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June 10th, 2010, 09:30 AM   #1
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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) S_n) \rightarrow \bigcup_n S_n" /> 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?
nandrolone is offline  
June 10th, 2010, 09:44 AM   #2
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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...
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June 10th, 2010, 10:07 AM   #3
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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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