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    The Russell class R cannot be a member of any class (i.e.... — Carmelics
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    321,452
    Perspectives
    108,905
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    42
    Home/Modality & Possibility
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    The Russell class R cannot be a member of any class (i.e., R must be a proper class).

    Modality & Possibility
    ?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.Sets are defined as members, while non-members are labeled 'proper classes'.
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    • 2.If R is assumed to be an element of a class A, then by one of von Neumann's axioms R is not equivalent to V.
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    • 3.R is equivalent to V.
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    Reasons Against

    2 perspectives
    Reason against 1 of 2
    ?
    • 1.In paraconsistent set theories (e.g., Routley's dialectical sets), R can consistently belong to itself and to a class without deriving explosion.
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    • 2.If contradictions do not entail all propositions, the inference from R∈R ↔ R∉R to R's exclusion from all classes is not logically compelled.
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    • 3.Therefore the proper-class conclusion rests on classical logic's ex contradictione quodlibet, which is a contested assumption, not a logical necessity.
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    Reason against 2 of 2
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    • 1.Quine's NF (New Foundations) permits a universal set V and stratified comprehension, dissolving the distinction between sets and proper classes entirely.
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    • 2.If R can be reconstructed under stratified comprehension as a legitimate set within NF, then labeling R a 'proper class' reflects one axiomatic choice, not an ontological necessity.
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    • 3.The von Neumann axiom invoked in P2 is specific to NBG/MK class theories and cannot ground a theory-neutral claim that R must be a proper class.
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    Topics

    Modality & PossibilityTruth & Knowledge

    Connections

    1 topic

    Philosophy of Language1 linked

    Related

    If R can be reconstructed under stratified comprehension as a legitimate set wit...If R is assumed to be an element of a class A, then by one of von Neumann's axio...If contradictions do not entail all propositions, the inference from R∈R ↔ R∉R t...In paraconsistent set theories (e.g., Routley's dialectical sets), R can consist...
    +6 moreShow less
    Quine's NF (New Foundations) permits a universal set V and stratified comprehens...R is equivalent to V.Sets are defined as members, while non-members are labeled 'proper classes'.The von Neumann axiom invoked in P2 is specific to NBG/MK class theories and can...Therefore R cannot be an element of A (for any class A).Therefore the proper-class conclusion rests on classical logic's ex contradictio...

    Similar

    Therefore R cannot be an element of A (for any class A).86%Sets are defined as members, while non-members are labeled 'proper cla...84%Problems complete for a class that properly contains P cannot be in P.82%It is widely believed that NP and coNP are distinct classes.81%

    Source

    AI-extracted1/3 agreementValid
    SEP: russell-paradox
    View source passageHide passage
    Sets are then defined as members, and non-members are labeled “proper classes.” So for example, the Russell class, \(R\), cannot be a member of any class, and hence it must be a proper class. If \(R\) is assumed to be an element of a class \(A\), then it follows from one of von Neumann’s axioms that \(R\) is not equivalent to \(V\). But \(R\) is equivalent to \(V\), and hence not an element of \(A\). Thus, von Neumann’s method is closely related to the result stated above about the set \(R_B\),
    Extraction notes

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

    Details

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