Fun with antifoundation

[sniffnoy]

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