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Given topological spaces $X$ and $Y$, we often consider the collection of continuous functions, $f: X \rightarrow Y$. My question is, given two sets $X$ and $Y$, and a sub-collection $\{g_{i}\}$ of the collection of all functions from $X$ to $Y$, do there exist topologies on $X$ and $Y$ so that the $g_{i}$ are precisely the continuous functions from $X$ to $Y$?

I have a few thoughts on this, if $\{g_{i}\}$ is the entire collection of functions from $X$ to $Y$, then we can give $X$ the discrete topology and give $Y$ any topology and the result follows. If $\{g_{i}\}$ is just the constant functions (which are always continuous) then we can give $Y$ the indiscrete topology and give $X$ any topology and the result follows. Aside from these two trivial cases, is there anything we can say about this question? Assume that $\{g_{i}\}$ contains the constant functions.

Dylan Yott
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    If you pick any topology on $Y$, then there is a smallest topology on $X$ such that the $g_i$ are all continuous (the one generated by all preimages of open sets in $Y$ under the $g_i$). As Cameron points out, there may then be extra continuous functions between $X$ and $Y$. What isn't immediately clear to me (but I'd be interested to know) is what conditions you need on the set ${g_i}$ such that this doesn't happen; this could easily depend on the choice of topology for $Y$. Alternatively, are there conditions on the $g_i$ giving a unique topology on $X$ and $Y$ so they're continuous. – mdp Mar 13 '13 at 17:00
  • A related question: http://math.stackexchange.com/q/382690/56914 –  May 23 '13 at 17:35
  • Also related: http://math.stackexchange.com/questions/427634/a-topology-such-that-the-continuous-functions-are-exactly-the-polynomials#427634 – Dominik Jun 23 '13 at 16:34

2 Answers2

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There need not be any such topologies in general. You've noted, for example, that all constant functions are continuous, so if your sub-collection of functions does not have all constant functions as elements, then we're out of luck in topologizing $X$ and $Y$ in the desired fashion.

There may be other necessary conditions on your subcollection to determine such topologies on $X$ and $Y$. I'll think on it.

Added: It occurred to me (very belatedly, of course) that this answer to a prior question of my own gives a very important necessary condition for such a topology to exist, if we are considering self-maps (meaning $X$ and $Y$ are the same topological space). Namely, the sub-collection of functions in question must be closed under finite compositions, i.e.: given elements $f,g$ of the sub-collection, $f\circ g$ is also an.element.

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Cameron Buie
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Assume $Y$ only consists of two points $Y=\{a,b\}$. There are effectively three choices for a topology on $Y$. Let us assume we don't have the indiscrete and not the discrete topology, so wlog the open sets are precisely $Y$, $\emptyset$ and $\{a\}$. Now every $g_i$ must be continuous, in other words $g_i^{-1}(a)$ has to be open. So the collection $\{g_i\}$ is the same as a subbasis of $X$. Conversely every open set $O$ in the topology of $X$ gives you a continuous function $g:X\to Y$ with $g(O)=\{a\}$ and $g(X-O)=\{b\}$.

So in this case you are asking if any subbasis of a topology is already a topology. I guess if $Y$ is bigger the situation will get worse.

Cameron Buie
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Simon Markett
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