• TheMadFool
    13.8k
    I recall someone else starting a thread in the same vein:

    1. Assume U = the set that contains everything
    2. If there's a that U contains everything, U must contain itself

    Line 2 might need some explaining. Suppose E = everything. U = {E}. But then U can't be {E} because we can think of {{E}, E} that's a better match for U. However, {{E}, E} ain't it either because there's {{{E}, E}, {E}, E} [nested sets]...the process reiterates ad infinitum and, I suppose, ad nauseum.

    There's also the matter of power sets. Suppose P(A) is the power set of set A, I believe a proven theorem in set theory is that the P(A) > A.

    1. There's a U that's the set of everything [assume for reductio ad absurdum]
    2. P(U) > U [The theorem I spoke of above]
    3 . If U is the set of everything then, any set A is such that A < U
    4. Any set A is such that A < U [from 1, 3 modus ponens]
    5. If any set A is such that A < U then, P(U) < U
    6. P(U) < A [from 4, 5 modus ponens]
    7. P(U) > A and P(U) < A [2, 6 conjunction, also a contradiction]
    Ergo,
    8. There's no U that's the set of everything [1 to 7 reductio absurdum]
  • TonesInDeepFreeze
    2.3k
    E = everythingTheMadFool

    In set theory, 'everything' doesn't name a thing. Rather, 'everything' is used for quantification.

    (1)

    Suppose ExAy yex. ("There exists an x such that every y is a member of x")

    Let Ay yeU.

    So UeU.

    'UeU' is not a contradiction (self membership is consistent with ZFC-regularity).

    (2) Cantor's paradox

    Suppose ExAy yex.

    Let Ay yeU.

    So PU is a subset U. ("The power set of U is a subset of U")

    So Ef f is an injection from PU into U.

    So Ef f is a surjection from U onto PU.

    Previously proved theorem: Ax ~Ef f is a surjection from x onto Px.

    So ~Ef f is a surjection from U onto PU.

    So ~EAy yex.
  • TheMadFool
    13.8k
    In set theory, 'everything' doesn't name a thing. Rather, 'everything' is used for quantification.

    (1)

    Suppose ExAy yex. ("There exists an x such that every y is a member of x")

    Let Ay yeU.

    So UeU.

    'UeU' is not a contradiction (self membership is consistent with ZFC-regularity).

    (2) Cantor's paradox

    Suppose ExAy yex.

    Let Ay yeU.

    So PU is a subset U. ("The power set of U is a subset of U")

    So Ef f is an injection from PU into U.

    So Ef f is a surjection from U onto PU.

    Previously proved theorem: Ax ~Ef f is a surjection from x onto Px.

    So ~Ef f is a surjection from U onto PU.

    So ~EAy yex.
    TonesInDeepFreeze

    Above my paygrade fellow forum member. Thanks for the effort. Much appreciated! When I do get round to studying set theory in earnest, I might just be able to have a intelligent conversation with those like you who know their stuff.
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