Redpill me on Russel's paradox. Why can't I just conclude

>>9134380
Well, that's the point of axiomatic set theory. In naive set theory if you can write a set then it exists. You can write R and therefore it exists. If you want R not to exist then you need to write an axiomatic system in which R cannot be written.

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Well, obviously no set can contain itself as an element.

For example,
[math]R = \{R\} \rightarrow R = \{\{\{\{\{ \ldots[/math]

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Naive set theory has

∀P. ∃X. X = {x|P(x)}

where "X = {x|P(x)}" is syntactic sugar for

∀x. (x ∈ X ⇔ P(x))

For P(x) := x ∈ x, you get a contradiction.

>>9134447
>obviously
Is that a "proof writing down half of a string"?
The claim follows from the wellfoundedness axiom and if you drop it, you get set theories that are just as valid
https://en.wikipedia.org/wiki/Non-well-founded_set_theory
and those have been studied a lot too.

And even if we consider naive set theory, from a formal logic standpoint having Russels paradox still makes it a perfectly nice theory. However, when you can proof any statement in a theory, it renders it useless. But it doesn't render it "wrong" in any strong sense of the word..

>>9134380
the problem is unrestricted comprehension.

Notice that you DEFINITELY want restricted comprehension, to guarantee that subsets are in fact sets (note that if we only used, say, the empty set axiom and extensionality, we have no guarantee that our universe of sets needs to be closed under subset-taking).

If we allowed unrestricted comprehension, by definition, R as constructed in pic MUST be a set. The problem is then that we get conflicting elementhood information, so an axiom system which allows unrestricted comprehension is inconsistent - there are theorems we can prove both true and false.

>>9134453
hm. I think this is a misleading response. Regularity axiom implies this fact, but you can have set theories which allow for sets like S\in S which do not run into Russell's paradox. OTOH, making an axiom that declares that you do not allow sets like S\in S doesn't absolve you of Russell's paradox either.

So you can have S\in S, and make other axioms that avoid Russell's paradox.

And you can have set theories which block S\in S, which still run into Russell's paradox.