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An uncountable countable set
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Virgil
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Joined: 24 Mar 2005
Posts: 5536

PostPosted: Mon Jun 19, 2006 6:43 pm    Post subject: Re: An uncountable countable set Reply with quote

In article <1150710209.401911.130270@i40g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Quote:
Virgil schrieb:


Note that M depends on the particular f that has been chosen.
We can indicate that dependence by writing M_f.

Oh, indeed? What is the number k mapped under f on the set K = {k e |N
: k /e f(k)} of this M_f ?

As soon as you tell us that M_f exists at all, YOU are telling us that
there must be such a set, K. The only other option is that there is no
such set K and no such set M_f and no such function f at all, in which
case the uncountable-countable-set becomes a non-existent
uncountable-countable-set.
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Virgil
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Joined: 24 Mar 2005
Posts: 5536

PostPosted: Mon Jun 19, 2006 6:47 pm    Post subject: Re: An uncountable countable set Reply with quote

In article <1150718841.726873.223390@g10g2000cwb.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Quote:
Dik T. Winter schrieb:

In article <1150654604.323294.139860@f6g2000cwb.googlegroups.com
mueckenh@rz.fh-augsburg.de writes:
An uncountable countable set

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

But is the set K in M? Pray give a proof.

Of course it is, by definition, for without K M would not be M.

Regards, WM

Give us an example of such an f, K_f and M_f, to show that such an f,
K_f anf M_f can actually exist.

If they can exist at all, it is easy enough to show that there is a
bijection beween |N and M_f, and if they can't then there is no M_f to
worry about.
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guenther.vonKnakspott@gmx
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Joined: 24 Apr 2005
Posts: 250

PostPosted: Mon Jun 19, 2006 6:54 pm    Post subject: Re: An uncountable countable set Reply with quote

Mueckenschleim wrote:
Quote:
Dik T. Winter schrieb:

In article <1150718841.726873.223390@g10g2000cwb.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:

Dik T. Winter schrieb:

In article <1150654604.323294.139860@f6g2000cwb.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
An uncountable countable set

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

But is the set K in M? Pray give a proof.

Of course it is, by definition, for without K M would not be M.

Ah, I see. The set M is defined depending on f. Well in that case f
is clearly not a bijection between N and M.

That is correct.

This does not tell us that
there is *no* bijection (say g) between N and M.

M depends on f. And when you take g, then M depends on g. That is the
essence of M.
Beautiful Mueckenschleim, really beautiful. This is you at your best! I

guess your admirer Blumstin must be in rapture.

Quote:
In order to show that
there is no bijection between N and M you are not allowed to change M
for each and every attempted mapping.

I do not change anything. I define K by exactly that mapping which is
assumed.

No, of course you don't. You can't even be consisten over a few lines
of text.

What is really disgusting is that this kook actually teaches at a
college and gets paid with the taxes earned by hard working people.

<snip crap>
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Virgil
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Joined: 24 Mar 2005
Posts: 5536

PostPosted: Mon Jun 19, 2006 7:01 pm    Post subject: Re: An uncountable countable set Reply with quote

In article <1150728187.749465.36710@g10g2000cwb.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Quote:
Dik T. Winter schrieb:

In article <1150718841.726873.223390@g10g2000cwb.googlegroups.com
mueckenh@rz.fh-augsburg.de writes:

Dik T. Winter schrieb:

In article <1150654604.323294.139860@f6g2000cwb.googlegroups.com
mueckenh@rz.fh-augsburg.de writes:
An uncountable countable set

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

But is the set K in M? Pray give a proof.

Of course it is, by definition, for without K M would not be M.

Ah, I see. The set M is defined depending on f. Well in that case f
is clearly not a bijection between N and M.

That is correct.

This does not tell us that
there is *no* bijection (say g) between N and M.

M depends on f. And when you take g, then M depends on g. That is the
essence of M.

Wrong! Given any two large sets are many functions for one to the other
and equality of cardinality depends on whether any one of them is a
bijection. Given two sets to compare, you cannot restrict things to only
one allowable function between them but must allow all possible
functions between them to be considered.
Quote:

In order to show that
there is no bijection between N and M you are not allowed to change M
for each and every attempted mapping.

I do not change anything. I define K by exactly that mapping which is
assumed.

Then there are other functions between N and M_f, one of which is a
bijection.
Quote:

Suppose f is a bijection between
N and S, where S is the set of finite subsets of N (such a bijection
does exist). Construct K(f) and next M(f). Clearly f is not a bijection
between N and M(f). However it is possible to construct a bijection
between N and M(f):
1. g(0) = K(f)
2. g(n) = f(n - 1) when n > 0.
So also M(f) is countable.

That is not at all the way a bijection has to be constructed between |N
and M. That is simply impossible!

That you do not like it does not at all mean that it is impossible.

There are two issues here. The first is whether a function of the sort
Meucken tries to create can exist at all. He has given no example of
such a function, nor proof of existence.

However if we allow the assumption that such a function f exists, it is
not hard to show that there are bijections between |N and M_f, though f
need not be one of them.
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Virgil
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Joined: 24 Mar 2005
Posts: 5536

PostPosted: Mon Jun 19, 2006 7:07 pm    Post subject: Re: An uncountable countable set Reply with quote

In article <1150728692.949790.258040@h76g2000cwa.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Quote:
Daryl McCullough schrieb:

mueckenh@rz.fh-augsburg.de says...

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

It's not exactly clear what you are talking about here. Do
you mean that M is the set of all finite subsets of N?

Above I spelled out clearly that M contains the set of all finite
subsets of |N and one more set K. What is unclear?

In
that case, there is definitely a bijection between M and N:

If M was the set of all finite subsets of N, that would be easy enough
to see.

If you let K = { k in N such that k is not an element of f(k) }
then we can see that

K = N

Such a mapping is possible with no doubt. There remains only one
question: what number is mapped on K?

That raises the question of whether such a set as M_f can exist at all.

If it cannot exist, then the issue of whether a non-existent set is
countable is irrelevant.

So that Meucken's first duty is to prove that a function having the sort
of codomain, M_f, he describes can exist at all.

If ever that is established, the construction of a bijection between it
and N is fairly trivial.
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Tim Peters
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Posts: 426

PostPosted: Mon Jun 19, 2006 10:53 pm    Post subject: Re: An uncountable countable set Reply with quote

[mueckenh@rz.fh-augsburg.de]
Quote:
An uncountable countable set

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

[Dik T. Winter]
Quote:
Ah, I see. The set M is defined depending on f. Well in that case f
is clearly not a bijection between N and M.

I'm not sure I'd be so generous here. K isn't a definite set before f is
specified (meaning that for K to provably exist by the axiom of separation,
f has to provably exist first), but to specify f K has to be a definite set
first (since K must be the image under f of some natural k, f can't be
proven to exist unless K can be proven to exist first).

IOW, I doubt this argument can be made in ZFC, just as it's not possible
under ZFC to prove that

F = {k e |N : k e F}
or
G = {k e |N : k /e G}

exist.

[mueckenh@rz.fh-augsburg.de]
Quote:
That is correct.

This does not tell us that there is *no* bijection (say g) between
N and M.

M depends on f. And when you take g, then M depends on g. That is the
essence of M.

And if Dik constructs g1, while I construct a distinct g2, the definition of
M frustrates both claimed bijections simultaneously? Cool ;-)

> ...
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W. Mueckenheim
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Joined: 23 Apr 2005
Posts: 934

PostPosted: Tue Jun 20, 2006 5:42 am    Post subject: Re: An uncountable countable set Reply with quote

Daryl McCullough schrieb:

Quote:
mueckenh@rz.fh-augsburg.de says...

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

Above I spelled out clearly that M contains the set of all finite
subsets of |N and one more set K. What is unclear?

Okay, so M is not a fixed set, but is a function of f. So to make
this precise, let's put in the functional dependence:

K(f) = { k in N | k is not an element of f(k) }
M(f) = { A in P(N) | A is finite, or A = K(f) }

where P(N) means the set of all subsets of N.

Then what you are saying is that

forall f: N -> P(N),
f is not a bijection between N and M(f)

That's true. But that doesn't mean that M(f) is uncountable.

If it is proven impossible to set up a mapping between M and |N, then M
is uncountable.
That is the definition of uncountability. Cp. Cantor's diagonal proof:
There it is even possible to set up a bijektion between {numbers of the
list & diagonal number}, (after the diagonal number has been
constructed) and |N. Nevertheless uncountablilty is claimed.

Quote:
We can prove

forall f: N -> P(N), exists g: N -> P(N)
g is a bijection between N and M(f)

No, it is not, because M(f) is incomplete without K. But K does not
exist. The reason for its non-existence is not lacking number of
elements of |N but an impredicable definition. I only wanted to point
out that the same impradicability is the reason Hessenberg' s proof
(non-surjectivity of f: |N --> P(|N)) seems to hold. It does not. In
fact, it has nothing to do with any number of elements of any set.

Regards, WM
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W. Mueckenheim
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Joined: 23 Apr 2005
Posts: 934

PostPosted: Tue Jun 20, 2006 5:46 am    Post subject: Re: An uncountable countable set Reply with quote

Virgil schrieb:


Quote:
For any given fNeutralN --> M_f there is a
h:M_f --> |N which is a bijection, and such an h is easy to constrruct.

No, it is not, because M_f is incomplete without K. But K does not
exist. The reason for its non-existence is not a lacking number of
elements of |N but an impredicable definition. I only wanted to point
out that the same impradicability is the reason why Hessenberg' s proof
(non-surjectivity of f: |N --> P(|N)) seems to hold. It does not. In
fact, it has nothing to do with any number of elements of any set.
Quote:


Now for any f: |N --> M_f, of all finite subsets of |N together with
K = {k e |N : k /e f(k)} as members:

(1) either K is finite, in which case
G = M_f and h = g is a bijection between M_f and |N

or

(2) K is not finite, in which case
we can define h so that h(K) = 0 and
for x in G h(x) = g(x) + 1.

Thus in any event, for every given f: |N --> M_f
there is an easily constricted bijection between |N and M_f.

The element K of M does not exist.

If it is proven impossible to set up a mapping between M and |N, then M
is uncountable.
That is the definition of uncountability. Cp. Cantor's diagonal proof:
There it is even possible to set up a bijektion between {numbers of the
list & diagonal number}, (after the diagonal number has been
constructed) and |N. Nevertheless uncountablilty is claimed.

Your arguing would establish countability for {numbers of the list &
diagonal number}, because the diagonal number increases the set only by
1 element.

Regards, WM
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Virgil
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Joined: 24 Mar 2005
Posts: 5536

PostPosted: Tue Jun 20, 2006 7:04 am    Post subject: Re: An uncountable countable set Reply with quote

In article <1150782120.872330.199120@c74g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Quote:
Daryl McCullough schrieb:

mueckenh@rz.fh-augsburg.de says...

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

Above I spelled out clearly that M contains the set of all finite
subsets of |N and one more set K. What is unclear?

Okay, so M is not a fixed set, but is a function of f. So to make
this precise, let's put in the functional dependence:

K(f) = { k in N | k is not an element of f(k) }
M(f) = { A in P(N) | A is finite, or A = K(f) }

where P(N) means the set of all subsets of N.

Then what you are saying is that

forall f: N -> P(N),
f is not a bijection between N and M(f)

That's true. But that doesn't mean that M(f) is uncountable.

If it is proven impossible to set up a mapping between M and |N, then M
is uncountable.

But you have not proven it impossible, you merely argue that ONE
function between N and M(f) is not a bijection.



WE have some f: N --> P(N), which we know is not a surjection since
there is no n in N for which one can have f(n) = K(f).

But that does no mean that there is no surjection between M(f) and N.

In fact, given a function f for which K(f) exists, it is not difficult
to build a bijection between M(f) and N as follows:

For each S in M(F) define g: M(f) --> N by:
If K is finite then
if 0 e N
then g(S) = sum_{s e S} 2^s
else g(S) = sum_{s e S} 2^(s-1)
endif
else (K being infinite)
g(K) = the first natural
and
if 0 e N
then g(S) = sum_{s e S} 2^s+1
else g(S) = sum_{s e S} 2^(s-1)+1
endif
endif

In all these cases, g will be a bijection from M(f) to N.



Quote:

We can prove

forall f: N -> P(N), exists g: N -> P(N)
g is a bijection between N and M(f)

No, it is not, because M(f) is incomplete without K. But K does not
exist.

But you have not proven it never exists, and for some functions, f, it
Does exist.

For example, let f(n) = {} for all n, then
K = {k in N : not k e f(k) } = N.

So that for SOME functions f, f, K(f) exists.

The reason that Meucken's argument fails here is that there is no
necessity for the function f to be a surjection onto M(f).

The desired bijection can be established by an entirely different
function between M(f) and N, as shown above.







Quote:
The reason for its non-existence is not lacking number of
elements of |N but an impredicable definition.

Wrong, as shown by a specific example above.
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Virgil
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Joined: 24 Mar 2005
Posts: 5536

PostPosted: Tue Jun 20, 2006 7:12 am    Post subject: Re: An uncountable countable set Reply with quote

In article <1150782414.064700.253320@p79g2000cwp.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Quote:
Virgil schrieb:


For any given fNeutralN --> M_f there is a
h:M_f --> |N which is a bijection, and such an h is easy to constrruct.

No, it is not, because M_f is incomplete without K. But K does not
exist.


That requires proof, and is false in at least one case.
Let f(n) = {} for all n, then K_f is quite well defined.


Quote:
The reason for its non-existence is not a lacking number of
elements of |N but an impredicable definition. I only wanted to point
out that the same impradicability is the reason why Hessenberg' s proof
(non-surjectivity of f: |N --> P(|N)) seems to hold. It does not. In
fact, it has nothing to do with any number of elements of any set.


Now for any f: |N --> M_f, of all finite subsets of |N together with
K = {k e |N : k /e f(k)} as members:

(1) either K is finite, in which case
G = M_f and h = g is a bijection between M_f and |N

or

(2) K is not finite, in which case
we can define h so that h(K) = 0 and
for x in G h(x) = g(x) + 1.

Thus in any event, for every given f: |N --> M_f
there is an easily constricted bijection between |N and M_f.

The element K of M does not exist.

That requires proof, and is false in at least one case.
Let f(n) = {} for all n, then K_f is quite well defined, and equals N.
Quote:

If it is proven impossible to set up a mapping between M and |N, then M
is uncountable.

Wrong!

If it is impossible to set up an injection from N to M_F or a surjection
from M_f to N, and the axiom of choice is assumed, only then is M_f
uncountable.

And if K does not exist, as Meucken asserts above, then neither does any
set which is required to have it as a member, so that no such set as M_f
nor any such function as f:N --> M_f exists either, and his whole
argument collapses.
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Rupert
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Joined: 18 May 2005
Posts: 372

PostPosted: Tue Jun 20, 2006 8:24 am    Post subject: Re: An uncountable countable set Reply with quote

mueckenh@rz.fh-augsburg.de wrote:
Quote:
Rupert schrieb:

mueckenh@rz.fh-augsburg.de wrote:
An uncountable countable set

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.


You're using the notation "f" in two ways.

No.

Yep. In one of the occurrences it occurs preceded by a universal
quantifier, in the other it occurs as a constant symbol.

Quote:

First you're denying that a
function f with certain properties exists, then you're defining M in
terms of some fixed function f,

f is not fixed by any prescription.


It doesn't make sense to talk about the set K={k e |N: k /e f(k)} unles
you've specified what f is.

Quote:
which it's not clear what it is. Use a
different letter for the two things, and the define the function

f is not restricted by any definition. Any mapping f: |N --> M is
allowed.


So you mean M is the set of all finite subsets of |N, together with all
sets of the form K={k e |N: k /e f(k)}, where f ranges over all
possible mappings |N->P(|N)? That set is clearly uncountable. There is
no problem there.

Quote:
in
terms of which you're defining M.

Let p and q be two natural numbers and let sqrt(2) = p/q.

Regards, WM
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W. Mueckenheim
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Joined: 23 Apr 2005
Posts: 934

PostPosted: Tue Jun 20, 2006 9:34 am    Post subject: Re: An uncountable countable set Reply with quote

Tim Peters schrieb:

Quote:
[mueckenh@rz.fh-augsburg.de]
An uncountable countable set

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

[Dik T. Winter]
Ah, I see. The set M is defined depending on f. Well in that case f
is clearly not a bijection between N and M.

I'm not sure I'd be so generous here. K isn't a definite set before f is
specified (meaning that for K to provably exist by the axiom of separation,
f has to provably exist first), but to specify f K has to be a definite set
first (since K must be the image under f of some natural k, f can't be
proven to exist unless K can be proven to exist first).

Of course f does not exist. But the reason has nothing at all to do
with cardinalities. This error, however, is the foundation of
Hessenberg's proof of Cantor's theorem.

A set is required which is the image of k if it is not the image of k.
A barber is required who shaves himself if he does not.
Quote:

IOW, I doubt this argument can be made in ZFC, just as it's not possible
under ZFC to prove that

F = {k e |N : k e F}
or
G = {k e |N : k /e G}

exist.

On the other hand, why should any mapping not exist? If you claim that
a surjective mapping |N --> P(|N) must contain the set K in any case,
then there is no arguing why my mapping should not exist.

Regards, WM
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W. Mueckenheim
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Joined: 23 Apr 2005
Posts: 934

PostPosted: Tue Jun 20, 2006 9:39 am    Post subject: Re: An uncountable countable set Reply with quote

Virgil schrieb:

Quote:
In article <1150782120.872330.199120@c74g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

Daryl McCullough schrieb:

mueckenh@rz.fh-augsburg.de says...

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

Above I spelled out clearly that M contains the set of all finite
subsets of |N and one more set K. What is unclear?

Okay, so M is not a fixed set, but is a function of f. So to make
this precise, let's put in the functional dependence:

K(f) = { k in N | k is not an element of f(k) }
M(f) = { A in P(N) | A is finite, or A = K(f) }

where P(N) means the set of all subsets of N.

Then what you are saying is that

forall f: N -> P(N),
f is not a bijection between N and M(f)

That's true. But that doesn't mean that M(f) is uncountable.

If it is proven impossible to set up a mapping between M and |N, then M
is uncountable.

But you have not proven it impossible, you merely argue that ONE
function between N and M(f) is not a bijection.



WE have some f: N --> P(N), which we know is not a surjection since
there is no n in N for which one can have f(n) = K(f).

But that does no mean that there is no surjection between M(f) and N.

In fact, given a function f for which K(f) exists, it is not difficult
to build a bijection between M(f) and N as follows:

For each S in M(F) define g: M(f) --> N by:
If K is finite then
if 0 e N
then g(S) = sum_{s e S} 2^s
else g(S) = sum_{s e S} 2^(s-1)
endif
else (K being infinite)
g(K) = the first natural
and
if 0 e N
then g(S) = sum_{s e S} 2^s+1
else g(S) = sum_{s e S} 2^(s-1)+1
endif
endif

In all these cases, g will be a bijection from M(f) to N.




We can prove

forall f: N -> P(N), exists g: N -> P(N)
g is a bijection between N and M(f)

No, it is not, because M(f) is incomplete without K. But K does not
exist.

But you have not proven it never exists, and for some functions, f, it
Does exist.

For example, let f(n) = {} for all n, then
K = {k in N : not k e f(k) } = N.

So that for SOME functions f, f, K(f) exists.

The reason that Meucken's argument fails here is that there is no
necessity for the function f to be a surjection onto M(f).

The desired bijection can be established by an entirely different
function between M(f) and N, as shown above.

Only if K(f) would exist.
Quote:







The reason for its non-existence is not lacking number of
elements of |N but an impredicable definition.

Wrong, as shown by a specific example above.

Your example ist wrong because K(f) does not exist.

Regards WM
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Dik T. Winter
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Joined: 25 Mar 2005
Posts: 1359

PostPosted: Tue Jun 20, 2006 11:55 am    Post subject: Re: An uncountable countable set Reply with quote

In article <1150728187.749465.36710@g10g2000cwb.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
Quote:
Dik T. Winter schrieb:
In article <1150718841.726873.223390@g10g2000cwb.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
....
An uncountable countable set

There is no bijective mapping f : |N --> M,
where M contains the set of all finite subsets of |N
and, in addition, the set K = {k e |N : k /e f(k)} of all natural
numbers k which are mapped on subsets not containing k.

But is the set K in M? Pray give a proof.

Of course it is, by definition, for without K M would not be M.

Ah, I see. The set M is defined depending on f. Well in that case f
is clearly not a bijection between N and M.

That is correct.

This does not tell us that
there is *no* bijection (say g) between N and M.

M depends on f. And when you take g, then M depends on g. That is the
essence of M.

In that case M is not a set. And in order to tell whether that is
uncountable or not you have to first define cardinality on such
non-sets.

Quote:
So also M(f) is countable.

That is not at all the way a bijection has to be constructed between |N
and M. That is simply impossible!

A bijection can be constructed, you only can not name if f.

Quote:
But if you like tricks of this kind, then you can biject |N and its
power set P(|N) \ K. Subsequently take
1. g(0) = K(f)
2. g(n) = f(n - 1) when n > 0
and we see that Cantor's theorem is not proven by Hessenberg's proof,
because this proof makes use of the impredicable definition of a set
which does not and cannot exist, as well as the barber who shaves all
men of his village.

Sorry, this is plain nonsense. You can not construct a bijection between
P(N) \ K. The difference between this and your construction with M(f)
is that the target M(f) depends on the bijection you are constructing.
P(N) is a properly defined set that does not depend on anything but N.
But you are wrong. For every mapping f, the set K does exist (and is
an element of P(N), which does not depend on f), but it is not in the
image of f, which it should be if f is a bijection.

Quote:
Of course, one may define a mapping from |N on P(|N), for instance n
--> {n}, where K is well defined. But the set {f, k, K}, essential for
Hessenberg's proof, does not exist. This shows that the customary
proof of Cantor's theorem makes use of a set which does not exist,
namely {f, k, K}. Therefore, its non-existence does not prove anything
about surjectivity of mappings.

The triple {f, k, K} does indeed not exist, but if f is a bijection it
should exist. It is just this contradiction that shows that f is not
a bijection.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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Dik T. Winter
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Joined: 25 Mar 2005
Posts: 1359

PostPosted: Tue Jun 20, 2006 12:16 pm    Post subject: Re: An uncountable countable set Reply with quote

In article <1150796061.943937.27660@p79g2000cwp.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
....
Quote:
[Dik T. Winter]
Ah, I see. The set M is defined depending on f. Well in that case f
is clearly not a bijection between N and M.

I'm not sure I'd be so generous here. K isn't a definite set before f is
specified (meaning that for K to provably exist by the axiom of separation,
f has to provably exist first), but to specify f K has to be a definite set
first (since K must be the image under f of some natural k, f can't be
proven to exist unless K can be proven to exist first).

Of course f does not exist. But the reason has nothing at all to do
with cardinalities. This error, however, is the foundation of
Hessenberg's proof of Cantor's theorem.

Oh, but I think I disagree with Tim. Given any f: N -> S where S is the
set of finite subsets of N, K(f) and M(f) do exist.

Quote:
A set is required which is the image of k if it is not the image of k.
A barber is required who shaves himself if he does not.

But here is your confusion.
1. Given a mapping f: N -> P(N), the set K(f) constructed according to
Hessenberg *does* exist. If f were a bijection it is required (by
the *definition* of bijection) that K(f) (because it is an element
of P(N)) is the image of some element of N, but it is not, showing
that f is not a bijection. P(N) does not depend on f, so there is
no mapping between N and P(N) that is a bijection.
2. Given a mapping f: N -> S, the sets K(f) and M(f) do exist. If f
were a bijection between N and M(f), K(f) should be in the image,
it is not, so f is not a bijection. From this you can *not*
conclude that M(f) is uncountable, because there can be a bijection
g: N -> M(f). You can at most conclude that the union of *all* M(f)'s
is uncountable. And that is easy, that union is just P(N).

Quote:
On the other hand, why should any mapping not exist? If you claim that
a surjective mapping |N --> P(|N) must contain the set K in any case,
then there is no arguing why my mapping should not exist.

You mapping does exist. See (2) above.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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