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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 $(S_n)_{n=1}^\infty$ be a sequence of nonempty sets, then $\prod_{n=1}^\infty S_n$ 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) $f:(S_n) \rightarrow \bigcup_n S_n$ such that $f(S_n) \rightarrow S_n$ for all $n=1\ldots$
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 $\displaystyle\prod_{n=1}^{\infty}S_n=\{f\in\mathca l{F}$$\mathbb{N},\displaystyle\bigcup_{n=1}^{\inft y}S_n$$:f(n)\in S_n\}$ where $\mathcal{F}$$\mathbb{N},\displaystyle\bigcup_{n=1} ^{\infty}S_n$$$ is the set of all functions from $\mathbb N$ to $\displaystyle\bigcup_{n=1}^{\infty}S_n$.

Supose $\displaystyle\prod_{n=1}^{\infty}S_n=\emptyset$. Then there would be no function from $\mathbb N$ to $\displaystyle\bigcup_{n=1}^{\infty}S_n$ such that $f(n)\in S_n$ for all $n\in\mathbb N$. But since $S_n\neq\emptyset$ for all $n\in\mathbb N$ the axiom of choice says that there's such a function and we have a contradiction. Hence $\displaystyle\prod_{n=1}^{\infty}S_n\neq\emptyset$.

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