Axiom of choice on function [duplicate]

Possible Duplicate:
Using a choice function to find an inverse for $F\colon A\to P(B)$

Let $F:A \rightarrow \mathcal P (B)$ be arbitary functions which covers $B$.

Use AC to show there is a function $\phi: B \rightarrow A$ such that $b \in F( \phi(b))$ for each $b \in B$ .

Do I use AC on the set B. I don't see what I'm choosing here.

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marked as duplicate by Asaf Karagila, Micah, rschwieb, Did, Martin ArgeramiDec 5 '12 at 19:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I think I answered this recently. On the iPhone, it's simpler to answer than to find duplicates... :-/ – Asaf Karagila Dec 5 '12 at 10:11

1 Answer

For every $b\in B$ there is some $a\in A$ such that $b\in F(a)$. You need to choose such $a$ for every $b$.

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I think I get it now. So $I=B$ and so we make a set $A_i = \{ a | i \in F(a)\}$. So we choice function is the $\phi$. – simplicity Dec 5 '12 at 10:08
@simplicity: Yes, that’s exactly right. – Brian M. Scott Dec 5 '12 at 10:26