Re: Question: What conditions are sufficient to prove a subset does not exist?

On Jul 25, 4:03 pm, Virgil <vir...@xxxxxxxxxxx> wrote:
The function, Phi, has domain of the set of functions from S to P(S),
and has for each f:S --> P(S) the value { x in S : x notin f(x) }, so
Phi's domain is a subset of P(S).

I.e., Phi( f:S --> P(S) ) = { x in S : x notin f(x) } for each function
f from S to P(S).
...
So, as far as I can see, a necessary condition for there not to be a
set, A, of the given form, { x in S : x notin f(x) }, is that there not
be a set S.

May I ask that you provide me with an example using the following? Let
N = { 0, 1, 2, 3, ... }. Suppose 0 is both odd and even. Let B =
{ xeN : ~even(x) }. Here 0eN and (0eB & ~0eB). Why must it be
necessary to prove N does not exist?

.