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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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Please forgive my ignorance of set theory, but working solely from the fact that $\mathsf{Set}$ is a category, my understanding is that the collection of all functions from a set $A$ to a set $B$ always forms a set, and the continuous functions between two topological spaces are of course just a subset of the set of all functions between them. Perhaps there is a different foundation you are working from where this is not the case? – Zev Chonoles May 15 '12 at 18:03
Yes, that is essentially what I was confirming. – Spyam May 15 '12 at 18:10
@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!). – Zhen Lin May 15 '12 at 19:20
up vote 5 down vote accepted

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. – Spyam May 15 '12 at 18:11

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