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As is well known, the General Set Comprehension Principle (any class is a set) leads to the Russell Paradox (the class $x \notin x$ cannot be a set). As a result, set theories must restrict the Comprehension Principle to avoid self-reference. For example, in the case of ZFC, this is done by enumerating a small list of "safe" comprehension schema such as Separation.

Does General Comprehension lead to other types of paradox than Russell's? Or is this basically the only thing that can go wrong?

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    $\begingroup$ The class of ordinals can't be a set: en.wikipedia.org/wiki/Burali-Forti_paradox $\endgroup$ May 2, 2012 at 15:32
  • $\begingroup$ The class of all singletons can't be a set. $\endgroup$ May 2, 2012 at 15:35
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    $\begingroup$ @Arturo, isn't that essentially the same paradox? If the class of singletons is a set, then the class of all sets is a set, then the class of all $x \not\in x$ is a set. $\endgroup$ May 2, 2012 at 15:42
  • $\begingroup$ @David: Of course, you can deduce many different contradictions from this problem; the proof I linked to uses the result that there can be no one-to-one function from $\mathcal{P}(X)$ to $X$, rather than going through Russell. $\endgroup$ May 2, 2012 at 15:51

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Early attempts to repair Russell's paradox tried simple patches, like forbidding the predicate $x\notin x$. But there are infinite families of predicates that all cause essentially the same problem. For example, let $P(x)$ be the predicate $\lnot\exists y. x\in y \wedge y\in x $. Then there is no set of all $x$ such that $P(x)$ holds. I think there is one of these for any cyclic directed graph; the original Russell predicate $x\notin x$ corresponds to the graph with one vertex and one directed edge.

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    $\begingroup$ There is a magnificent construction of models with ZF without regularity which essentially says that almost any exetensional relationship can be embedded into $\in$ of some model of the theory. $\endgroup$
    – Asaf Karagila
    May 2, 2012 at 16:08
  • $\begingroup$ Where can I read more about that? $\endgroup$
    – MJD
    Mar 12, 2015 at 13:22
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    $\begingroup$ Can you read Hebrew? $\endgroup$
    – Asaf Karagila
    Mar 12, 2015 at 14:04
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    $\begingroup$ Not so far.${}{}$ $\endgroup$
    – MJD
    Mar 12, 2015 at 14:07
  • $\begingroup$ So I'll try to find something else, other than the handwritten notes of my professor. $\endgroup$
    – Asaf Karagila
    Mar 12, 2015 at 14:28
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Curry's paradox is somewhat different. It considers the set $X_Y = \{ x : x\in x \implies Y \}$. One can show that if this set exists, then $Y$ is true. (See the Wikipedia article for the simple proof.) So if your theory allows the $X_Y$ to exist for all $Y$, then all $Y$ are true and the theory is inconsistent.

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