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I would like to show precisely what I have stated in the title (assuming that it is correct; I have reason to suspect it is, thanks to a tricky past exam paper I'm trying to surmount); namely, that given any 2 compact Hdf topological spaces $T_1$ and $T_2$, the class of continuous functions between them forms a set (rather than, say, a class which is too large to be a set).

I suspect this result might be obvious in a much more general sense, and just wanted to check my thinking: is it actually valid to say that for any function between 2 fixed sets $S_1$ and $S_2$, the function is simply a set of ordered pairs $(a,f(a))$ and therefore contained in the power set of the union $S_1 \cup S_2$, so is again a set (by standard set-theoretic axioms for power sets and unions)?

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    @Zev: The only possible way your argument could fail is if the set theory in question doesn't have a sufficiently strong separation axiom or if it doesn't have function spaces. The former is highly unlikely in any reasonable set theory, and it's very hard to do any point set topology in the latter (since there would be no powersets!).2012-05-15

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tl;dr You are correct.

One usually takes the ordered pair $(a, b)$ to be an abbreviation for the set $\{\{a\}, \{a, b\}\}$, so if $a\in A$ and $b\in B$, then $\def\pow#1{{\mathcal P}{(#1)}} (a,b)\in\pow{\pow{A\cup B}}$, and the cartesian product $A\times B$ is a subset of $\pow{\pow{A\cup B}}$, which is a set by the union and power set axioms.

A function $f:A\to B$ is a subset of $A\times B\subset \pow{\pow{A\cup B}}$, and so is a set and also is an element of $\pow{\pow{\pow{A\cup B}}}$. The set $\mathcal F$ of all functions $f:A\to B$ is therefore a subset of $\pow{\pow{\pow{A\cup B}}}$ and is a set. The set of all continuous functions is a subset of $\mathcal F$, and is therefore a set, an element of $\pow{\pow{\pow{\pow{A\cup B}}}}$.

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    Great, thankyou.2012-05-15