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Given spaces $X,Y$, where $X$ is Hausdorff, and the topology on $Y$ has basis $\mathcal{U}$, I would like to show that the set $\mathcal{S} := \{ S(K,U) \mid K \subset X, K \text{ compact, } U \in \mathcal{U} \}$, where $S(K,U) = \{ f \in C(X,Y) \mid f(K) \subset U\}$ defines a sub-basis for the compact-open topology on $C(X,Y)$.

I know that $\mathcal{S}$ is contained inside the usual subbasis for $C(X,Y)$. If I were dealing with bases, I would just pick some $f \in C(X,Y)$ and some basis element $B$ of $C(X,Y)$ containing $f$, then prove that there exists some $S \in \mathcal{S}$ with $f \in S \subset B$ (a la Lemma 13.3 in Munkres). But I don't know that such a result works for just subbases. If anyone could shed some light on a result like this, I'd appreciate it.

Other than this potential approach, I'm really not used to work with subbasis, and this is my first time seeing the compact-open topology so I'm not sure what I should do here. Any hints would be nice! Thanks!

Michael Hardy
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user121410
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2 Answers2

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Let us assume either $X$ or $Y$ has the property that every compact subspace is normal. This holds in regular spaces, for example. In a normal space, every finite open cover $(U_i)_i$ has a shrinking, that is an open cover $(V_i)_i$ satisfying $\overline V_i \subseteq U_i$.

Given a map $f:X\to Y$ sending the compact set $K$ to the open set $U$, the image $f(K)$ is covered by finitely many basic sets $\mathcal B = \{B_1,\dots, B_n\}$ contained in $U$, and $K$ is covered by $\mathcal B' = \{f^{-1}(B_1)\cap K,\dots,f^{-1}(B_n)\cap K\}$. By assumption, either $\cal B'$ has a shrinking $\mathcal A' = \{A'_1,\dots,A'_n\}$, or $\mathcal B \cap f(K)$ has a shrinking $\mathcal A = \{A_1,\dots,A_n\}$ in which case $(A'_i = f^{-1}(A_i)\cap K)_1^n$ is a shrinking of $\cal B'$.

Let $C_i$ be the closure of $A'_i$ in $K$. Then these sets are compact and cover $K$, and $f(C_i)\subseteq B_i$, thus $f\in S(C_i,B_i)$. It follows that $f \in \bigcap_{i=1}^n S(C_i,B_i) \subseteq S(K,U)$. We conclude that the sets $S(K,B)$ for compact $K$ and basic $B$ constitute a basis for the compact-open topology in $C(X,Y)$, at least for spaces $Y$ with the property stated above.

Stefan Hamcke
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Hint: $\mathcal B$ is a subbase for a space $X$ if and only if the collection $\mathcal C$ given by:

$$ \mathcal C = \{W\subset X:W\textrm{ can be written as a finite intersection of elements of }\mathcal B\}\cup\{X,\emptyset\} $$

is a basis of open sets for $X$.

Now you should be able to apply your knowledge of bases.

John Gowers
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