Re: ping GhostInTheMachine



The Ghost In The Machine wrote:
> > This article does not mention the diagonalization issue.
> If T = {s in S: s not in f(s)}, then one can define
> g(s,t) : S x S -> boolean : g(s,t) = (s not in f(t)).
> Then T = {s in S: !g(s,s)}. I am not familiar with
> Quine's "New Foundations" set theory.
>

Sorry, but I don't understand the above. Where's the diagonal?
How is it used? How do you derive T = {s in S: !g(s,s)}?

Ken

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