Compact open topology and nets If $X$ and $Y$ are topological spaces we can form a topology on $Y^X$, which has as subbasis sets of the form $B(T,U) := \{f \in Y^X : f(T) \subset U \}$ where $T$ compact and $U$ open.
Is there a nice way to describe convergence of nets in this topology ? Maybe for more specific spaces $X$ and $Y$ ? 
 A: Fix $f\in Y^X$, and let $\nu:D\to Y^X$ be a net based on the directed set $\langle D,\preceq\rangle$. Then $\nu\to f$ iff for each finite family $\{B(F_k,U_k):k=1,\ldots,n\}$ of subbasic open nbhds of $f$, $\nu$ is eventually in $\bigcap_{k=1}^nB(F_k,U_k)$. Since $D$ is directed, this is equivalent to saying that $\nu$ is eventually in $B(F,U)$ for each subbasic open set $B(F,U)$ containing $f$: 

$\nu\to f$ iff whenever $F\subseteq X$ is compact, and $U$ is an open nbhd of $f[F]$ in $Y$, there is a $d\in D$ such that $\nu(d')\in B(F,U)$ for all $d'\in D$ such that $d\preceq d'$.

This is an exact analogue of convergence in the product topology. To see this, give $Y^X$ the product topology, and for each $x\in X$ and open $U\subseteq Y$ let $B(x,U)=\{f\in Y^X:f(x)\in U\}$; the sets $B(x,U)$ are of course a subbase for the product topology. Let $\nu$ be as above. Then

$\nu\to f$ iff whenever $x\in X$, and $U$ is an open nbhd of $f(x)$ in $Y$, there is a $d\in D$ such that $\nu(d')\in B(x,U)$ for all $d'\in D$ such that $d\preceq d'$.

In other words, we’ve just replaced convergence at each point of $X$ (i.e., pointwise convergence) with convergence at each compact subset of $X$ — compactwise convergence, to coin a phrase.
I don’t know how useful this is, but it does seem to be the sort of test suggested by your comment.
