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    L cannot recognize that κ is a measurable cardinal — Carmelics
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    L cannot recognize that κ is a measurable cardinal

    Modality & PossibilityTruth & Knowledge
    ?Rate how convincing each reason is below to see the overall strength.
    1 reason for
    2 reasons against

    Reasons For

    1 perspective
    Reason for
    ?
    • 1.Recognizing measurability requires the existence of an ultrafilter witnessing the measurability of κ
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    • 2.L is too 'thin' to contain the ultrafilter that witnesses the measurability of κ
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    Reasons Against

    2 perspectives
    Reason against 1 of 2
    ?
    • 1.L's 'thinness' is relative to a background universe; from within L, all cardinals satisfy L's own definability conditions.
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    • 2.Scott's theorem shows measurability implies V≠L, but this is an external judgment requiring a richer metatheory, not a failure of L's internal recognition.
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    • 3.What L 'recognizes' depends on the model-theoretic framework assumed; conflating external and internal truth predicates smuggles in a privileged universe.
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    Reason against 2 of 2
    ?
    • 1.Ultrafilters witnessing measurability need not be absent from L if one adopts a liberal notion of class-theoretic resources, as in Kelley-Morse set theory applied internally.
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    • 2.The argument assumes the standard fine-structural account of L, but Sy Friedman's inner model program shows enriched L-like models can accommodate large cardinal structure while retaining constructibility-style definability.
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    Related

    L is too 'thin' to contain the ultrafilter that witnesses the measurability of κL's 'thinness' is relative to a background universe; from within L, all cardinal...Recognizing measurability requires the existence of an ultrafilter witnessing th...Scott's theorem shows measurability implies V≠L, but this is an external judgmen...
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    The argument assumes the standard fine-structural account of L, but Sy Friedman'...Ultrafilters witnessing measurability need not be absent from L if one adopts a ...What L 'recognizes' depends on the model-theoretic framework assumed; conflating...

    Similar

    ZFC + 'There is a measurable cardinal' proves ¬V=L, making measurable ...86%A measurable cardinal κ cannot be the least strongly inaccessible card...85%The theory ZFC + 'There is a measurable cardinal' proves ¬V=L84%Measurable cardinals cannot exist in Gödel's constructible universe (V...79%

    Source

    AI-extracted1/3 agreementValid
    SEP: independence-large-cardinals
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    In fact, Scott showed that (in contrast to the small large cardinals) measurable cardinals cannot exist in Gödel's constructible universe. Let us be precise about this. Let V=L be the statement that asserts that all sets are constructible. Then for each small large cardinal axiom φ (to be precise, those listed above) if the theory ZFC+φ is consistent then so is the theory ZFC+φ+V=L. In contrast, the theory ZFC + “There is a measurable cardinal” proves ¬V=L. This may seem somewhat counterintu
    Extraction notes

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

    Details

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