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nandrolone June 10th, 2010 08:30 AM

Axiom of Choice problem
Hi guys of the forum, can you help me with this:

(*) Let be a sequence of nonempty sets, then is non empty.
(Hint: prove the statement without using the Zorn's Lemma).

I have though that what I need to show is that there is a function (choice) such that for all
That would be trivial using the axiom of choice, but the The assertion (*) is the Axiom of Choice itself, right? so I'm running out of ideas.
Any suggestions?

cknapp June 10th, 2010 08:44 AM

Re: Axiom of Choice problem
If the sets have no extra structure (are they well-ordered?), it's actually the axiom of countable choice-- a weaker form of the axiom of choice... Which nevertheless is outside of ZF. I think the idea of the exercise is to point out that from a choice function, you get an element of the cartesian product. It probably should be simple...

parasio June 10th, 2010 09:07 AM

Re: Axiom of Choice problem
By definition where is the set of all functions from to .

Supose . Then there would be no function from to such that for all . But since for all the axiom of choice says that there's such a function and we have a contradiction. Hence .

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