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| Virgil |
 Posted: Tue Jun 20, 2006 7:16 pm |
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Guest
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In article <1150835899.493068.116800@g10g2000cwb.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:67a4ff7f25]Virgil schrieb:
In article <1150710209.401911.130270@i40g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
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.
Of course K exists for every mapping f (if |N exists). But the set {f,
k, K} is a paradox set.
[/quote:67a4ff7f25]
If K_f exists it is not paradoxical and if it is paradoxical, it does
not exist. The paradox occurs if one has any function f with domain N
and image some subset of P(N) containing K_f.
If the image of f is not required to contain K_f, there re no problems.
[quote:67a4ff7f25]It is not suitable to show that a bijection |N
--> {all finite subsets of P(|N) plus one further set} does not exist.
Why do you believe it could be capable of showing that |N --> P(|N)
does not exist?
[/quote:67a4ff7f25]
It is your example!
Work it out for yourself why f(n) = {m in N: m not in f(m)} doesn't work. |
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| Virgil |
| Posted: Tue Jun 20, 2006 7:23 pm |
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Guest
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In article <1150836068.607034.205340@y41g2000cwy.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:d1ae1ad75e]Virgil schrieb:
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 the set {f, k, K} could actually exist, then M would be countable.
But it is not, isn't it?
[/quote:d1ae1ad75e]
If M_f exists at all it is countable, and if it doesn't exist the
question of its countability is disappears.
[quote:d1ae1ad75e]
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.
In order to define K_f you first have map all natural numbers n of |N
on the finite subsets of |N. Then consider K_f. But there is no element
k of |N remaining, which could be mapped on K.
[/quote:d1ae1ad75e]
Then try a different mapping, say g:N --> M_f.
I have previously shown that if K_f and M_f exist for a non-bijective
f:N --> M_f, then M_f must be countable.
I have also shown that if f itself is required to be a bijection neither
it not K_f nor M_f even exist, so the question of countability of a
non-set is nonsense. |
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| Virgil |
| Posted: Tue Jun 20, 2006 8:08 pm |
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Guest
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In article <1150837129.824319.182500@i40g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:06497029a4]Rupert schrieb:
mueckenh@rz.fh-augsburg.de wrote:
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.
Could you say what you mean?
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.
f is not fixed by any specification.
[/quote:06497029a4]
its domain, codomain and range have been specified, but that codomain
and range have been define circularly in a way which prohibits the
existence of an F, K_f and M_f satisfying those conditions.
The easiest condition to relax is the one requiring that f be a
bijection, in particular that it have K_f as a value.
If this condition is dropped then, and only then, can an {f,K_f, M_f} as
described actually exist, and then M_f is easily countable.
[quote:06497029a4]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)?
No. K is that single set which belongs to the function f.
[/quote:06497029a4]
Thus you need to have that there to exist an n in N such that
f(n) = {x in N: x not in f(x)} = K_f
But if that n is in K_f then n is not in f(n) = K_f
and if it is not in K_f then it is automatically in f(n) = K_f
so that no such n can exist such that f(n) = K_f. |
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| Rupert |
| Posted: Wed Jun 21, 2006 12:14 am |
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mueckenh@rz.fh-augsburg.de wrote:
[quote:30d6957798]Rupert schrieb:
mueckenh@rz.fh-augsburg.de wrote:
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.
Could you say what you mean?
[/quote:30d6957798]
Are you familiar with existential quantifiers and universal
quantifiers? You are making a statement of the form "There does not
exist an f such that...", or, alternatively "For all f it is not the
case that..." Then later on you use f to refer to a specific function.
You should use different letters on these two occasions.
[quote:30d6957798]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.
f is not fixed by any specification. Therefore my arguing holds for any
f. K is that subset of |N which contains all natural numbers which are
not mapped under f on sets containing themselves.
[/quote:30d6957798]
Okay. So let me have a shot in translating what you've said into
something coherent.
"Let f be an arbitrary function. There does not exist a bijective
mapping g:N->M, where M is the set consisting of all the finite subsets
of N together with the set K={k e N: k /e f(k)}."
This is false. There always will exist such a bijection g. It will of
course be different to f.
[quote:30d6957798]
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)?
No. K is that single set which belongs to the function f.
Regards, WM[/quote:30d6957798] |
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| Rupert |
| Posted: Wed Jun 21, 2006 12:17 am |
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mueckenh@rz.fh-augsburg.de wrote:
[quote:f339899a9c]Rupert schrieb:
mueckenh@rz.fh-augsburg.de wrote:
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.
Could you say what you mean?
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.
f is not fixed by any specification. Therefore my arguing holds for any
f. K is that subset of |N which contains all natural numbers which are
not mapped under f on sets containing themselves.
[/quote:f339899a9c]
An alternative interpretation for what you're trying to say would be
"It is not the case that there exists a set M of subsets of N and a
bijection f:N->M, such that M consists of all the finite subsets of N
together with the set K={k e N: k /e f(k)}."
This is quite correct, but it does not amount to identifying an
uncountable countable set. What you've done is observe that there does
not exist a pair (M,f) with certain properties. To identify an
uncountable set you have to first identify a set M and then observe
that there does not exist an f with certain properties.
[quote:f339899a9c]
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)?
No. K is that single set which belongs to the function f.
Regards, WM[/quote:f339899a9c] |
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| Virgil |
| Posted: Wed Jun 21, 2006 12:45 am |
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Guest
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In article <1150870468.791288.14540@p79g2000cwp.googlegroups.com>,
"Rupert" <rupertmccallum@yahoo.com> wrote:
[quote:3c162c6719]mueckenh@rz.fh-augsburg.de wrote:
Rupert schrieb:
mueckenh@rz.fh-augsburg.de wrote:
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.
Could you say what you mean?
Are you familiar with existential quantifiers and universal
quantifiers? You are making a statement of the form "There does not
exist an f such that...", or, alternatively "For all f it is not the
case that..." Then later on you use f to refer to a specific function.
You should use different letters on these two occasions.
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.
f is not fixed by any specification. Therefore my arguing holds for any
f. K is that subset of |N which contains all natural numbers which are
not mapped under f on sets containing themselves.
Okay. So let me have a shot in translating what you've said into
something coherent.
"Let f be an arbitrary function. There does not exist a bijective
mapping g:N->M, where M is the set consisting of all the finite subsets
of N together with the set K={k e N: k /e f(k)}."
This is false. There always will exist such a bijection g. It will of
course be different to f.
[/quote:3c162c6719]
Except that if the function is require to have K as a value, the f and K
and M cannot exist:
if f(n) = K = {k e N: k /e f(k)} is n a member of K or not?
[quote:3c162c6719]
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)?
No. K is that single set which belongs to the function f.
Regards, WM[/quote:3c162c6719] |
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| Guest |
| Posted: Wed Jun 21, 2006 1:55 am |
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Rupert schrieb:
[quote:3b1cc4ccc8]Are you familiar with existential quantifiers and universal
quantifiers? You are making a statement of the form "There does not
exist an f such that...", or, alternatively "For all f it is not the
case that..." Then later on you use f to refer to a specific function.
You should use different letters on these two occasions.
[/quote:3b1cc4ccc8]
No. "For all f it is not the case" means the same as "there does not
exist any f for which it is the case". A f (f =/= surj) <==> ~E f (f =
sur).
And I let f be any function you like.
[quote:3b1cc4ccc8]
"Let f be an arbitrary function. There does not exist a bijective
mapping g:N->M, where M is the set consisting of all the finite subsets
of N together with the set K={k e N: k /e f(k)}."
[/quote:3b1cc4ccc8]
That is what I said.
[quote:3b1cc4ccc8]
This is false. There always will exist such a bijection g. It will of
course be different to f.
[/quote:3b1cc4ccc8]
The same is true is you take Hessenberg's mapping of f:N --> P(N). It
is not surjective because of K(f). But if you define g(n+1) = f(n) and
g(1) = K(f), the proof is no longer a proof.
Regards, WM |
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| Rupert |
| Posted: Wed Jun 21, 2006 2:02 am |
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Guest
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Virgil wrote:
[quote:44555cf1dd]In article <1150870468.791288.14540@p79g2000cwp.googlegroups.com>,
"Rupert" <rupertmccallum@yahoo.com> wrote:
mueckenh@rz.fh-augsburg.de wrote:
Rupert schrieb:
mueckenh@rz.fh-augsburg.de wrote:
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.
Could you say what you mean?
Are you familiar with existential quantifiers and universal
quantifiers? You are making a statement of the form "There does not
exist an f such that...", or, alternatively "For all f it is not the
case that..." Then later on you use f to refer to a specific function.
You should use different letters on these two occasions.
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.
f is not fixed by any specification. Therefore my arguing holds for any
f. K is that subset of |N which contains all natural numbers which are
not mapped under f on sets containing themselves.
Okay. So let me have a shot in translating what you've said into
something coherent.
"Let f be an arbitrary function. There does not exist a bijective
mapping g:N->M, where M is the set consisting of all the finite subsets
of N together with the set K={k e N: k /e f(k)}."
This is false. There always will exist such a bijection g. It will of
course be different to f.
Except that if the function is require to have K as a value, the f and K
and M cannot exist:
if f(n) = K = {k e N: k /e f(k)} is n a member of K or not?
[/quote:44555cf1dd]
This is a perfectly good proof that K cannot be in the range of f. But
as far as I can tell, there would still be a bijection g:N->M, and K
would be in the range of g. g would of course be different from f. No?
[quote:44555cf1dd]
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)?
No. K is that single set which belongs to the function f.
Regards, WM[/quote:44555cf1dd] |
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| Rupert |
| Posted: Wed Jun 21, 2006 2:05 am |
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mueckenh@rz.fh-augsburg.de wrote:
[quote:610948005b]Rupert schrieb:
Are you familiar with existential quantifiers and universal
quantifiers? You are making a statement of the form "There does not
exist an f such that...", or, alternatively "For all f it is not the
case that..." Then later on you use f to refer to a specific function.
You should use different letters on these two occasions.
No. "For all f it is not the case" means the same as "there does not
exist any f for which it is the case". A f (f =/= surj) <==> ~E f (f =
sur).
[/quote:610948005b]
I know.
[quote:610948005b]And I let f be any function you like.
[/quote:610948005b]
The question is, is the second occurrence of f within the scope of the
quantifier? I have given you two interpretations of what you said, one
on the assumption that the second occurrence of f is not within the
scope of the quantifier, and the other on the assumption that it is.
And I have responded to what you said on both interpretations.
[quote:610948005b]
"Let f be an arbitrary function. There does not exist a bijective
mapping g:N->M, where M is the set consisting of all the finite subsets
of N together with the set K={k e N: k /e f(k)}."
That is what I said.
This is false. There always will exist such a bijection g. It will of
course be different to f.
The same is true is you take Hessenberg's mapping of f:N --> P(N). It
is not surjective because of K(f). But if you define g(n+1) = f(n) and
g(1) = K(f), the proof is no longer a proof.
[/quote:610948005b]
Yes, it is. For each mapping f:N->P(N), you have a bijection g:N->M. If
you change the mapping, you've got to change the bijection. But there
still is one. The proof still works.
> Regards, WM |
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| Guest |
| Posted: Wed Jun 21, 2006 7:41 am |
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Rupert schrieb:
[quote:3a9fe1b568]Except that if the function is require to have K as a value, the f and K
and M cannot exist:
if f(n) = K = {k e N: k /e f(k)} is n a member of K or not?
This is a perfectly good proof that K cannot be in the range of f. But
as far as I can tell, there would still be a bijection g:N->M, and K
would be in the range of g. g would of course be different from f. No?
[/quote:3a9fe1b568]
2) 0.12324389
3) 0.23123123
4) 0.85348714
...
1) 0.244...
This is a perfectly good proof, that a surjective mapping g: |N -->
[0,1] does exist, isn't it?
Regards, WM |
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| Virgil |
| Posted: Wed Jun 21, 2006 2:00 pm |
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Guest
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In article <1150876556.902172.6800@i40g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:a27ad61551]Rupert schrieb:
Are you familiar with existential quantifiers and universal
quantifiers? You are making a statement of the form "There does not
exist an f such that...", or, alternatively "For all f it is not the
case that..." Then later on you use f to refer to a specific function.
You should use different letters on these two occasions.
No. "For all f it is not the case" means the same as "there does not
exist any f for which it is the case". A f (f =/= surj) <==> ~E f (f =
sur).
And I let f be any function you like.
[/quote:a27ad61551]
Let it be any function which does not have K_f as a value.
[quote:a27ad61551]
"Let f be an arbitrary function. There does not exist a bijective
mapping g:N->M, where M is the set consisting of all the finite subsets
of N together with the set K={k e N: k /e f(k)}."
That is what I said.
[/quote:a27ad61551]
Then f is not arbitrary, since you insist on restricting the function f
to have some n in N such that f(n) = {x in n: x not in f(x)}.
With that restriction, no such f can exist, absent that restriction,
lots of them do. |
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| Virgil |
| Posted: Wed Jun 21, 2006 2:03 pm |
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Guest
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In article <1150897288.785127.87000@g10g2000cwb.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:52a0e772c3]Rupert schrieb:
Except that if the function is require to have K as a value, the f and K
and M cannot exist:
if f(n) = K = {k e N: k /e f(k)} is n a member of K or not?
This is a perfectly good proof that K cannot be in the range of f. But
as far as I can tell, there would still be a bijection g:N->M, and K
would be in the range of g. g would of course be different from f. No?
2) 0.12324389
3) 0.23123123
4) 0.85348714
..
1) 0.244...
This is a perfectly good proof, that a surjective mapping g: |N --
[0,1] does exist, isn't it?
[/quote:52a0e772c3]
No. And it totally ignores the fact that Muecken's alleged example of a
n uncountable countable set is fatally flawed, as is shown above. |
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| Daryl McCullough |
| Posted: Wed Jun 21, 2006 4:24 pm |
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Guest
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In article <1150876556.902172.6800@i40g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de says...
[quote:2988aa54df]
Rupert schrieb:
Are you familiar with existential quantifiers and universal
quantifiers? You are making a statement of the form "There does not
exist an f such that...", or, alternatively "For all f it is not the
case that..." Then later on you use f to refer to a specific function.
You should use different letters on these two occasions.
No. "For all f it is not the case" means the same as "there does not
exist any f for which it is the case". A f (f =/= surj) <==> ~E f (f =
sur).
And I let f be any function you like.
"Let f be an arbitrary function. There does not exist a bijective
mapping g:N->M, where M is the set consisting of all the finite subsets
of N together with the set K={k e N: k /e f(k)}."
That is what I said.
This is false. There always will exist such a bijection g. It will of
course be different to f.
The same is true is you take Hessenberg's mapping of f:N --> P(N). It
is not surjective because of K(f). But if you define g(n+1) = f(n) and
g(1) = K(f), the proof is no longer a proof.
Regards, WM
[/quote:2988aa54df] |
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| Daryl McCullough |
| Posted: Wed Jun 21, 2006 4:35 pm |
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mueckenh@rz.fh-augsburg.de says...
[quote:1c9978648d]Rupert schrieb:
"Let f be an arbitrary function. There does not exist a bijective
mapping g:N->M, where M is the set consisting of all the finite subsets
of N together with the set K={k e N: k /e f(k)}."
That is what I said.
This is false. There always will exist such a bijection g. It will of
course be different to f.
The same is true is you take Hessenberg's mapping of f:N --> P(N). It
is not surjective because of K(f). But if you define g(n+1) = f(n) and
g(1) = K(f), the proof is no longer a proof.
[/quote:1c9978648d]
Sure it is. f is not a surjection from N -> P(N), because
K(f) is not in the image of f. Similarly, g is not a surjection
from N to P(N) because K(g) is not in the image of g.
Let's go through this one more time.
Let f be any function from N to P(N).
Let image(f) =
{ A in P(N) | exists x in N: f(x) = A }
Let K(f) =
{ x in N | x is not in f(x) }
Let M(f) =
{ A in P(N) | A is in image(f) or A = K(f) }
Finally, let g be defined by
g(0) = K(f)
g(x+1) = f(x)
Okay, so we have the following facts:
1. f is not a surjection from N to M(f)
2. f is not a surjection from N to P(N)
3. g is a surjection from N to M(f).
4. g is not a surjection from N to M(g).
5. g is not a surjection from N to P(N).
So, M(f) is countable, since there is a surjection g from
N to M(f). But there is no surjection from N to P(N). So
P(N) is *not* countable.
--
Daryl McCullough
Ithaca, NY |
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| Rupert |
| Posted: Wed Jun 21, 2006 5:33 pm |
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Guest
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mueckenh@rz.fh-augsburg.de wrote:
[quote:9913ca3332]Rupert schrieb:
Except that if the function is require to have K as a value, the f and K
and M cannot exist:
if f(n) = K = {k e N: k /e f(k)} is n a member of K or not?
This is a perfectly good proof that K cannot be in the range of f. But
as far as I can tell, there would still be a bijection g:N->M, and K
would be in the range of g. g would of course be different from f. No?
2) 0.12324389
3) 0.23123123
4) 0.85348714
..
1) 0.244...
This is a perfectly good proof, that a surjective mapping g: |N --
[0,1] does exist, isn't it?
[/quote:9913ca3332]
No, that's absolute nonsense.
> Regards, WM |
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