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    A symmetric function defined on a set of pairs cannot be ... — Carmelics
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    Home/Modality & Possibility
    HistoryEditSee Inverse

    A symmetric function defined on a set of pairs cannot be a choice function on that set

    Modality & Possibility
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    1 reason for
    2 reasons against

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    Reasons For

    1 perspective
    Reason for
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    • 1.For a symmetric function f defined on P, there exists a finite list L of pairs from P such that fixing all elements of pairs in L suffices to fix f and all its values
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    • 2.For any pair U in P but not in L, a permutation π can be found that fixes all elements of pairs in L but does not fix the members of U
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    • 3.Because π fixes all elements of pairs in L, π must fix the value of f at U
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    Reasons Against

    2 perspectives
    Reason against 1 of 2
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    • 1.The argument presupposes that symmetry over all permutations is required, but choice functions need only be definable relative to a fixed well-ordering of the domain.
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    • 2.Given a well-ordering of any set (provable from AC itself), a canonical choice function exists that is not symmetric yet satisfies all formal requirements of a choice function.
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    • 3.Thus the argument establishes only that no *symmetric* procedure suffices, which is precisely what motivates AC as an independent axiom, not a refutation of choice functions per se.
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    Reason against 2 of 2
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    • 1.Fraenkel's permutation models show symmetry-based impossibility results hold in set theories with urelements but do not straightforwardly transfer to pure ZF set theory.
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    • 2.In ZF without urelements, sets have no 'intrinsic' symmetric indistinguishability of the kind the argument exploits, since all sets are distinguished by their membership structure.
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    • 3.Therefore the permutation argument, following Fraenkel-Mostowski methodology, demonstrates the independence of AC from weaker systems rather than the impossibility of choice functions in standard set theory.
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    Topics

    Modality & PossibilityTruth & Knowledge

    Related

    Because π does not fix the members of U, the value of f at U cannot lie in UBecause π fixes all elements of pairs in L, π must fix the value of f at UFor a symmetric function f defined on P, there exists a finite list L of pairs f...For any pair U in P but not in L, a permutation π can be found that fixes all el...
    +7 moreShow less
    Fraenkel's permutation models show symmetry-based impossibility results hold in ...Given a well-ordering of any set (provable from AC itself), a canonical choice f...If f cannot choose an element of U for some U in P, then f cannot be a choice fu...In ZF without urelements, sets have no 'intrinsic' symmetric indistinguishabilit...The argument presupposes that symmetry over all permutations is required, but ch...Therefore the permutation argument, following Fraenkel-Mostowski methodology, de...Thus the argument establishes only that no *symmetric* procedure suffices, which...

    Similar

    The symmetric universe Sym(V) contains no choice function on the parti...79%If f cannot choose an element of U for some U in P, then f cannot be a...76%For a symmetric function f defined on P, there exists a finite list L ...75%A choice function exists in constructive mathematics73%

    Source

    AI-extracted1/3 agreementValid
    SEP: axiom-choice
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    The point here is that for a symmetric function \(f\) defined on \(P\) there is a finite list \(L\) of pairs from \(P\) the fixing of all of whose elements suffices to fix \(f\), and hence also all the values of \(f\). Now, for any pair \(U\) in \(P\) but not in \(L\) , a permutation \(\pi\) can always be found which fixes all the elements of the pairs in \(L\), but does not fix the members of \(U\). Since \(\pi\) must fix the value of \(f\) at \(U\), that value cannot lie in \(U\). Therefore \(
    Extraction notes

    Validity: Extracted via Max plan + API grounding/validity checks

    Details

    Type
    claim
    Perspectives
    3 (1 for, 2 against)
    Edits
    1 edit