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

In article <1186161736.517279.72240@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Scott <ToaTerra@xxxxxxxxx> wrote:
On Aug 2, 4:59 pm, G. Frege <nomail@invalid> wrote:
Sure, man. Post it! :-)

[1] Potter, M., Set Theory and its Philosophy, 2004.

The context for my original question was an email:

Let S be any non-empty set.

Def 1: The cartesian product, denoted PxQ, is { (x,y) : x in P and y
in Q }.

Lemma 2: M = S x P(S) exists.
Proof: From Proposition 4.7.1 in [1], if P and Q are sets, then PxQ is
a set. Since S and P(S) are sets, SxP(S) is a set. QED.

Fine...

Prop 3: (As in S)(Ap in P(S))( (s,p) in M ).

Sigh. This is the DEFINITION of S x P(S). So why are you "proving"
this? Worse, why are you proving it ->by contradiction<-?

Proof: Suppose (Es in S)(Ep in P(S))~( (s,p) in M). Then from Def 1
(Es in S)(Ep in P(S))~(s in S and p in P(S)) which can be rewritten as
(Es in S)(Ep in P(S))( s notin S or p notin P(S) ). But s in S and p
in P(S). Contradiction. Therefore, (As in S)(Ap in P(S))( (s,p) in
M ). QED.


Note the following sentence:

Let f be a function f:S -> P(S).

Now note what you do immediately afterwards:

Prop 4: p in P(S) -> (Ef)( p in range(f) ).

Sigh. You are already using f for a (presumably) fixed function form S
to P(S). Here you confuse matters by making f a bound variable inside
of Prop 4, which means it is something completely separate from the f
in the immediate previous sentence.

That is lousy notation, prone to confusion.

So, I'm going to ignore the "let f be a fucntion f:S -> P(S)". It is
not even there.

So, you are saying in Prop 4: if p is an element of P(S), then there
is a function (with domain where?) whose image includes p.

Better:

If p in P(S), then there exists f(p) : S->P(S) such that p is in
the range of f(p).

Note that f(p) depends on p, hence the notation.

Proof: If p in P(S), then (As in S)( (s,p) in M ) [Prop 3]. Let f(s) =
p <-> (As in S)( (s,p) in M ).

This is lousy. Are you trying to define f as "constant p"? Then DO
SO. What you write here is pretty near unintelligible, close to
nonsense. Since (s,q) in M for all s in S and all q in S(p), your
condition on the right ALWAYS holds, so it does not define a
function. Your "function" is in fact "assign to each element of S
every element of P(S)" and that is not a function.

f is functional

No, it is note. Your definition makes f = M, and that is in general
not a function.

Perhaps you meant:

Let f(p) = { (s,p) in M : s in S}.

This would make f(p) into the function that sends everything to p.

because, by inspection,
for all s in S and x,y in P(S) it holds that f(s) = x and f(s) = y
then x = y.

False. See above. Your condition on the right of the definition fails
to define a single function.

f is total because, by inspection, for all s in S there
exists a p in P(S) such that f(s) = p. Since (As in S)( (s,p) in M ),
then p in range(f). Therefore, p in P(S) -> (Ef)( p in range(f) ).
QED.

See above for an easy proof of what you probably meant to say.

Let A = { x in S : x notin f(x) }.

What f? The one you define in Prop 4 for a given p, or the one we
started with? See the problem with your lousy notation?

So, let f be an arbitrary function, f:S->P(S). It need not be equal to
f(p) for any p in P(S), i.e., it may not be the conditional one we
have from Proposition 4.

Given f, we define A_f to be

A_f = { x in S : x notin f(x) }.

So, for example, if you were given p in P(S), you could take f(p) from
Prop 4, and then consider A_{f(p)} = {x in S : x notin f(x)}
= {x in S : x notin p}
= S\p.

Prop 5: (Ef)( A_f subset S and ~( A_f subset S) ).

Proof: A exists by the axiom schema of separation so (Af)( A_f subset
S ). Suppose (Ef)( A_f in range(f) ). Then (Ex)( f(x) = A_f ). Since
A_f subset S, then (As in S)( s in A_f <-> s notin f(x) ). In
particular, when s = x then ( x in A_f <-> x notin A_f ).
Contradiction. Thus, ~(Ef)( A_f in range(f) ).

This is simply Cantor's Argument.

From Prop 4 and the
contrapositive, ~(Ef)( A_f in range(f) ) -> A_f notin P(S).

Wrong. Proposition 4 was incorrect and does not refer to the same f as
proposition 5. Here is your mistake, born out of that LOUSY notation
where you use f to mean many different things at the same time.

No wonder you are confused.

Proposition 4 says:

(Ap) (p in S(p) ->
(Ef(p))(f(p):S->P(S), f(p) is a function,
and (Ex)(x in S and f(p)(x)=p )) )


What is the contrapositive of this? It is:

(Ap)( (A f(p)) f(p) is not a function, or f(p) is not a function from
S to P(S), or for all x in S, f(p)(x) =/= p)
--> (p not in S(p))

Nothing like what you assert above.

The commenters response was: "The set M exists but is of course
empty. Hence your selection of f isn't possible."

The commenter was thrown by your conflation of several different
things.

So my question: Is is possible for a cartesian product to be empty
when both sets are non-zero?

No, it is not possible. But then, that is not what the commenter
asserted. That is what you have misunderstood, as usual.

What this is, just another instance of you making assertion A thinking
you are really making assertion B; someone comments on A, and then you
take the answer and think that it refers to B. In other words, you
still think we can read your mind, and interpret our comments that
way. Most of the time, you conflate this by changing B into C before
applying the comment, and then wondering why someone would assert that
the answer to A applies to C, given that you were thinking about B
originally.

--
======================================================================
"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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