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Math Help - Axiom of choice on finite sets

  1. #1
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    Question Axiom of choice on finite sets

    Hello all,

    First: Have a wonderful new year ahead!!

    I have a fundamental question about axiom of choice. Let me quote wiki over a variant of axiom of choice

    Given any set X of pairwise disjoint non-empty sets, there exists at least one set C that contains exactly one element in common with each of the sets in X
    or
    Given any set X of pairwise disjoint non-empty sets, there exists at least one set C that contains exactly one element in common with each of the sets in X
    Is this not a variation or a special case of axiom of schema which states that

    Given any set A, there is a set B such that, given any set x, x is a member of B if and only if x is a member of A and φ holds for x.
    where φ is a predicate.

    My question is that it appears axiom of choice is a special case of the axiom of schema where predicate is a choice function. But axiom of choice is considered as a basic axiom and it is quoted that it cannot be proved in ZF. Can some one clarify this contradiction??


    Best Regards,

    Sudeep
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  2. #2
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    Re: Axiom of choice on finite sets

    Quote Originally Posted by skanur View Post
    First: Have a wonderful new year ahead!!
    Thanks, and the same to you!

    Quote Originally Posted by skanur View Post
    My question is that it appears axiom of choice is a special case of the axiom of schema where predicate is a choice function.
    But the existence of a choice function is what is guaranteed by the axiom of choice (AC). If you can construct a choice function without AC, then AC is indeed unnecessary and you can use the axiom schema of specification, or restricted comprehension, to construct the set that contains exactly one element in common with each of the sets in X. A choice function can be constructed, in particular, when X is finite.

    This page about AC is useful.
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  3. #3
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    Thumbs up Re: Axiom of choice on finite sets

    Ah! I get it, your answer can also be paraphrased something like this

    To assert that a mathematical object "exists," even when you cannot give an example of it, is a little bit like this: Suppose that one day you go to a football game by yourself. There are thousands of other people in the stadium, but you don't know the names of any of them. (And let's suppose you're shy, so you're not about to ask anyone their name.) Then you know those people have names, but you cannot give any of those names. (Admittedly, this is only a metaphor, and not a perfect one; don't make too much of it.)
    The above quote was from your link. Thanks for the link and the answer.

    Sudeep

    PS: Sorry for the repeated definition of axiom of choice in quotes, that was unintentional
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