Re: Question: Given |X|>0 and |Y|>0, can X x Y be empty?

On Aug 3, 5:19 pm, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
Better is to use f(p), like I did. Or f_p. Or something that makes the
connection between f and p.

I will use f_p as f(p) is too confusing with f(x) where x is the
argument.

As a constant function - yes. I thought that is what I was defining:
for any p in P(S), there exists a constant function f(s) = p. I was
showing this function exists because its a subset of M.

But you did not define it correctly; that's the problem, not whether
"it exists". You defined it incoherently.

Okay, I see the problem. I meant each function is relative only to a
single p, but what I wrote meant the function was relative to all p.
So a subscript tying each function to the subset is what is needed.

Prop 5 says: given a function f:S->P(S), we can construct a set A_f, THAT
DEPENDS ON f, with the property that f(x)=/=A_f for any x in S.

Prop 4 says: if S is not empty, then given an element p in P(S), we
can construct a function g(p):S->P(S) (which I will call
g(p) to emphasize the separation between Prop 4 and 5), such that p is
in the range of g; that is, there exists s in S such that g(s)=q.

So, take a function f. Apply Prop. 5 to obtain the set A_f in
P(S). Now apply Prop. 4 to the set A_f to obtain a NEW function (not
necessarily equal to the f you started with), g(A_f):S->P(S), such
that there exists s in S for which g(A_f)(s) = A_f.

Can we now apply Prop 5 to g(A_f)? Yes, we can. But g(A_f) is not the
same as f, and so there is no reason to assume or suspect that
A_(g(A_f)) will be the same as A_f. In fact, it won't. Our function

Okay. Understand.

.

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