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| Guest |
 Posted: Sun Jun 18, 2006 12:16 pm |
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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.
This shows M to be uncountable.
Regards, WM |
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| Virgil |
| Posted: Sun Jun 18, 2006 1:30 pm |
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Guest
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In article <1150654604.323294.139860@f6g2000cwb.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:74f1b41cc8]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.
This shows M to be uncountable.
Regards, WM
[/quote:74f1b41cc8]
If M consists exactly of all finite subsets of }N, Meuckenh is wrong.
Using the 0 origin naturals for |N, every finite subset, S, of N
bijects with a unique natural number under the mapping
g:M -> |N: g(S) = sum_{x in S} 2^s
so that the inverse of the f function is precisely the function that
Muecknh has declared not to exist.
If one uses the 1 origin naturals, the function is
g(S) = sum_{x in S} 2^(s-1) |
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| Rupert |
| Posted: Sun Jun 18, 2006 5:19 pm |
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Guest
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mueckenh@rz.fh-augsburg.de wrote:
[quote:0eb045db73]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.
[/quote:0eb045db73]
You're using the notation "f" in two ways. First you're denying that a
function f with certain properties exists, then you're defining M in
terms of some fixed function f, which it's not clear what it is. Use a
different letter for the two things, and the define the function in
terms of which you're defining M.
[quote:0eb045db73]This shows M to be uncountable.
Regards, WM[/quote:0eb045db73] |
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| Guest |
| Posted: Mon Jun 19, 2006 1:43 am |
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Virgil schrieb:
[quote:13d2dec9c5]In article <1150654604.323294.139860@f6g2000cwb.googlegroups.com>,
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.
This shows M to be uncountable.
Regards, WM
If M consists exactly of all finite subsets of }N, Meuckenh is wrong.
[/quote:13d2dec9c5]
But it does not. The set M is uncountable while the set of all finite
subsets of |N is countable.
Regards, WM |
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| Guest |
| Posted: Mon Jun 19, 2006 1:49 am |
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Rupert schrieb:
[quote:2af5f2e5e6]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.
[/quote:2af5f2e5e6]
No.
[quote:2af5f2e5e6]First you're denying that a
function f with certain properties exists, then you're defining M in
terms of some fixed function f,
[/quote:2af5f2e5e6]
f is not fixed by any prescription.
[quote:2af5f2e5e6]which it's not clear what it is. Use a
different letter for the two things, and the define the function
[/quote:2af5f2e5e6]
f is not restricted by any definition. Any mapping f: |N --> M is
allowed.
[quote:2af5f2e5e6]in
terms of which you're defining M.
[/quote:2af5f2e5e6]
Let p and q be two natural numbers and let sqrt(2) = p/q.
Regards, WM |
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| Virgil |
| Posted: Mon Jun 19, 2006 2:47 am |
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Guest
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In article <1150703016.762443.286860@h76g2000cwa.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:09ba61e97d]Virgil schrieb:
In article <1150654604.323294.139860@f6g2000cwb.googlegroups.com>,
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.
This shows M to be uncountable.
Regards, WM
If M consists exactly of all finite subsets of }N, Meuckenh is wrong.
But it does not. The set M is uncountable while the set of all finite
subsets of |N is countable.
Regards, WM
[/quote:09ba61e97d]
Note that M depends on the particular f that has been chosen.
We can indicate that dependence by writing M_f.
Now for every f: |N --> M_f there is a bijection g N --> M_f.
And that is enough to prove every M_f countable. |
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| Guest |
| Posted: Mon Jun 19, 2006 3:43 am |
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Virgil schrieb:
[quote:7f3c129818]In article <1150703016.762443.286860@h76g2000cwa.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
Virgil schrieb:
In article <1150654604.323294.139860@f6g2000cwb.googlegroups.com>,
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.
This shows M to be uncountable.
Regards, WM
If M consists exactly of all finite subsets of }N, Meuckenh is wrong.
But it does not. The set M is uncountable while the set of all finite
subsets of |N is countable.
Regards, WM
Note that M depends on the particular f that has been chosen.
We can indicate that dependence by writing M_f.
[/quote:7f3c129818]
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 ?
[quote:7f3c129818]
Now for every f: |N --> M_f there is a bijection gN --> M_f.
And that is enough to prove every M_f countable.
[/quote:7f3c129818]
If you consider the mapping |N --> P(|N), there is the same remedy by
showing that the mapping f : |N --> P(|N) \ K cannot be used to prove
P(|N) uncountable.
Regards, WM |
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| Dik T. Winter |
| Posted: Mon Jun 19, 2006 4:44 am |
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Guest
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In article <1150654604.323294.139860@f6g2000cwb.googlegroups.com> mueckenh@rz.fh-augsburg.de writes:
[quote:83e3d2af1c]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.
[/quote:83e3d2af1c]
But is the set K in M? Pray give a proof.
--
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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| Guest |
| Posted: Mon Jun 19, 2006 6:07 am |
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Dik T. Winter schrieb:
[quote:220d576fee]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.
[/quote:220d576fee]
Of course it is, by definition, for without K M would not be M.
Regards, WM |
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| Virgil |
| Posted: Mon Jun 19, 2006 12:43 pm |
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Guest
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In article <1150710209.401911.130270@i40g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:109fda05f7]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 ?
[/quote:109fda05f7]
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 |
| Posted: Mon Jun 19, 2006 12:47 pm |
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Guest
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In article <1150718841.726873.223390@g10g2000cwb.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
[quote:3aa6be18ca]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
[/quote:3aa6be18ca]
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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| Rupert |
| Posted: Tue Jun 20, 2006 2:24 am |
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Guest
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mueckenh@rz.fh-augsburg.de wrote:
[quote:b6b9f183a5]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.
[/quote:b6b9f183a5]
Yep. In one of the occurrences it occurs preceded by a universal
quantifier, in the other it occurs as a constant symbol.
[quote:b6b9f183a5]
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.
[/quote:b6b9f183a5]
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:b6b9f183a5]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.
[/quote:b6b9f183a5]
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:b6b9f183a5]in
terms of which you're defining M.
Let p and q be two natural numbers and let sqrt(2) = p/q.
Regards, WM[/quote:b6b9f183a5] |
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| Guest |
| Posted: Tue Jun 20, 2006 2:38 pm |
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Virgil schrieb:
[quote:3035a37f6d]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.
[/quote:3035a37f6d]
Of course K exists for every mapping f (if |N exists). But the set {f,
k, K} is a paradox set. 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?
Regards, WM |
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| Guest |
| Posted: Tue Jun 20, 2006 2:41 pm |
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Virgil schrieb:
[quote:37ea322329]
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.
[/quote:37ea322329]
If the set {f, k, K} could actually exist, then M would be countable.
But it is not, isn't it?
[quote:37ea322329]
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.
[/quote:37ea322329]
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. Hence the set M is
uncountable. It is about the same as Cantors diagonal proof. The
diagonal number depends on all numbers of his list. Moreover, after
constructing the diagonal number, it turns out to belong to a countable
set. My M remains uncountable in any instance.
Regards, WM |
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| Guest |
| Posted: Tue Jun 20, 2006 2:58 pm |
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Rupert schrieb:
[quote:f466551485]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.
[/quote:f466551485]
Could you say what you mean?
[quote:f466551485]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:f466551485]
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:f466551485]
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)?
[/quote:f466551485]
No. K is that single set which belongs to the function f.
Regards, WM |
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