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May I refer you to:

Lemma 4.1.6 in:

Page $4$

OK so I want to understand why that family forms a subbasis for the compact open topology. By definition of subbasis we must show that the each basic open set is a finite intersection of sets of elements of the form $[K,V]$ right? My question is, it seems that the author is just showing that each element of $[K,U]$ is a finite union of finite intersections of elements of the collection, why this implies that it is a subbasis?

I'm really confused with the statement. Can you please explain what we really have to prove?

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See the second bullet point underneath the "Theorems" heading at the Wikipedia article on topological bases. – Dylan Moreland Aug 11 '11 at 2:08
up vote 5 down vote accepted

Let $\mathcal{T}$ be the compact-open topology on $C(X,Y)$. Let $$\mathcal{S}_0 = \{[K,U]:K\subseteq X\text{ is compact and }U\subseteq Y\text{ is open}\},$$ the standard subbase for $\mathcal{T}$, and let $\mathcal{B}_0$ be the associated base, $$\mathcal{B}_0 = \{\bigcap\mathcal{F}:\mathcal{F}\subseteq \mathcal{S}_0\text{ and }\mathcal{F}\text{ is finite}\}.$$

In Lemma $4.1.6$ we’re given a subbase $\mathcal{S}$ of $Y$, and the claim is that $$\mathcal{S}_1 = \{[K,V]:V \in \mathcal{S},K \subseteq X\text{ compact}\}$$ is a subbase for $\mathcal{T}$. Stripped of its details, the argument given there shows that if $$\mathcal{B}_1 = \{\bigcap\mathcal{F}:\mathcal{F}\subseteq \mathcal{S}_1\text{ and }\mathcal{F}\text{ is finite}\}$$ is the base generated by $\mathcal{S}_1$, then for any $[K,U]\in \mathcal{S}_0$ and any $\gamma\in [K,U]$ there is an $H\in \mathcal{B}_1$ such that $\gamma \in H \subseteq [K,U]$. From this it follows immediately that each member of $\mathcal{S}_0$ is a union of members of $\mathcal{B}_1$, and you’re wondering why this is sufficient to show that $\mathcal{S}_1$ is a subbase for $\mathcal{T}$.

To show that $\mathcal{S}_1$ is a subbase for $\mathcal{T}$, we need only show that $\mathcal{B}_1$ is a base for $\mathcal{T}$, i.e., that every member of $\mathcal{T}$ is a union of members of $\mathcal{B}_1$. This will certainly follow if we can show that every member of the base $\mathcal{B}_0$ is a union of members of $\mathcal{B}_1$. Fix $B \in \mathcal{B}_0$, say $$B = [K_1,U_1]\cap [K_2,U_2]\cap\dots\cap [K_n,U_n],$$ where $[K_1,U_1],[K_2,U_2],\dots,[K_n,U_n] \in \mathcal{S}_0$. Suppose that $\gamma\in B$. Then $\gamma\in [K_i,U_i]$ for $i=1,2,\dots,n$, so the proof of Lemma $4.1.6$ guarantees that there are sets $H_i \in \mathcal{B}_1$ such that $\gamma \in H_i \subseteq [K_i,U_i]$ for $i=1,2,\dots,n$. Let $H = \bigcap\limits_{i=1}^n H_i$; then $H\in \mathcal{B}_1$, and $\gamma \in H \subseteq B$. It follows that $B$ is a union of members of $\mathcal{B}_1$, which is exactly what we needed.

(I suppose that I should mention in passing that since $\mathcal{S}_1$ is clearly a subset of $\mathcal{T}$, it cannot generate a topology finer than $\mathcal{T}$; the only question is whether it generates all of $\mathcal{T}$.)

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