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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.

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    $\begingroup$ 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. $\endgroup$
    – mdp
    Commented Mar 13, 2013 at 17:00
  • $\begingroup$ A related question: math.stackexchange.com/q/382690/56914 $\endgroup$
    – user56914
    Commented May 23, 2013 at 17:35
  • $\begingroup$ Also related: math.stackexchange.com/questions/427634/… $\endgroup$
    – Dominik
    Commented Jun 23, 2013 at 16:34

2 Answers 2

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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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  • $\begingroup$ Of course, I meant to say that. I've edited the question appropriately. $\endgroup$
    – Dylan Yott
    Commented Mar 13, 2013 at 17:02
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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.

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  • $\begingroup$ What do you mean when you say the $\{g_{i}\}$ are a subbasis of $X$. Do you mean $\{g_{i}^{-1} (a)\}$? $\endgroup$
    – Dylan Yott
    Commented Mar 15, 2013 at 23:04
  • $\begingroup$ Yes, there is a one to one correspondence of the family of functions $\{g_i\}$ and the collection of sets $\{g_i^{-1}(a)\}$. $\endgroup$ Commented Mar 16, 2013 at 14:08

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