On May 11, 10:56 am, Zeit Geist <tucsond...@me.com> wrote: > On Friday, May 10, 2013 4:57:08 PM UTC-7, Graham Cooper wrote: > > 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. > > > CANTORS POWERSET PROOF > > > | CARDINALITY | > | INFINITY | > > > IF SET1 has 1 - then MYSET skips 1 > > > or > > > IF SET1 skips 1 - then MYSET has 1 > > > AND > > > IF SET2 has 2 - then MYSET skips 2 > > > or > > > IF SET2 skips 2 - then MYSET has 2 > > > AND > > > IF SET3 has 3 - then MYSET skips 3 > > > or > > > IF SET3 skips 3 - then MYSET has 3 > > > AND > > > IF SET4 has 4 - then MYSET skips 4 > > > or > > > 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? > > > Herc > > Actually it's a prof the "set of all sets" is not a set. > If it were it would to a contradiction. > > ZG
No it doesn't.
It's just p(X)<->true
in Naive Set theory.
E(S) A(X) XeS <-> p(X)
All your contradictions are self imposed and lead to worse atrocities such as E(X) X>oo