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:
What would have been better is if you had kept the difference between
functions you get from subsets per Prop 4 separate from subsets you
get from functions per Prop 5.

Okay, so keeping them separate and using subscripts:

Prop 4A: (AS != emptyset)(Ap)( p in P(S) -> (Eg_p)(g_p:S -> P(S) & p
in range(g_p) ) ).
Proof: Let S be any non-empty set. If p in P(S), then let g_p be the
constant function g_p: x |-> p where x in S. g_p exists because g_p
subset S x P(S) and, by inspection, g_p is functional and total.
Furthermore, p in range(g_p). QED.

Prop 5A: (AS != emptyset)(Ep)( p in P(S) -> ~(Ef_p)(f_p:S -> P(S) & p
in range(f_p) ) ).
Proof: Let S be any non-empty set. Let A_{f_A} = { x in S : x notin
f_A(x) } so A_{f_A} in P(S). Suppose (f_A: S -> P(S) & A_{f_A} in
range(f_A)). Then (Ey)(f_A(y) = A_{f_A}). Since A_{f_A} subseteq S,
then (Ax in S)( x in A_{f_A} <-> x notin f_A(x)). In particular, when
x = y, ( y in A_{f_A} <-> y notin A_{f_A} ). Contradiction. QED.

Now (hopefully) Prop 4A is a "machine" that, given a subset p, gives
me a function g_p with property P and Prop 5A is a "machine" that,
given a specific subset A, gives me the result that there are no
functions f_A with property P.

Comments/feedback welcome.

.

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