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I want to prove "There is no set to which every function belongs." Can I approach it as follows?

Attempt No. 1: Let $$F: A\rightarrow B$$ Since $$F\subset A\times B,$$ it follows that $$F\in \mathcal{P}(A\times B).$$ Now, let $$\mathcal{P}(A\times B)$$ be the set of all functions from A into B. Since $$\mathcal{P}(A\times B)\subseteq \mathcal{PP}(A\times B),$$ it follows that $$\mathcal{PP}(A\times B)$$ is also a set of all functions from A into B. Therefore $$\mathcal{P}(A\times B)=\mathcal{PP}(A\times B).$$

Attempt No. 2 is to approach it by the concept of Russel Paradox. Then I must build a set of all functions that are not in itself, but, honestly, I have not a clue of what constitute a function that is not in itself.

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    Maybe this is too naive, but isn't the subset of all identity functions in bijection with the class of all sets?2011-06-04
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    @Theo: That seems true (I'm naive also), but what do you do next? Does this fact, that there's a bijection to a "smaller" set, help in some way?2011-06-04
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    @ShreevatsaR: I'd say: If the collection of all functions were a set then the subcollection of all identity functions would be a set as well, but it isn't by Russell.2011-06-04

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