Fun with antifoundation

Throw out the axiom of foundation and instead assume the axiom of antifoundation, because otherwise what I'm about to do is totally, totally, invalid, for obvious reasons.

Let Ω be the unique set st Ω={Ω}. (Like I said, there goes the axiom of foundation.)

Then note that (Ω,Ω)={Ω,{Ω,Ω}}={Ω,{Ω}}={Ω,Ω}={Ω}=Ω and furthermore that ({Ω},{(Ω,Ω})=({Ω},{Ω})=(Ω,Ω)=Ω.

Therefore, Ω is a directed graph with itself as the only vertex and itself as the only edge, which goes from itself to itself.

...isn't antifoundation great? :D

-Sniffnoy

--
"He may look like an idiot, and he may sound like an idiot, but don't
let him fool you. He really is an idiot."
-Groucho Marx

...this is ridiculous.

Can somebody please send me a copy of this year's PROMYS contact list? It seems all attempts to send me one so far have failed.