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- - **Axiom of Choice problem**
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Axiom of Choice problemHi 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? |

Re: Axiom of Choice problemIf 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... |

Re: Axiom of Choice problemBy 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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