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No set to which all functions belong

I'm having trouble proving that there is no set of all functions. I tried the approach similar to the approach used to show that there is no set of all sets, but was not able to make this work.

First I assumed the set $A$ of all functions, and then constructed the subset $B$ where each member of the subset is not a member of itself. I could then prove that the subset was not a member of $A$. To complete the proof I need to show that $B$ is a function, but $B$ doesn't seem to be a function.

Would like some guidance on the remaining steps, or a better approach to take.

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marked as duplicate by t.b., Asaf Karagila, Zev Chonoles Jul 13 '12 at 5:19

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1 Answer

up vote 6 down vote accepted

For every set $X$ there exists a (unique) function $\{X\}\to \{1\}$.

Now, if $A$ were the set of all functions, then what would $\bigcup_{f\in A} \operatorname{Dom} f$ be?

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Beat me to it! :-) –  Asaf Karagila Dec 3 '11 at 23:37
    
This gets particularly annoying if "functions" is replaced by "nets on a particular set". –  dfeuer Dec 4 '11 at 4:37
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