Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How does ZF fixes Russell's paradox?

  1. Jan 1, 2009 #1


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In naive set theory, Russell's paradox shows that the "set" [itex]S:=\{X:X \in X\}[/itex] satisfies the weird property [itex]S \in S[/itex] and [itex]S\notin S[/itex].

    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. jcsd
  3. Jan 1, 2009 #2
    It's not that there's an axiom in ZF explicitly preventing Russell's Paradox, but rather ZF *doesn't* include the axiom of comprehension, which is the cause of all the trouble. Instead, it only allows for sets to be "built from" other sets, by union, by image under a function, by power set, and so on, and so on.

    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!)
  4. Jan 2, 2009 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    To put things differently... Cantor's set theory said [itex]S:=\{X:\mid X \in X\}[/itex], and Russell's paradox proves S is not a set. Thus contradiction.

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook