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    Carmelics

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    Home/Original/inverse
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    Inverse View

    It is not the case that L cannot recognize that κ is a measurable cardinal

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    Reasons For

    2 perspectives
    Reason for 1 of 2
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    • 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 for 2 of 2
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    • 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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    Reasons Against

    1 perspective
    Reason against
    ?
    • 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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