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

In article <1186512440.514432.38300@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Scott <ToaTerra@xxxxxxxxx> wrote:
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.

Note that your proof is stronger than your proposition. In your
proposition, you are merely saying that there exist s in S such that
g_p(s)=p. In your proof, however, the function satisfies that for all
t in S, g_p(t)=p. If you want the latter, then you might want to say
it in the proposition.

Not that it will help. Your flailings will still not avail you.

Prop 5A: (AS != emptyset)(Ep)( p in P(S) -> ~(Ef_p)(f_p:S -> P(S) & p
in range(f_p) ) ).

This proposition is false.


Proof: Let S be any non-empty set. Let A_{f_A} = { x in S : x notin
f_A(x) }

No. This is nonsense and incoherent. There is no f_A defined, so you
cannot do this at this stage.

Comments/feedback welcome.

You are still beating your head against a very solid wall, and rather
stupidly. Why not LEARN some basic propositional logic before
continuing?

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

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