical definition.
> To see this you cannnot assume an
> actual infinity assembly of such segments, because for every segments
> you csan prove that there are only n-1 segments that are smaller.
But it is the ones which are larger than any given one that provide the
infiniteness.
> So
> you can do the proof by letting n start from 1 and run through all
> natural numbers. At no n you have to consider infinitely many
> segments
Every n divides N into two parts, the smaller part being the set of
members up to n and the larger being the set of members beyond n, and
you cannot see all of N by looking only at smaller parts.
> or you do not accept that a proof covering all
> finite initial segments of natural numbers covers all finite initial
> segments of natural numbers.
Every FISON (finite initial segment of natural numbers) omits more
naturals than it contains.
WM
> On Feb 4, 2:51 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
> > On 4 Feb., 17:04, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > On Feb 4, 10:17 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > That need not be shown because I do not consider the union at all. I
> > > > consider only every natural number. The only thing of interest is: Is
> > > > there a natural number in N that requires or causes or even allows N
> > > > to be actually infinite. This is disproven.
>
> > > There is no single natural number in N that makes
> > > N infinite. This is correct and totally beside the point. There is
> > > no single natural number that makes N infinite, but this does not
> > > mean that N is not infinite.
>
> > My proof dhows that every natural number and all smaller natural
> > numbers are within a finite set.
>
> However, you want to show that something is true about N. Showing
> that something is true about all elements of N is not the same thing,
I want to show something about all natural numbers.
>
>
>
> > > If a union of elements has property
> > > P, this does not mean there is a single element of the union
> > > that means that the union has property P (e.g. P is "has no
> > > largest element").
>
> > Obviously you imagine some union N that is different from the assembly
> > of all natural numbers and in particular is not covered by the
> > assembly of all finite initial segments {1, 2, 3, ..., n}.
>
> No. This is exaclty the union I do imagine. Call it N.
Then you should know that a finite uinon of finite initial segments is
the same as its largest initial segment. And it is shown that it is
impossible to have an infinite union of initial segments, when you
form the unions as I do it. Is my method in error? If so, why and
where?
>
> > The assembly of all finite initial segments {1, 2, 3, ..., n} of natural
> > numbers is not actually infinite.
> > To see this you cannnot assume an
> > actual infinity assembly of such segments, because for every segments
> > you csan prove that there are only n-1 segments that are smaller. So
> > you can do the proof by letting n start from 1 and run through all
> > natural numbers. At no n you have to consider infinitely many
> > segments, because at no n you have infinitely many such segments.
>
> Indeed, *at no n*. You have shown that something is true for every
> element
> of N. However, you need to show that something is true
> for N. Showing that something is true for every element of N is
> not the same thing.
Remember: I use "all n". Nothing else, nothing more.
>
> > This
> > proof has the advantage that is covers all (infinitely many, if
> > available) natural numbers but need not and cannot run into the
> > problem of starting with the assumption of an infinitude.
>
> > A n : {1, 2, 3, ..., n} is finite
> > and
> > A n : U[k =< n] {1, 2, 3, ..., k} is finite
>
> > If you feel that
> > for all n the union U[k =< n] {1, 2, 3, ..., k}
> > is different from N,
>
> Since there is no n for which the union U[k =< n] {1, 2, 3, ..., k}
> is equal to N we have:
> for all n the union U[k =< n] {1, 2, 3, ..., k} is
> different from n.
>
> > then you have only two possible logical excuses
> > for lack of comprehension:
> > Either you assume that N is not covered by the union of all finite
> > initial segments,
>
> No, it is obvious that N is covered by the union of all finite
> initial segments. However for no n is
> the union U[k =< n] {1, 2, 3, ..., k}
> equal to the union of all finite segments.
This is obviously a contradiction, because every initial segment {1,
2, 3, ..., k} covers all n from 1 to k. No chance to need more than
one - in infinity. Further my proof does not suffer from needing
infinitely many. It applies only finitely many segments, infinitely
ofthen though, but that does not allow someone to reasonably require
the "change of infinity".
>
> > or you do not accept that a proof covering all
> > finite initial segments of natural numbers covers all finite initial
> > segments of natural numbers.
>
> No, a proof covering all finite intial segments of the natural numbers
> covers all finite intial segments of the natural numbers. However, it
> does not cover all initial segments of the natural numbers (it
> does not cover the intitial segment with no largest element). The
> fact that
> P is true for every element of a set which covers N does not mean that
> P is true for N.
But sometimes it is true for all n and sufficient to satisfy set
theorists:
With a_n the n-th line of Cantor's list and d its antidiagonal number
set theory accepts
1) forall n in N : a_n =/= d
Set theory also accepts
2) forall n in N : E m > n (with nothing known about a_m)
and even
3) forall n in N : {1, 2, 3, ..., n} is followed by infinitely many m
in N.
So
4) forall n in N : {1, 2, 3, ..., n} =/= N
Therefore we have the problem that
"forall n in N" does not generally mean forall n in N but at most in
certain instances convenient for set theorists.
Is there a general rule when "forall" means forall, or is it necessary
to have studied 10 semesters set theory in order to know the correct
procedure?
Regards, WM
Virgil
<58ee4200-667d-4fc0...@t39g2000prh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 5 Feb., 03:59, William Hughes <wpihug...@hotmail.com> wrote:
> > On Feb 4, 2:51 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 4 Feb., 17:04, William Hughes <wpihug...@hotmail.com> wrote:
> >
> > > > On Feb 4, 10:17 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > That need not be shown because I do not consider the union at all. I
> > > > > consider only every natural number. The only thing of interest is: Is
> > > > > there a natural number in N that requires or causes or even allows N
> > > > > to be actually infinite. This is disproven.
> >
> > > > There is no single natural number in N that makes
> > > > N infinite. This is correct and totally beside the point. There is
> > > > no single natural number that makes N infinite, but this does not
> > > > mean that N is not infinite.
> >
> > > My proof dhows that every natural number and all smaller natural
> > > numbers are within a finite set.
> >
> > However, you want to show that something is true about N. Showing
> > that something is true about all elements of N is not the same thing,
>
> I want to show something about all natural numbers.
What you show about each natural, or even to each FISON, only applies to
those individual naturals or FISONs, but not to the IISON N.
> >
> >
> >
> > > > If a union of elements has property
> > > > P, this does not mean there is a single element of the union
> > > > that means that the union has property P (e.g. P is "has no
> > > > largest element").
> >
> > > Obviously you imagine some union N that is different from the assembly
> > > of all natural numbers and in particular is not covered by the
> > > assembly of all finite initial segments {1, 2, 3, ..., n}.
> >
> > No. This is exaclty the union I do imagine. Call it N.
>
> Then you should know that a finite uinon of finite initial segments is
> the same as its largest initial segment.
It is not a finite union when there is no end of FISONs.
> And it is shown that it is
> impossible to have an infinite union of initial segments, when you
> form the unions as I do it. Is my method in error?
Your method assumes what it wants to prove, that there are only finitely
many FISONs.
> If so, why and
> where?
See above.
> >
> > > The assembly of all finite initial segments {1, 2, 3, ..., n} of natural
> > > numbers is not actually infinite.
It certainly is not actually finite, and unless your definition of an
"assembly" is totally non-mathematical, that makes it infinite.
> > > To see this you cannnot assume an
> > > actual infinity assembly of such segments, because for every segments
> > > you csan prove that there are only n-1 segments that are smaller.
It is the infinitely many larger than any given one that interests us
more.
> > > So
> > > you can do the proof by letting n start from 1 and run through all
> > > natural numbers.
A process which cannot end if it is to include all of them.
> > > At no n you have to consider infinitely many
> > > segments, because at no n you have infinitely many such segments.
But at no n have you exhausted all such segments either.
You presume some sort of stopping point in a process that cannot have a
stopping point.
> >
> > Indeed, *at no n*. You have shown that something is true for every
> > element
> > of N. However, you need to show that something is true
> > for N. Showing that something is true for every element of N is
> > not the same thing.
>
> Remember: I use "all n". Nothing else, nothing more.
Except that you have presumed an n for which no n+1 exists.
> >
> > No, it is obvious that N is covered by the union of all finite
> > initial segments. However for no n is
> > the union U[k =< n] {1, 2, 3, ..., k}
> > equal to the union of all finite segments.
>
> This is obviously a contradiction, because every initial segment {1,
> 2, 3, ..., k} covers all n from 1 to k.
But ignores the infinitely many more naturals from k+1 onwards.
> Further my proof does not suffer from needing
> infinitely many.
Only because at each step it ignores infinitely many.
>
> With a_n the n-th line of Cantor's list and d its antidiagonal number
> set theory accepts
> 1) forall n in N : a_n =/= d
> Set theory also accepts
> 2) forall n in N : E m > n (with nothing known about a_m)
> and even
> 3) forall n in N : {1, 2, 3, ..., n} is followed by infinitely many m
> in N.
> So
> 4) forall n in N : {1, 2, 3, ..., n} =/= N
> Therefore we have the problem that
> "forall n in N" does not generally mean forall n in N but at most in
> certain instances convenient for set theorists.
> Is there a general rule when "forall" means forall, or is it necessary
> to have studied 10 semesters set theory in order to know the correct
> procedure?
In each case above, and quite generally in mathematics, "for all" may be
replaced by "for each" with no change in meaning or truth.
Those less mathematically incompetent than WM are usually aware of this.
WM
> In article
> What you show about each natural, or even to each FISON, only applies to
> those individual naturals or FISONs, but not to the FISON N.
Why not?
>
> Your method assumes what it wants to prove, that there are only finitely
> many FISONs.
You are wrong. I consider all (infinitely many) FISONs. Here you can
learn why this is so:
I consider FISON {1, 2, 3, ..., n} and all smaller FISONs {1, 2,
3, ..., k} for k < n. The union of these FISONs is FISON {1, 2,
3, ..., n}. This is so for every union of finitely many FISONs. And in
fact there are not infinitely many FISONs for any n. But of course I
consider every n in N, and there is always the same result: The union
of all these FISONs U[k =< n] {1, 2, 3, ..., k} is one FISON, namley
FISON {1, 2, 3, ..., n}.
And now try to explain where in *this* infinite chain of unions we
have to jump from the reasonable result that a union of FISONs is the
largest FISON to your claim that a union of FISONs is not the largest
FISON. Or try to explain why my proof does not cover all n (and all
smaller natural numbers) in N. Or admit that set theory is
selfcontradictory.
>
> But at no n have you exhausted all such segments either.
> You presume some sort of stopping point in a process that cannot have a
> stopping point.
No. There is no stopping point. My proof holds for all n in N as
Cantor's proof holds for all n in N. His proof presumes the same sort
of stopping point as mine, namely none. Otherwise you should be able
to say at what points our proofs stop.
>
> > This is obviously a contradiction, because every initial segment {1,
> > 2, 3, ..., k} covers all n from 1 to k.
>
> But ignores the infinitely many more naturals from k+1 onwards.
So does Cantor's proof too.
>
> > Further my proof does not suffer from needing
> > infinitely many.
>
> Only because at each step it ignores infinitely many.
So does Cantor's proof too.
>
>
>
>
>
>
>
> > With a_n the n-th line of Cantor's list and d its antidiagonal number
> > set theory accepts
> > 1) forall n in N : a_n =/= d
> > Set theory also accepts
> > 2) forall n in N : E m > n (with nothing known about a_m)
> > and even
> > 3) forall n in N : {1, 2, 3, ..., n} is followed by infinitely many m
> > in N.
> > So
> > 4) forall n in N : {1, 2, 3, ..., n} =/= N
> > Therefore we have the problem that
> > "forall n in N" does not generally mean forall n in N but at most in
> > certain instances convenient for set theorists.
> > Is there a general rule when "forall" means forall, or is it necessary
> > to have studied 10 semesters set theory in order to know the correct
> > procedure?
>
> In each case above, and quite generally in mathematics, "for all" may be
> replaced by "for each" with no change in meaning or truth.
But not vice versa?
Regards, WM
WM
It does exist. But it may differ for different mathematical realities
of different persons. this strange form of existence is the only
chance to have a consistent concept of natural numbers. The reason is
this:
I consider FISON {1, 2, 3, ..., n} and all smaller FISONs {1, 2,
3, ..., k} for k < n. The union of these FISONs is FISON {1, 2,
3, ..., n}. This is so for every union of finitely many FISONs. And in
fact there are not infinitely many FISONs for any n. But of course I
consider every n in N, and there is always the same result: The union
of all these FISONs U[k =< n] {1, 2, 3, ..., k} is one FISON, namley
FISON {1, 2, 3, ..., n}.
And now try to explain where in *this* infinite chain of unions we
have to jump from the reasonable result that a union of FISONs is the
largest FISON to the claim that a union of FISONs is not the largest
FISON. Or try to explain why my proof does not cover all n (and all
smaller natural numbers) in N. Or admit that my concept is
unavoidable.
Or try to find another resolution.
Regards, WM
WM
> On Feb 4, 11:46 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > But all sets consisting of natural numbers only are of this form
> > {1,2,3,...,n}.
> > You seem to claim that N is not covered by a proof that is correct
> > forall n.
>
> > A n : U[k =< n] {1,2,3,...,k} is finite.
>
> If you claim to truly have a proof of:
>
> A n : U[k =< n] {1,2,3,...,k} is finite
>
> which holds for /all/ n,
Do you think this claim is wrong? Please give an n as a counter
example.
> then I would like to know what you mean, for
> example, by saying that "k is a natural and k =< n" when n = {f : f is
> a function from R^3 to C and for all x in R^3, |f(x)| < 15.5}.
>
I would not say that. Why should I?
Regards, WM
WM
In case you have not yet understood my proof, here it is again:
I consider FISON {1, 2, 3, ..., n} and all smaller FISONs {1, 2,
3, ..., k} for k < n. The union of these FISONs is FISON {1, 2,
3, ..., n}. This is so for every union of finitely many FISONs. And in
fact there are not infinitely many FISONs for any n. But of course I
consider every n in N, and there is always the same result: The union
of all these FISONs U[k =< n] {1, 2, 3, ..., k} is one FISON, namley
FISON {1, 2, 3, ..., n}.
And now try to explain where in *this* infinite chain of unions we
have to jump from the reasonable result that a union of FISONs is the
largest FISON to your claim that a union of FISONs is not the largest
FISON. Or try to explain why my proof does not cover all n (and all
smaller natural numbers) in N.
Regards, WM
Virgil
<f8c59100-fb6c-4390...@p20g2000yqi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 5 Feb., 09:29, Virgil <Vir...@gmale.com> wrote:
> > In article
>
> > What you show about each natural, or even to each FISON, only applies to
> > those individual naturals or FISONs, but not to the FISON N.
>
> Why not?
> >
>
> > Your method assumes what it wants to prove, that there are only finitely
> > many FISONs.
>
> You are wrong. I consider all (infinitely many) FISONs. Here you can
> learn why this is so:
If there are infinitely many FISONs then there are also infinitely many
naturals, a different natural as the terminal member of each different
FISON.
> And now try to explain where in *this* infinite chain of unions we
> have to jump from the reasonable result that a union of FISONs is the
> largest FISON to your claim that a union of FISONs is not the largest
> FISON.
Every FISON has a last element, but if there are infinitely many FISONs
with successively larger last elements, there is no last/largest FISON
with a last/largest last element.
> Or try to explain why my proof does not cover all n (and all
> smaller natural numbers) in N. Or admit that set theory is
> selfcontradictory.
Or admit that WM is all screwed up.
> >
> > But at no n have you exhausted all such segments either.
> > You presume some sort of stopping point in a process that cannot have a
> > stopping point.
>
> No. There is no stopping point. My proof holds for all n in N as
> Cantor's proof holds for all n in N.
For each of infinitely many n's in N, at least for Cantor.
His proof presumes the same sort
> of stopping point as mine, namely none. Otherwise you should be able
> to say at what points our proofs stop.
Your "proof" is false without a stop. To say that something holds for
each n in N, or for each FISON does not meqan that it holds for the set
of all n's or the set of all FISONs.
And once you concede the non-finiteness of the number of FISON's, as you
have done, you automatically concede the same for the number of naturals.
> > > With a_n the n-th line of Cantor's list and d its antidiagonal number
> > > set theory accepts
> > > 1) forall n in N : a_n =/= d
> > > Set theory also accepts
> > > 2) forall n in N : E m > n (with nothing known about a_m)
> > > and even
> > > 3) forall n in N : {1, 2, 3, ..., n} is followed by infinitely many m
> > > in N.
> > > So
> > > 4) forall n in N : {1, 2, 3, ..., n} =/= N
> > > Therefore we have the problem that
> > > "forall n in N" does not generally mean forall n in N but at most in
> > > certain instances convenient for set theorists.
> > > Is there a general rule when "forall" means forall, or is it necessary
> > > to have studied 10 semesters set theory in order to know the correct
> > > procedure?
> >
> > In each case above, and quite generally in mathematics, "for all" may be
> > replaced by "for each" with no change in meaning or truth.
>
> But not vice versa?
Only if one is a good deal more careful about interpreting "for all"
than WM is.
Virgil
<aeeb796e-9fca-4ef4...@p2g2000prf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 5 Feb., 02:22, David R Tribble <da...@tribble.com> wrote:
> > Virgil wrote:
> > >> Do you claim that N I one of its own initial segments?
> > >> If so, you should be able to name the largest member of N, or at least
> > >> prove that there is one.
> >
> > WM wrote:
> > > I cannot name it - among other reasons, because it is not fixed
> >
> > Is saying "it is not fixed" the same as saying "it does not exist"?
>
> It does exist.
WM's varying "it" has the problem of almost always being larger than
itself.
:
>
> I consider FISON {1, 2, 3, ..., n} and all smaller FISONs
It is primarily the larger ones which mathematics considers.
>
> And now try to explain where in *this* infinite chain of unions we
> have to jump from the reasonable result that a union of FISONs is the
> largest FISON
What evidence do you have that there is any such thing as a largest
FISON when for any given FISON one can name a larger one?
> to the claim that a union of FISONs is not the largest
> FISON.
Where is your proof that a union of FISONs is a FISON at all?
Or try to explain why my proof does not cover all n (and all
> smaller natural numbers) in N. Or admit that my concept is
> unavoidable.
>
> Or try to find another resolution.
I will take that other resolution, that there can be no largest FISON
nor largest natural until they can be explicitly named.
And I suspect that even your own unfortunate students will agree with me
on that point whenever you are unable to look over their shoulders.
Virgil
<5f4f8a89-3f84-4750...@k19g2000yqg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 5 Feb., 04:57, cbr...@cbrownsystems.com wrote:
> > On Feb 4, 11:46 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > But all sets consisting of natural numbers only are of this form
> > > {1,2,3,...,n}.
> > > You seem to claim that N is not covered by a proof that is correct
> > > forall n.
> >
> > > A n : U[k =< n] {1,2,3,...,k} is finite.
> >
> > If you claim to truly have a proof of:
> >
> > A n : U[k =< n] {1,2,3,...,k} is finite
> >
> > which holds for /all/ n,
>
> Do you think this claim is wrong? Please give an n as a counter
> example.
Any set of naturals with a known maximal member is known to be finite,
but there are sets of naturals not known to have a maximal member
because the assumption of its having a maximal member contradicts itself.
Virgil
<6f83dd60-3247-4121...@r38g2000vbi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
But you ignore all infinitely many larger ones.
The union of these FISONs is FISON {1, 2,
> 3, ..., n}.
And the union of the ones you have been ignoring is the infinite
non-fison, or IISON, N.
> This is so for every union of finitely many FISONs.
But that leaves out the vast majority of them.
> And in
> fact there are not infinitely many FISONs for any n.
There are infinitely many FISONS containing any n. All but finitely many
of them, in fact.
> But of course I
> consider every n in N, and there is always the same result: The union
> of all these FISONs U[k =< n] {1, 2, 3, ..., k} is one FISON, namley
> FISON {1, 2, 3, ..., n}.
Which still omits most FISONS and most naturals
>
> And now try to explain where in *this* infinite chain of unions we
> have to jump from the reasonable result that a union of FISONs is the
> largest FISON to your claim that a union of FISONs is not the largest
> FISON.
A union of finitely many FISONs is a FISON but the union of ALL FISON's
is an IISON, not a FISON.
> Or try to explain why my proof does not cover all n (and all
> smaller natural numbers) in N.
The problem is that, for some n, your non-proof ignores everything from
n+1 onwards.
WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 5 Feb., 09:29, Virgil <Vir...@gmale.com> wrote:
> > > In article
>
> > > What you show about each natural, or even to each FISON, only applies to
> > > those individual naturals or FISONs, but not to the FISON N.
>
> > Why not?
>
> > > Your method assumes what it wants to prove, that there are only finitely
> > > many FISONs.
>
> > You are wrong. I consider all (infinitely many) FISONs. Here you can
> > learn why this is so:
>
> If there are infinitely many FISONs then there are also infinitely many
> naturals, a different natural as the terminal member of each different
> FISON.
I do not claim that there is an actual infinitude of natural numbers.
But even if there is an actual infinitude, then my proof holds for all
of them. Do you oppose to this simple truth?
>
> > And now try to explain where in *this* infinite chain of unions we
> > have to jump from the reasonable result that a union of FISONs is the
> > largest FISON to your claim that a union of FISONs is not the largest
> > FISON.
>
> Every FISON has a last element, but if there are infinitely many FISONs
> with successively larger last elements, there is no last/largest FISON
> with a last/largest last element.
I do not assume that there is a last/largest FISON. I only use the
fact that every FISON has a last element and, therefore, there is only
a finitude of smaller FISONs. The result of my proof is that there is
a largest FISON. But I would not like to hear your opinion about this
result, but about the proof and where it disobeys logic.
Unless you can, your implication
"if there are infinitely many FISONs with successively larger last
elements, there is no last/largest FISON
with a last/largest last element"
shows that the axiom of actual infinity has been contradicted.
>
> > Or try to explain why my proof does not cover all n (and all
> > smaller natural numbers) in N. Or admit that set theory is
> > selfcontradictory.
>
> Or admit that WM is all screwed up.
Put your finger on the point where my proof lacks logic.
>
>
>
> > > But at no n have you exhausted all such segments either.
> > > You presume some sort of stopping point in a process that cannot have a
> > > stopping point.
>
> > No. There is no stopping point. My proof holds for all n in N as
> > Cantor's proof holds for all n in N.
>
> For each of infinitely many n's in N, at least for Cantor.
But not for me?
>
> His proof presumes the same sort
>
> > of stopping point as mine, namely none. Otherwise you should be able
> > to say at what points our proofs stop.
>
> Your "proof" is false without a stop. To say that something holds for
> each n in N, or for each FISON does not mean that it holds for the set
> of all n's or the set of all FISONs.
Therefore I do not only say "for each FISION" but for each FISON and
all smaller FISONs. Is there an n outside of these unions?
>
> And once you concede the non-finiteness of the number of FISON's, as you
> have done, you automatically concede the same for the number of naturals.
I do not assume that the number of FISONs is finite. I prove it.
>
>
>
>
>
> > > > With a_n the n-th line of Cantor's list and d its antidiagonal number
> > > > set theory accepts
> > > > 1) forall n in N : a_n =/= d
> > > > Set theory also accepts
> > > > 2) forall n in N : E m > n (with nothing known about a_m)
> > > > and even
> > > > 3) forall n in N : {1, 2, 3, ..., n} is followed by infinitely many m
> > > > in N.
> > > > So
> > > > 4) forall n in N : {1, 2, 3, ..., n} =/= N
> > > > Therefore we have the problem that
> > > > "forall n in N" does not generally mean forall n in N but at most in
> > > > certain instances convenient for set theorists.
> > > > Is there a general rule when "forall" means forall, or is it necessary
> > > > to have studied 10 semesters set theory in order to know the correct
> > > > procedure?
>
> > > In each case above, and quite generally in mathematics, "for all" may be
> > > replaced by "for each" with no change in meaning or truth.
>
> > But not vice versa?
>
> Only if one is a good deal more careful about interpreting "for all"
> than WM is
Such "careful interpretation" reminds me of theology. In my opinion
"all" means all. That's how I use it.
Regards, WM
Don Stockbauer
5 Argue about infinity GO TO 5
(potential, actual)
WM
> > I consider FISON {1, 2, 3, ..., n} and all smaller FISONs
>
> It is primarily the larger ones which mathematics considers.
>
Except in case a set theorists says "all n". Then the larger ones are
not of interest.
>
>
> > And now try to explain where in *this* infinite chain of unions we
> > have to jump from the reasonable result that a union of FISONs is the
> > largest FISON
>
> What evidence do you have that there is any such thing as a largest
> FISON when for any given FISON one can name a larger one?
My proof that shows that every set of FISONs is covered (can be
replaced) by the greatest FISON.
>
> > to the claim that a union of FISONs is not the largest
> > FISON.
>
> Where is your proof that a union of FISONs is a FISON at all?
I consider FISON {1, 2, 3, ..., n} and all smaller FISONs {1, 2,
3, ..., k} for k < n. The union of these FISONs is FISON {1, 2,
3, ..., n}. This is so for every union of finitely many FISONs. And in
fact there are not infinitely many FISONs for any n. But of course I
consider every n in N, and there is always the same result: The union
of all these FISONs U[k =< n] {1, 2, 3, ..., k} is one FISON, namley
FISON {1, 2, 3, ..., n}.
>
> > smaller natural numbers) in N. Or admit that my concept is
> > unavoidable.
>
> > Or try to find another resolution.
>
> I will take that other resolution, that there can be no largest FISON
> nor largest natural until they can be explicitly named.
That would imply that my proof repeated above is false. Pleas state
where that is the case in your opinion. I enumerated the steps for
your convenience:
1) I consider FISON {1, 2, 3, ..., n} and all smaller FISONs {1, 2,
3, ..., k} for k < n.
2) The union of these FISONs is FISON {1, 2, 3, ..., n}. This is so
for every union of finitely many FISONs.
3) And in fact there are not infinitely many FISONs for any n.
4) But of course I consider every n in N, and there is always the same
result:
5) The union of all these FISONs U[k =< n] {1, 2, 3, ..., k} is one
FISON, namley FISON {1, 2, 3, ..., n}.
>
> And I suspect that even your own unfortunate students will agree with me
> on that point whenever you are unable to look over their shoulders.
You are wrong. One week ago I got a letter from a mathematician,
personally unknown to me, who works together with a former student of
mine and who reported that this student is completely convinced of my
arguments on infinity. I have the impression that in an extended
correspondence I have also induced the mathematician to think it over
again.
And a fortnight ago I had the pleasure to see the exams of 27
students. Each and every one of them had understood the binary tree in
the correct way.
Say, why should my students try to follow your misleading theory? You
are unable to detect any error in my argument. You only state that the
result is in contradiction with your beloved axiom of infinity. That
is no reason for a sober mind to believe in set theory.
Regards, WM
WM
> > I consider FISON {1, 2, 3, ..., n} and all smaller FISONs {1, 2,
> > 3, ..., k} for k < n.
>
> But you ignore all infinitely many larger ones.
>
> The union of these FISONs is FISON {1, 2,
>
> > 3, ..., n}.
>
> And the union of the ones you have been ignoring is the infinite
> non-fison, or IISON, N.
Which FISON do I have ignored?
>
> > This is so for every union of finitely many FISONs.
>
> But that leaves out the vast majority of them.
Every union leaves out the majority?
>
> > And in
> > fact there are not infinitely many FISONs for any n.
>
> There are infinitely many FISONS containing any n.
That is not interesting for my proof.
> > But of course I
> > consider every n in N, and there is always the same result: The union
> > of all these FISONs U[k =< n] {1, 2, 3, ..., k} is one FISON, namley
> > FISON {1, 2, 3, ..., n}.
>
> Which still omits most FISONS and most naturals
For all n?
>
>
>
> > And now try to explain where in *this* infinite chain of unions we
> > have to jump from the reasonable result that a union of FISONs is the
> > largest FISON to your claim that a union of FISONs is not the largest
> > FISON.
>
> A union of finitely many FISONs is a FISON but the union of ALL FISON's
> is an IISON, not a FISON.
Please do not exclaim you weird ideas but discuss my proof.
>
> > Or try to explain why my proof does not cover all n (and all
> > smaller natural numbers) in N.
>
> The problem is that, for some n, your non-proof ignores everything from
> n+1 onwards
But I do not stop at some n! I claim that my proof covers all n. What
you claim is simply that it is impossible to treat all n, even if I
say that I don't not want to ignore any n. So either my proof covers
all n, or it is impossible to cover all n and therefore meaningless to
talk about all n.
Regards, WM
Dik T. Winter
> On 3 Feb., 22:42, David R Tribble <da...@tribble.com> wrote:
...
> > > So, if there should be an actually infinite set (different from a
> > > potentially infinite set that is always finite but not fixed), then it
> > > is forced to contain elements that are not subject of my theorem.
> >
> > Correction: If there should be an actually infinite set, then it is
> > forced to be of a form other than {1,2,3,...,n}, because all the
> > sets of this form are finite initial segments, and thus an actually
> > infinite set is not subject to your theorem.
>
> But all sets consisting of natural numbers only are of this form
> {1,2,3,...,n}.
No. There are sets of natural numbers that are *not* of that form. And N
is one of them.
> You seem to claim that N is not covered by a proof that is correct
> forall n.
If something is correct for all n, it is not necessarily correct for the
set of all n.
> A n : U[k =< n] {1,2,3,...,k} is finite.
Right. And U[k in N] {1, 2, 3, ..., k} is not of that form.
> I cannot imagine what you understand by (a set that has) "no last
> element". This "no last element" obviously is not a natural number.
It is nonsense to state such things about something that does not exist.
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
> On 3 Feb., 22:11, Virgil <Vir...@gmale.com> wrote:
...
> > Your "theorem" speaks only of finite intitial segments of N, so until
> > you prove that N is one of its own initial segments, you have failed to
> > prove anything about N itself.
>
> My theorem holds for every n, for every finite initial segments and
> for every union
Those are only unions of *finitely many* objects.
> Unless there is something in N that does not belong to such a union,
> my proof holds for N.
Each element of N belongs to such a union, but N itself is not such a union,
it is an *infinite* union.
> > Do you claim that N I one of its own initial segments?
>
> Of course. That is the most basic property of nartural numbers.
But not a finite initial segment.
> But the proof has been given:
> Consider all natural numbers by letting n run through 1, 2, 3, ...,
> n, ....
> Obviously this does not leave out any natural number.
Indeed. It holds for each natural number n. But if you proof something
about {1, 2, 3, ..., n} for each natural nuber n, this does not necessarily
hold for N itself, because N is not of that form.
> And obviously you will never get in trouble with infinity, if you only
> consider all segments that are not larger than {1, 2, 3, ..., n}.
Hm? So {1, 2, 3, ..., n+1} does not exist?
> But if this does not leave out any natural number, then it holds for
> all of them, doesn't it?
Yes, but not for the set of all of them.
> > But the successor rule for naturals makes such a proof impossible, so
> > that WM's arguments are refuted by the very nature of naturals.
>
> But if my proof does not leave out any initial segment of natural
> numbers, then it holds for all of them, doesn't it?
It does not leave out any finite initial segment, so it holds for all finite
initial segments, and N itself is not a finite initial segment.
> Otherwise you must claim that a proof that does not leave out any
> finite initial segment of natural numbers, does not hold for all
> natural numbers.
It holds for each finite initial segment, but not for N, because that is *not*
a finite initial segment.
> And that is obviously nonsense, if N is nothing but
> the assembly of all its finite initial segments.
Right, and is itself not a finite initial segment.