General Topology/The compact-open topology

Definition (compact-open topology):

Let $X$ and $Y$ be two topological spaces, and let $F(X,Y)$ be the set of all functions from $X$ to $Y$. The compact-open topology on $F(X,Y)$ is defined to be the topology a subbasis for which is given by the sets

$M(K,U):=\{f\in F(X,Y)|f(K)\subseteq U\}$,

where $K$ ranges over all compact subsets of $X$ and $U$ ranges over all open subsets of $Y$.

Proposition (the topology of uniform convergence on compact sets is at least as fine as the compact-open topology on spaces of continuous functions):

Let $X$ be a topological space and let $Y$ be a uniform space. Let $S$ be the collection of all compact subsets of $X$. Then the subspace topology on $C(X,Y)$ induced by the topology of $S$-convergence is at least as fine as the subspace topology on $C(X,Y)$ that is induced by the compact-open topology.

Proof: We prove that any neighbourhood of an arbitrary $f\in C(X,Y)$ in the compact-open topology contains a neighbourhood of $f$ in the topology of uniform convergence on compact subsets of $X$. Thus, let $f\in C(X,Y)$ be arbitrary. Thus, suppose that $f\in M(K,U)$, where $K\subseteq X$ is compact and non-empty and $U\subseteq Y$ is open; any neighbourhood of $f$ with respect to the compact-open topology will be the finite intersection of sets of this form. Let now $x\in K$ be arbitrary. By the definition of the topology induced by a uniform space, the set of those entourages $V_{x}$ of $Y$ such that $V_{x}(f(x))\subseteq U$ is nonempty. Moreover, for each such $V_{x}$, we may choose an entourage $W_{x}$ of $Y$ such that $W_{x}\circ W_{x}\subseteq V_{x}$. For each such entourage, let $U_{x}$ be an open neighbourhood of $x$ such that $f(U_{x})\subseteq W_{x}(f(x))$. We shall denote the collection of all such $U_{x}$ by $E(x)$. Then the union of all these $E(x)$, ie. the collection

$\bigcup _{x\in K}E(x)$,

constitutes an open cover of $K$, because each $E(x)$ is nonempty and hence contains an open set that contains $x$. But $K$ is compact, so that we may choose a finite subcover $U_{1},\ldots ,U_{n}$. By definition, each $U_{j}$ is identical to one of the previously defined $U_{x}$'es, so that there is an entourage $W_{j}$ and a point $x_{j}\in X$ such that $f(U_{j})\subseteq W_{j}(f(x_{j}))$ and $(W_{j}\circ W_{j})(f(x_{j}))\subseteq U$. Now define

$W:=W_{1}\cap \cdots \cap W_{n}$.

We claim that $W(K,f)$ is a neighbourhood of $f$ that is contained within $M(K,U)$. Indeed, let $g\in W(K,f)$. If $y\in K$ is arbitrary, there exists a $j\in \{1,\ldots ,n\}$ such that $y\in U_{j}$. From the definition of $W(K,f)$, we infer that $(f(y),g(y))\in W\subseteq W_{j}$. Yet we also know that $(f(x_{j}),f(y))\in W_{j}$, so that $(f(x_{j}),g(y))\in W_{j}\circ W_{j}\subseteq V_{j}$, whence $g(y)\in V_{j}(f(x_{j}))\subseteq U$. Since $y\in K$ was arbitrary, $g(K)\subseteq U$ and $g\in M(K,U)$. $\Box$

Proposition (the compact-open topology and the uniform convergence on compact sets coincide on continuous functions on locally compact spaces):

Let $X$ be a locally compact space and let $Y$ be a uniform space. Let $S$ be the collection of all compact subsets of $X$. Then the subspace topologies on $C(X,Y)$ induced by the topology of $S$-convergence and the compact-open topology resp., coincide.

Proof: We prove that both topologies generate the same neighbourhood systems. In view of the fact that the topology of uniform convergence on compact sets on spaces of continuous functions is at least as fine as the compact-open topology, it is sufficient to show that any neighbourhood of an arbitrary $f\in C(X,Y)$ with respect to the topology of uniform convergence on compact subsets contains a neighbourhood of $f$ with respect to the compact-open topology. Hence, let $V$ be any entourage of $Y$ and let $K\subseteq X$ be compact, so that $V(K,f)$ represents an arbitrary element of the canonical neighbourhood basis of $f$ with respect to the topology of uniform convergence on compact sets. We choose an entourage $W$ of $Y$ such that $W\circ W\subseteq V$. Now $X$ is locally compact, so that for each point $x\in K$, the collection of compact neighbourhoods $K_{x}$ of $x$ such that $f(K_{x})\subseteq W(f(x))$ is non-empty. The collection of all those $K_{x}$ we shall denote by $E(x)$. Now the collection of all ${\overset {\circ }{K_{x}}}$ (where $x$ ranges over all of $K$) is an open cover of $K$, whence we may choose a finite subcover ${\overset {\circ }{K_{1}}},\ldots ,{\overset {\circ }{K_{n}}}$. Since the interior is a subset of its original set, the sets $K_{1},\ldots ,K_{n}$ cover $K$. Moreover, by definition, each $K_{j}$ has an $x_{j}\in K_{j}$ such that $f(K_{j})\subseteq W(f(x_{j}))$. We claim that

$M(K_{1},W(f(x_{1})))\cap \cdots \cap M(K_{n},W(f(x_{n})))$

is contained within $W(K,f)$. Indeed, suppose that $g\in M(K_{1},W(f(x_{1})))\cap \cdots \cap M(K_{n},W(f(x_{n})))$, and let $x\in K$. Let $j\in \{1,\ldots ,n\}$ such that $x\in K_{j}$. Since $g(K_{j})\subseteq W(f(x_{j}))$, in particular $g(x)\in W(f(x_{j}))$. But $f(K_{j})\subseteq W(f(x_{j}))$ as well, so that $f(x)\in W(f(x_{j}))$. Thus,

$(f(x_{j}),f(x))\in W\wedge (f(x_{j}),g(x))\in W\Rightarrow (f(x),g(x))\in V$,

and since $x\in K$ was arbitrary, $g\in V(K,f)$. $\Box$