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Topic: Matheology � 261
Replies: 7   Last Post: May 10, 2013 11:39 PM

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Graham Cooper

Posts: 4,360
Registered: 5/20/10
Re: Matheology § 261
Posted: May 10, 2013 7:57 PM
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On May 10, 11:10 am, Zeit Geist <tucsond...@me.com> wrote:
> ...
> I somewhat agree with the above assessment by G & N.  Finding the solution and
> correcting the problem by avoidance are two different procedures.
> ...
> When Hilbert sought to formalize set theory, and thus all of mathematics, the project
> did not include all of logic but only that was mathematical.  This set, U, constructed
> from unions of powersets of union of powersets of unions of powersets of ... and then
> the union of all of that, must be everything "producable" from those operations.
> Since U is "producable", U must be a set. Hence, we can take the powerset of U and
> "produce" something not in U.

this is a naive copycat proof of |PS(N)| > |N|

before you said such a proof is irrelevant.



IF SET1 has 1 - then MYSET skips 1
IF SET1 skips 1 - then MYSET has 1

IF SET2 has 2 - then MYSET skips 2
IF SET2 skips 2 - then MYSET has 2

IF SET3 has 3 - then MYSET skips 3
IF SET3 skips 3 - then MYSET has 3

IF SET4 has 4 - then MYSET skips 4
IF SET4 skips 4 - then MYSET has 4

> Does this mean there is a problem with the theory? Not really, what is says is
> that U can not be treated as a set.  The problem only arises, when we treat U as
> A set.

Is there a predicate that defines U?

E(U) A(S) SeU

Can you formulaically infer values of set membership of U?


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