compact convergence topology If $Y$ is a topological space, $Z$ a metric space. We can define the topology of compact convergenc  and we have the compact open topology on $Z^Y$. Why are these two the same topology on $Z^Y$ ?
Thanks in advance 
 A: Allow me to start with a short recapitulation of compact convergence and the compact-open topology:  

Let $\cal K$ denote the set of all compact subsets of $X$. For any $K\in\cal K$ there is a restriction mapping $r_K:Y^X\to Y^K$, sending $f:X \to Y$ to its restriction $f|_K$. If the spaces $Y^K$ are given the topology of uniform convergence, then the topology of compact convergence is the coarsest topology on $Y^X$ such that all the $r_K$'s are continuous.
  The compact-open topology on $Y^X$ has a subbase formed by the sets
  $W(K,U)=\{g\in Y^X\mid g(K)⊆U\}$ where $K$ is compact and $U$ is open.

If the compact-open topology on $Y^X$ makes all these restriction mappings continuous, then it is finer than the topology of compact convergence. To show this, consider a compact set $K\subseteq X$ and a map $f:X\to Y$. The mapping $r_K:f\mapsto f|_K$ is continuous at $f$ if given an entourage $V$ of $Y$ (so $V(f|_K)$ is a neighborhood of $f|_K$) there is an open set  $B$ in the compact-open topology such that $f\in B$ and $g|_K\in V(f)$ for all $g\in B$.
There exists a symmetric entourage $W$ such that $W^6\subseteq V$. Now for every point $x\in K$ there is an open neighborhood $U_x$ in $K$ such that $f(U_x)\subseteq W(f(x))$. These open sets cover $K$, so by compactness there are $x_1,...,x_n$ such that $K=U_1\cup..\cup U_n$. Note that each relative closure $\overline U_i$ is compact and 
$$f\left(\overline U_i\right)⊆\overline{W(f(x_i))}⊆\overline W(f(x_i))
⊆W^3(f(x_i))$$
So it seems adequate to consider the basic set
$B:=\bigcap_{i=1}^n W\left(\overline U_i,W^3(f(x_i))\right)$. Clearly $f\in B$. Suppose that $g\in B$. For any $y\in K$ there is a $k$ so that $y\in U_k$. Then $g(y)\in W^3(f(x_k))$. But since $f\in B$ we also know that $f(y)\in W^3(f(x_k))$. As $W$ is symmetric if follows that $(f(y),g(y))\in W^6\subseteq V$. Hence $g|_K\in V(f|_K)$.
For the reverse direction, it suffices to show that whenever $f:X\to Y\in W(K,U)$, then there is an entourage $V$ such that $\{g:X\to Y\mid g|_K\in V(f|_K)\}⊆W(K,U)$. The set $f(K)$ is a compact subset of the open set $U$. There is thus an entourage $V$ such that $V(f(K))⊆U$. It follows that if $g:X\to Y$ has a restriction in $V(f|_K)$, then $g\in W(K,U)$. So each $W(K,U)$ is actually open in the topology of compact convergence.
