Jack Sarfatti remember that guy ?

JSH got gone in 2005

But that is when WM shows up in Google files

May 14, 2005, 11:41:01 AM
........................................................

CANTOR's theorem
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muec...@rz.fh-augsburg.de
unread,
May 14, 2005, 11:41:01 AM
to
All those naturals n, which are mapped upon sets S(n) which do not
contain n, form a set M. Call them non-generators. If e.g. 2 --> {1,3}
then 2 is a non-generator. It is in M. M is in P(M). Hence there must
be a natural m which is mapped upon M. M contains only the
non-generators. If m is not in M, then it is a non-generator, and m
must be included into M. But if m is in M, then it is not a
non-generator and it must be removed. This dilemma cannot be solved.
Therefore, no surjective mapping N --> P(N) is possible.

That is the reasoning of the Cantorians. But it is wrong. Not the
mapping is proven impossible but the set M cannot exist. It is an
impossible set.

Here is he proof. Consider the bijection P(N) --> P(N) which certainly
does not suffer from too small a cardinality on either side. For
instance consider the identical mapping id: P(N) --> P(N).
Introduce the condition that the set {a,b,c,...} of all non-generators
of the form {n} be the image of a set {m}. This condition which already
is implicite in the mapping N --> P(N) cannot be satisfied.

Examples:

If we have the identical mapping, then the set of non-generators is
empty:
{ } --> { }
{1} --> {1}
{1,2} --> {1,2}
....
In order to satisfy the condition, we can exchange the images of {1}
and {}:
{1} --> { }
{ } --> {1}
{1,2} --> {1,2}
....
Now 1 must be included in the (empty) set M. Obviously we have the same
dilemma as in the mapping N --> P(N).

If there is any other mapping with a non-empty set M of non-generators
of the form {n}, then it cannot be the image of a set of the form {m},
because m cannot be in M and m cannot be outside M. Obviously it is
impossible to satisfy our condition. And obviously this fact is
independent of the cardinality.

Regards, WM

On Thursday, 12 August 2021 at 23:15:41 UTC-3, Sergio wrote:
> Jack Sarfatti remember that guy ?
>
> JSH got gone in 2005
>
> But that is when WM shows up in Google files
>
> May 14, 2005, 11:41:01 AM
> .......................................................
>
> CANTOR's theorem
> 2 views
> muec...@rz.fh-augsburg.de's profile photo
> muec...@rz.fh-augsburg.de
> unread,
> May 14, 2005, 11:41:01 AM
> to
> All those naturals n, which are mapped upon sets S(n) which do not
> contain n, form a set M. Call them non-generators. If e.g. 2 --> {1,3}
> then 2 is a non-generator. It is in M. M is in P(M). Hence there must
> be a natural m which is mapped upon M. M contains only the
> non-generators. If m is not in M, then it is a non-generator, and m
> must be included into M. But if m is in M, then it is not a
> non-generator and it must be removed. This dilemma cannot be solved.
> Therefore, no surjective mapping N --> P(N) is possible.
>
> That is the reasoning of the Cantorians. But it is wrong. Not the
> mapping is proven impossible but the set M cannot exist. It is an
> impossible set.
>
> Here is he proof. Consider the bijection P(N) --> P(N) which certainly
> does not suffer from too small a cardinality on either side. For
> instance consider the identical mapping id: P(N) --> P(N).
> Introduce the condition that the set {a,b,c,...} of all non-generators
> of the form {n} be the image of a set {m}. This condition which already
> is implicite in the mapping N --> P(N) cannot be satisfied.
>
> Examples:
>
> If we have the identical mapping, then the set of non-generators is
> empty:
> { } --> { }
> {1} --> {1}
> {1,2} --> {1,2}
> ...
> In order to satisfy the condition, we can exchange the images of {1}
> and {}:
> {1} --> { }
> { } --> {1}
> {1,2} --> {1,2}
> ...
> Now 1 must be included in the (empty) set M. Obviously we have the same
> dilemma as in the mapping N --> P(N).
>
> If there is any other mapping with a non-empty set M of non-generators
> of the form {n}, then it cannot be the image of a set of the form {m},
> because m cannot be in M and m cannot be outside M. Obviously it is
> impossible to satisfy our condition. And obviously this fact is
> independent of the cardinality.

WM, ignoramus since 2005

Sometimes WM also proved countable sets to be uncountable:

> An uncountable countable set - 18.06.2006, 20:16:44
>
> 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
https://groups.google.com/g/sci.math/c/UnbIbO-_vcE/m/3EPHlC9_zHcJ

But P_fin(N) is countable, provable by Ackerman encoding for example:
https://en.wikipedia.org/wiki/Hereditarily_finite_set

Gus Gassmann schrieb am Freitag, 13. August 2021 um 13:23:05 UTC+2:
> On Thursday, 12 August 2021 at 23:15:41 UTC-3, Sergio wrote:
> > Jack Sarfatti remember that guy ?
> >
> > JSH got gone in 2005
> >
> > But that is when WM shows up in Google files
> >
> > May 14, 2005, 11:41:01 AM
> > .......................................................
> >
> > CANTOR's theorem
> > 2 views
> > muec...@rz.fh-augsburg.de's profile photo
> > muec...@rz.fh-augsburg.de
> > unread,
> > May 14, 2005, 11:41:01 AM
> > to
> > All those naturals n, which are mapped upon sets S(n) which do not
> > contain n, form a set M. Call them non-generators. If e.g. 2 --> {1,3}
> > then 2 is a non-generator. It is in M. M is in P(M). Hence there must
> > be a natural m which is mapped upon M. M contains only the
> > non-generators. If m is not in M, then it is a non-generator, and m
> > must be included into M. But if m is in M, then it is not a
> > non-generator and it must be removed. This dilemma cannot be solved.
> > Therefore, no surjective mapping N --> P(N) is possible.
> >
> > That is the reasoning of the Cantorians. But it is wrong. Not the
> > mapping is proven impossible but the set M cannot exist. It is an
> > impossible set.
> >
> > Here is he proof. Consider the bijection P(N) --> P(N) which certainly
> > does not suffer from too small a cardinality on either side. For
> > instance consider the identical mapping id: P(N) --> P(N).
> > Introduce the condition that the set {a,b,c,...} of all non-generators
> > of the form {n} be the image of a set {m}. This condition which already
> > is implicite in the mapping N --> P(N) cannot be satisfied.
> >
> > Examples:
> >
> > If we have the identical mapping, then the set of non-generators is
> > empty:
> > { } --> { }
> > {1} --> {1}
> > {1,2} --> {1,2}
> > ...
> > In order to satisfy the condition, we can exchange the images of {1}
> > and {}:
> > {1} --> { }
> > { } --> {1}
> > {1,2} --> {1,2}
> > ...
> > Now 1 must be included in the (empty) set M. Obviously we have the same
> > dilemma as in the mapping N --> P(N).
> >
> > If there is any other mapping with a non-empty set M of non-generators
> > of the form {n}, then it cannot be the image of a set of the form {m},
> > because m cannot be in M and m cannot be outside M. Obviously it is
> > impossible to satisfy our condition. And obviously this fact is
> > independent of the cardinality.
> WM, ignoramus since 2005

herpes boy at that time was also on acid, like he is today:

Ross A. Finlayson wrote in 28.09.2006, 17:31:14
> Cantor's (Georg Cantor, mathematician's) expectation
> that there is a universe, is called here the domain principle.
> In ZF, which is taught to a lot of people, there is no universe.
> So if you use a universe in a set theory, then it is not ZF.
https://groups.google.com/g/sci.math/c/fdSdl_dkmGE/m/-4zgkg5ItMYJ

Of course there is an universe in ZF, for example if you say:

forall x (x = {} v x ≠ {})

Your forall quantifier refers to the domain of discourse.
Maybe herpes boy was confused because the domain is not a set?

Mostowski Collapse schrieb am Freitag, 13. August 2021 um 13:46:56 UTC+2:
> Sometimes WM also proved countable sets to be uncountable:
>
> > An uncountable countable set - 18.06.2006, 20:16:44
> >
> > 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
> https://groups.google.com/g/sci.math/c/UnbIbO-_vcE/m/3EPHlC9_zHcJ
>
> But P_fin(N) is countable, provable by Ackerman encoding for example:
> https://en.wikipedia.org/wiki/Hereditarily_finite_set
> Gus Gassmann schrieb am Freitag, 13. August 2021 um 13:23:05 UTC+2:
> > On Thursday, 12 August 2021 at 23:15:41 UTC-3, Sergio wrote:
> > > Jack Sarfatti remember that guy ?
> > >
> > > JSH got gone in 2005
> > >
> > > But that is when WM shows up in Google files
> > >
> > > May 14, 2005, 11:41:01 AM
> > > .......................................................
> > >
> > > CANTOR's theorem
> > > 2 views
> > > muec...@rz.fh-augsburg.de's profile photo
> > > muec...@rz.fh-augsburg.de
> > > unread,
> > > May 14, 2005, 11:41:01 AM
> > > to
> > > All those naturals n, which are mapped upon sets S(n) which do not
> > > contain n, form a set M. Call them non-generators. If e.g. 2 --> {1,3}
> > > then 2 is a non-generator. It is in M. M is in P(M). Hence there must
> > > be a natural m which is mapped upon M. M contains only the
> > > non-generators. If m is not in M, then it is a non-generator, and m
> > > must be included into M. But if m is in M, then it is not a
> > > non-generator and it must be removed. This dilemma cannot be solved.
> > > Therefore, no surjective mapping N --> P(N) is possible.
> > >
> > > That is the reasoning of the Cantorians. But it is wrong. Not the
> > > mapping is proven impossible but the set M cannot exist. It is an
> > > impossible set.
> > >
> > > Here is he proof. Consider the bijection P(N) --> P(N) which certainly
> > > does not suffer from too small a cardinality on either side. For
> > > instance consider the identical mapping id: P(N) --> P(N).
> > > Introduce the condition that the set {a,b,c,...} of all non-generators
> > > of the form {n} be the image of a set {m}. This condition which already
> > > is implicite in the mapping N --> P(N) cannot be satisfied.
> > >
> > > Examples:
> > >
> > > If we have the identical mapping, then the set of non-generators is
> > > empty:
> > > { } --> { }
> > > {1} --> {1}
> > > {1,2} --> {1,2}
> > > ...
> > > In order to satisfy the condition, we can exchange the images of {1}
> > > and {}:
> > > {1} --> { }
> > > { } --> {1}
> > > {1,2} --> {1,2}
> > > ...
> > > Now 1 must be included in the (empty) set M. Obviously we have the same
> > > dilemma as in the mapping N --> P(N).
> > >
> > > If there is any other mapping with a non-empty set M of non-generators
> > > of the form {n}, then it cannot be the image of a set of the form {m},
> > > because m cannot be in M and m cannot be outside M. Obviously it is
> > > impossible to satisfy our condition. And obviously this fact is
> > > independent of the cardinality.
> > WM, ignoramus since 2005