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]

E = everything — TheMadFool

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.

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.