Skip to content
Carmelics
TopicsThinkersChangesContributorsLoading account…

    Carmelics

    A reasoning platform. Break down any belief into clear reasons, explore both sides, and weigh the evidence honestly.

    Navigate

    • Topics
    • Search
    • Recent Changes
    • Contribute
    • How It Works
    • Glossary
    • Thinkers
    • Contributors
    • About
    • Statistics
    • Terms
    • Privacy

    Database

    Statements
    —
    Perspectives
    —
    Topics
    —

    Press ? for keyboard shortcuts

    LoyalLoyalJusticeJustice
    Made withinDC&Austin
    Statements
    321,452
    Perspectives
    108,905
    Topics
    42
    G is the group of permutations of A that fix all pairs in P. — Carmelics
    Home/Modality & Possibility
    HistoryEditSee Inverse

    Part of a larger discussion

    Supports→The symmetric universe Sym(V) contains no choice function on the partition P of pairs.

    G is the group of permutations of A that fix all pairs in P.

    Modality & PossibilityTruth & Knowledge
    ?Rate how convincing each reason is below to see the overall strength.

    No one has weighed in yet. Be the first to share reasons for or against this statement.

    Sign in or register to share your perspective on this statement.

    Topics

    Modality & PossibilityTruth & Knowledge

    Related

    Because π fixes the support of f, π also fixes f (i.e., πf = f).If a choice function f on P were in Sym(V), then f would have a finite support o...P is a necessarily infinite mutually disjoint set of pairs partitioning A.Since P is infinite, there exists a pair {c, d} = U in P distinct from all {ai, ...

    Next step

    Based on where you are in your exploration

    Browse more in Modality & Possibility
    Related propositions within the same area of thought.
    +8 moreShow less
    Since f is a choice function on P and U is in P, f(U) must equal c or d.Since π interchanges c and d, π(f(U)) ≠ f(U).Since π is an automorphism, it preserves function application, so π(f(U)) = πf(π...The conclusions π(f(U)) ≠ f(U) and π(f(U)) = f(U) are contradictory.The symmetric universe Sym(V) contains no choice function on the partition P of ...There exists a permutation π in G that fixes each ai and bi and interchanges c a...π(U) = U because π only interchanges c and d, which are both in U, so U as a set...πf = f, therefore π(f(U)) = f(U).

    Similar

    For any pair U in P but not in L, a permutation π can be found that fi...82%Because π fixes all elements of pairs in L, π must fix the value of f ...76%For a symmetric function f defined on P, there exists a finite list L ...75%Since PH is defined as the union of all Σ^P_k classes, TWO PLAYER SAT ...73%

    Source

    AI-extracted
    SEP: axiom-choice
    View source passageHide passage
    Now suppose \(A\) to be partitioned into a (necessarily infinite) mutually disjoint set \(P\) of pairs. Take \(G\) to be the group of permutations of \(A\) which fix all the pairs in \(P\). Then \(P\in Sym(V)\); it can now be shown that \(Sym(V)\) contains no choice function on \(P\). For suppose \(f\) were a choice function on \(P\) and \(f \in Sym(V)\). Then \(f\) has a finite support which may be taken to be of the form \(\{a_{1}, \ldots, a_{n}, b_{1},\ldots,b_{n}\} with each pair \{a_{i}, b_

    Details

    Type
    premise
    Perspectives
    0 (0 for, 0 against)
    Edits
    1 edit

    Open for perspectives

    This idea is waiting for its first supporting or challenging perspective.

    Share the first perspective