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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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The class of ordinals can't be a set: en.wikipedia.org/wiki/Burali-Forti_paradox –  Qiaochu Yuan May 2 '12 at 15:32
    
The class of all singletons can't be a set. –  Arturo Magidin May 2 '12 at 15:35
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@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. –  David Harris May 2 '12 at 15:42
    
@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. –  Arturo Magidin May 2 '12 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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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. –  Asaf Karagila May 2 '12 at 16:08
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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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