TonesInDeepFreeze

I can't imagine you'd have issues so severe that you couldn't see that self-membership does not imply impossibility of being in another set.

Set theory does not have a predicate "inside braces".

Then the demon is not allowing the subset operation. So the collection would not be one recognizable as serving an ordinary set theoretic role. But yo...

You thought about it for at least half a minute?

If there is a set of all sets, then that set has a subset that is the set of all sets that are not members of themselves, which implies a contradictio...

So you figured why this is not the case?: without regularity Ax({x}e{x} -> Ay(~y={x} -> ~{x}ey))

I should qualify that remark and others I made along the same lines. We prove (though not in the object system) that the Godel-sentence is true on the...

That is so blazingly incorrect that it scorches the core of this planet.

Right.

When we're talking about plain arithmetical truths, I don't know why we would have to go down the road of wondering about realism. I mean, non-realist...

Yes, the theorem itself, as you quoted it, does not mention truth. But from the theorem, we do go on to remark that the undecided sentence is true. An...

I don't think he is. The distinction he's making is very important.

That is exactly the most salient oversimplification that causes misunderstanding. You know the following, but it bears emphasizing: There is no mathem...

I don't know what issue you mean when you ask what the issue is about. But for incompleteness, it's not just a matter of having to assume things to pr...

~ExAy(yex <-> ~yey) is proven from first order logic alone; we don't need any set theory axiom for that. Everything east is in the class, not set, of ...

I'm all for self-scrutiny, but not so overzealous that I chase demons that there is no reason to think exist. I have plenty of faults, but being a cra...

Because the reckoning itself is not necessarily in a formal context, so it is not formal proof, though it could be.

{P} is a member of other sets. Meanwhile, as I already mentioned, without regularity, it is not inconsistent that P = {P} = {{P}} = ... for finitely m...

By the axioms, there is no set x such that every set y is a member of x. That's not childishness; it's axiomatic mathematics.

Circular. I asked what is the operative import of "agree to disagree". Your answer is that we "agree to disagree". It's not just what I believe, it's ...

You are to think that "doing something" is not a set theoretic predicate. And you skipped my counterexamples to your incorrect reasoning.

With regularity, we may prove: E!yAx(xey <-> xex) thus justifying abstraction notation: {x | xex} and the theorem: {x | xex} = 0

What you think reflects a profound ignorance of logic. Namely, that the logic is monotonic. Adding the theorem that no set is a member of itself does ...

"something was done to it" is not a set theoretic predicate. / S = SuS With S, "nothing was done to" S. With SuS, "something was done to S". / 2 = 2*1...

(1) Set theory is incomplete, therefore set theory is consistent. (2) Any consistent, recursively axiomatized, arithmetically adequate theory is incom...

I'm not real knowledgeable myself, and, of course, one shouldn't expect that everyone is equally knowledgeable. But I don't get why someone would post...

It depends on the definition of 'mathematically proven'. It is easy to see that there are theories that are proper extensions of ZFC . But that doesn'...

The general idea you expressed is okay, but I suggest some clarifications and context (much of which you likely know already). We are concerned not ju...

what is the mathematical definition of 'abritrary'? Anyway, I don't think we need 'inherent', 'actual' or 'non-arbitrary'. It is enough to observe tha...

There is. It can be any set whatsoever. Here's a proof: AyExAz(zex <-> (zey & ~z=z)) instance of axiom schema of separation (zex <-> (zey & ~z=z)) ?UI...

It is typical that cranks confront the implications of Russell's paradox with hostility (also the other antinomies, incompleteness, the halting proble...

I don't just think it. I prove it from the axioms of Z set theory. You are welcome to present your own system and axioms in which there is a set of al...

Please stop using '=' to stand for the biconditional. (1) N = {N} premise (2) N e N <-> {N} e {N} from (1) (3) ~ {N} e {N} non sequitur Why do you was...

First, do you understand my explanation that you just quoted?

It is unprovable in the system being discussed. It is provably true in the mathematics used to discuss that system. We have a sure example. It's as su...

That's just another undefined term by you as is 'inherent order'. It adds nothing to your incorrect argument. There are many orders. You have not defi...

That's not phrasing I've ever seen in set theory. I already told you I don't know what that means. I don't know why you present it again without sayin...

You're talking about Russell's paradox in context of sets, and about set membership, and using extensional braces. But your posting shows that you hav...

No, you skipped my points. You see, right there, you skipped my point, posted at least three times now, that "member of itself twice" has no apparent ...

Look at the word "IF" I wrote. IF there is a set of all sets, then it has a subset that is the set of all sets that are not members of themselves. So,...

How is it logically necessary? (With an ordinary understanding of 'logically necessary'.) I have said: (1) "member of itself twice" has no apparent se...

The equal sign there and the ones following it are not syntactical. Perhaps you mean the biconditional. If so, rewrite to see whether your argument st...

" a self-referential sentence which “says of itself” Such figures of speech may be heuristically useful, but they are also easily misleading and sugge...

I don't think it's so over your head. Give it a bit of thought and you'll see it pretty clearly.

N = {N} is such a set. I don't know what you mean by that. / Wrong. N = {x N}, so N is not a proper subset of {x N}. I'll stop there.

Just to be clear, ZR-R+~R is relatively consistent with ZF-R. And to underline your point: The conjunction of "ZF-R+~R is relatively consistent with Z...

I'm looking at 'The Joy Of Sets' by Devlin. He has a cute example (though it steps our of formal set theory by using 'explicitly referred to' and 'thi...

No, it is a non-sequitur to infer ZFC-R |- ~Ex xex -> regularity from it is not the case that ZFC-R |- ~Ex xex Now, clearly ZFC-R |- regularity -> ~Ex...

You're not reading my posts. Without the axiom of regularity, we cannot prove ~Ex xex. And the rest of your post is more of your misunderstanding of h...

That's not the sense of the word 'order' we're talking about! You have yourself even being using 'order' in a sense not expressed as "in its right pla...