Petra
Posts:
8
Registered:
5/7/12
|
|
Re: 10^7-ZFC
Posted:
May 4, 2012 9:58 PM
|
|
On May 5, 12:36 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On May 5, 8:34 am, MoeBlee <modem...@gmail.com> wrote:
>
> > On May 4, 4:22 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > On May 5, 3:41 am, MoeBlee <modem...@gmail.com> wrote:
> > > > ZFC proves that there are no X and Y such XeYeX.
>
> > > No it doesn't.
>
> > ZFC includes the axiom of regularity, and so ZFC proves that there are
> > no X and Y such that XeYeX.
>
> > Indeed ZFC proves that there is no finite sequence x1...xn such that
> > x1 e x2 ... e xn e x1.
>
> > MoeBlee
>
> Nope!
>
> http://en.wikipedia.org/wiki/Axiom_of_regularity
>
> No infinite descending sequence of sets exists
> Suppose, to the contrary, that there is a function, f, on the natural
> numbers with f(n+1) an element of f(n) for each n. Define S = {f(n): n
> a natural number}, the range of f, which can be seen to be a set from
> the axiom schema of replacement. Applying the axiom of regularity to
> S, let B be an element of S which is disjoint from S. By the
> definition of S, B must be f(k) for some natural number k. However, we
> are given that f(k) contains f
have a cup of tea..
(k+1) which is also an element of S. So
> f(k+1) is in the intersection of f(k) and S. This contradicts the fact
> that they are disjoint sets. Since our supposition led to a
> contradiction, there must not be any such function, f.
>
> ****
>
> This is not proof of what you are claiming!
>
> Classic programming mistake!
>
> AOR stipulates a break in the transitive chain, not a condition of
> directed acyclic graphs.
>
> AGAIN: you need a TRANSITIVE ELEMENT Relation.
>
> A(X,Z) XtZ <-> E(Y) XeY ^ YtZ
>
> Then you can define Regularity using that!
>
> Herc
|
|
