# for which A is AC necessary to prove A X A ~A?

• Jul 19th 2011, 02:07 AM
for which A is AC necessary to prove A X A ~A?
One doesn't need the full AC in order to prove (in ZF) N X N ~ N (~ being equinumerous), whereas one does need the full AC in order to prove that for all sets A, AXA~A. So, what determines, for a particular A, whether one can prove that AXA~A? For example, does one need the full AC to prove that RXR~R? (I suspect the answer to this last question is no.) So where is the "cut-off point"?
• Jul 19th 2011, 06:53 AM
MoeBlee
Re: for which A is AC necessary to prove A X A ~A?
Quote:

Originally Posted by nomadreid
all sets A, AXA~A

You mean that for any INFINITE set A, we have A equinumerous with AXA.

Quote:

Originally Posted by nomadreid
So, what determines, for a particular A, whether one can prove that AXA~A? For example, does one need the full AC to prove that RXR~R? (I suspect the answer to this last question is no.) So where is the "cut-off point"?

Personally, I don't see how to make your question precise. What is meant by "what determines whether"? I mean, if for a certain A, whatever axioms prove that A is equinumerous with AXA, then that determines it, but, of course, that doesn't preclude that one could provide different axioms to make the proof.

As to needing the full axiom, in any given case, we could use a weaker form of the axiom, even if just by stating the axiom except only with regard to the particular set, such as R or whatever set. So, at least to me, it's not clear what you mean by a "cut off point".
• Jul 19th 2011, 06:55 AM
MoeBlee
Re: for which A is AC necessary to prove A X A ~A?
Quote:

Originally Posted by nomadreid
One doesn't need the full AC in order to prove (in ZF) N X N ~ N

Not only do we not need full AC, we don't need any choice principle at all for that.

Also, NO use of AC MUST have full AC, since we could make whatever proof using just an instantiation of AC to the particular set we're talking about. And that holds for any axiom.
• Jul 19th 2011, 08:26 AM
Re: for which A is AC necessary to prove A X A ~A?
Quote:

You mean that for any INFINITE set A, we have A equinumerous with AXA.
Yes, sorry.

Quote:

make your question precise. What is meant by "what determines whether"?
Granted, that was pretty vague. I'll try again.

Suppose you were to partition the universe V into three classes:
(C), in which AXA ~A could be deduced only from ZF & B, where B is the (perhaps redundant) sentence asserting A's existence,
(D), in which it could not be deduced as in (1), but could be deduced from ZF & B & DC
(E), in which it could not be deduced as in (1) or (2), but could be deduced from ZF & B & AC

Now, suppose we were to again partition the universe into three classes C, D, E , using three one-place formulas f,g, and h defined on sets
S $\displaystyle \epsilon$ W iff f(S) & not-g(S) & not-h(S)
S $\displaystyle \epsilon$ X iff not- f(S) & g(S) & not-h(S)
S $\displaystyle \epsilon$ Y iff not- f(S) & not- g(S) & h(S)

So, questions:
(i) Are there any f,g,h which could be formulated independently of conditions (C)-(E) to make C=W, D=X, E=Y?
(ii) Would the choice of B (say, a large cardinal axiom) influence the ability to find such properties?
(iii) Is there any model (other than defining the model in terms of these partitions) that would enable such properties to be found? For example, if V=L, would that make a difference?
(iv) If I restrict S to ordinals, or perhaps cardinals, would that help?

Quote:

NO use of AC MUST have full AC
Er, AC implies the well-ordering theorem….(Blush) OK, I could adjust AC and DC above to AC* and DC*, being these principles applied to A.
• Jul 19th 2011, 09:07 AM
MoeBlee
Re: for which A is AC necessary to prove A X A ~A?
I should have been more precise. When I say no use requires full AC, I meant no use for proving about any PARTICULAR sets.

I'll read over the rest of your post later.
• Jul 19th 2011, 09:09 AM
MoeBlee
Re: for which A is AC necessary to prove A X A ~A?
Quote:

Originally Posted by nomadreid
Suppose you were to partition the universe V into three classes:
(C), in which AXA ~A could be deduced only from ZF & B, where B is the (perhaps redundant) sentence asserting A's existence,

Set theory has theorems that assert that there exist certain sets having certain properties. But it is not clear what it means for a sentence B to assert "A exists". Given ANY term t of the langauge we have the theorem:

Ex x = t.

I'll read the rest of your post later.