# General Topology/The compact-open topology

**Definition (compact-open topology)**:

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

- ,

where ranges over all compact subsets of and ranges over all open subsets of .

**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 be a topological space and let be a uniform space. Let be the collection of all compact subsets of . Then the subspace topology on induced by the topology of -convergence is at least as fine as the subspace topology on that is induced by the compact-open topology.

**Proof:** We prove that any neighbourhood of an arbitrary in the compact-open topology contains a neighbourhood of in the topology of uniform convergence on compact subsets of . Thus, let be arbitrary. Thus, suppose that , where is compact and non-empty and is open; any neighbourhood of with respect to the compact-open topology will be the finite intersection of sets of this form. Let now be arbitrary. By the definition of the topology induced by a uniform space, the set of those entourages of such that is nonempty. Moreover, for each such , we may choose an entourage of such that . For each such entourage, let be an open neighbourhood of such that . We shall denote the collection of all such by . Then the union of all these , ie. the collection

- ,

constitutes an open cover of , because each is nonempty and hence contains an open set that contains . But is compact, so that we may choose a finite subcover . By definition, each is identical to one of the previously defined 'es, so that there is an entourage and a point such that and . Now define

- .

We claim that is a neighbourhood of that is contained within . Indeed, let . If is arbitrary, there exists a such that . From the definition of , we infer that . Yet we also know that , so that , whence . Since was arbitrary, and .

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

Let be a locally compact space and let be a uniform space. Let be the collection of all compact subsets of . Then the subspace topologies on induced by the topology of -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 with respect to the topology of uniform convergence on compact subsets contains a neighbourhood of with respect to the compact-open topology. Hence, let be any entourage of and let be compact, so that represents an arbitrary element of the canonical neighbourhood basis of with respect to the topology of uniform convergence on compact sets. We choose an entourage of such that . Now is locally compact, so that for each point , the collection of compact neighbourhoods of such that is non-empty. The collection of all those we shall denote by . Now the collection of all (where ranges over all of ) is an open cover of , whence we may choose a finite subcover . Since the interior is a *subset* of its original set, the sets cover . Moreover, by definition, each has an such that . We claim that

is contained within . Indeed, suppose that , and let . Let such that . Since , in particular . But as well, so that . Thus,

- ,

and since was arbitrary, .