- #1

quasar987

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How does the set theory of Zermelo and Fraenkel get rid of this "paradox"? I.e., which axioms or theorem prohibit S above to be a set?

Thank you.

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- Thread starter quasar987
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- #1

quasar987

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How does the set theory of Zermelo and Fraenkel get rid of this "paradox"? I.e., which axioms or theorem prohibit S above to be a set?

Thank you.

- #2

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In particular, the expression [tex] \{ x \mid P(x) \} [/tex] for any old first-order predicate P is no longer necessarily a set, as it would be with comprehension.

Instead, we have the axiom of separation, which guarantees that [tex]\{ x \in A \mid P(x) \}[/tex] is a set, given any existing set A and predicate P.

So [tex]\{ x \in \mathbb{R} \mid x \notin x \}[/tex], [tex]\{ x \in 2^{\mathbb N} \mid x \notin x \}[/tex], and so on are sets, but none of these leads to contradictions (assuming ZF is consistent, of course!)

- #3

Hurkyl

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In ZF, Russell's paradox still works to prove S is not a set. However, ZF does not (seem to) have a way to prove S is a set, and so there is no contradiction.

Incidentally, don't forget replacement which gives another ZF-legal form of set-builder notation:

[tex]\{ f(x) \mid x \in A \}[/tex]

The way in which all of this works is remarkedly similar to how formal logic (e.g. first-order, second-order, and so forth) avoids the liar's paradox.

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