June 10th, 2010, 09:30 AM  #1 
Newbie Joined: Jun 2010 Posts: 4 Thanks: 0  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? 
June 10th, 2010, 09:44 AM  #2 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: Axiom of Choice problem
If the sets have no extra structure (are they wellordered?), 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...

June 10th, 2010, 10:07 AM  #3 
Senior Member Joined: Feb 2009 Posts: 172 Thanks: 5  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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