Compact Sets, Compact Open Topology

Compact Open Topology

Let X and Y be topological spaces and consider the set of continuous functions from X into Y. This is a valid set, but we would like to turn it into a topological space. In other words, certain collections of these functions will be open and/or closed.

For every compact set K in X, and for every open set U in Y, let B(K,U) be the collection of functions that map K into U. If you want to consider the empty set as compact, which I suppose it is, then B(∅,Y) includes every function. That's all right; the entire collection is suppose to be open. Similarly, B(X,∅) is empty, whence the empty collection of functions is open. (If X is not compact then choose any compact set in X, such as the point x.)

However, B(K,U) is not a base for the topology. Let's look at a simple example. Let X consist of two points a and b with the discrete topology, and let Y be the real line. Let G consist of functions that map a into the open interval (0,2), and let H map b into (2,4). Now G∩H is the set of functions, and there certainly are continuous functions like this, that map a into (0,2) and b into (2,4). Let f map a and b to 1 and 3 respectively. This should be contained in a base set, which is contained in the intersection. Restricting either a or b alone won't do, hence K is the entire domain X. Both images must be contained in an open set U. Clearly U contains 1 and 3. This brings in the function that maps both a and b to 1, and that is not part of G∩H. So these collections do not act as a base. However, they do act as a subbase, as we shall see below.

Consider all finite intersections of the sets B(K,U). This will act as the base for the compact open topology. (It is called "compact open" because K is compact and U is open.) The problem we ran into earlier, where G∩H was not covered by base sets, is now solved, because G∩H is automatically a base set. In other words, the intersection of two finite intersections is another finite intersection. So we have a base, and the arbitrary union of base sets is open in the compact open topology.

One Point

When X is a single point, a function from X into Y is the same as a point of Y. Furthermore, the point X is the one and only compact set K. Thus an open set in the compact open topology corresponds to an open set in Y. The spaces are homeomorphic.

Hausdorff

If Y is hausdorff then so is the space of functions from X into Y. Let f and g be two different functions, that differ on a point p. Let K = p. The functions that map p into f(p) form an open set, and the functions that map p into g(p) form another, disjoint open set. Thus the function space is hausdorff.

Components

If X is connected, then the image of X is connected, and lies entirely in one of the components of Y. The functions into Y are merely the functions into the components of Y. Select a point p in X, and a component Y₀ of Y, and the functions that map p into Y₀ form an open set. These are all the functions from X into Y₀. Therefore the functions from X into Y are the functions from X into the components of Y, and these become separate components in the function space.

Compact Convergence Topology

Let Y be a metric space. If f is a continuous function from X into Y, and K is compact in X, the functions that are within ε of f, on the domain K, form an open set. This is similar to an open ball in a metric space. These open balls of functions form a base for the compact convergence topology.

Of course we need to prove this is a base. Start with two open balls based on f₁ K₁ and r₁, and f₂ K₂ and r₂. Let g be a function in the intersection. Since K₁ is compact, there is a distance d₁ between f₁ and g, across K₁, and similarly there is a distance d₂ between f₂ and g across K₂. Note that d₁ < r₁, and d₂ < r₂.

Let K₃ be the union of K₁ and K₂. This is a compact set. Let ε be smaller than r₁-d₁ and r₂-d₂. The set of functions within ε of g across K₃ is contained in our original open sets. This holds for every g in the intersection, hence we have a valid base, and a valid topology.

When Both Topologies Coincide

If X is locally compact and Y is a metric space, compact open and compact convergence are equivalent.

Consider a continuous function f from X into Y, and a distance ε. This defines an open ball of functions, a base set under compact convergence. We want to cover this ball with open sets under the compact open topology.

Let g be any function that is uniformly within ε of f. In other words, g is a point in our open ball.

Let d be the distance between g and f. Note that d < ε. Choose γ so that it is less than half the difference between ε and d.

Let a be an arbitrary point in K. Choose an open neighborhood about a whose image is within γ of g(a). Then restrict O to a (possibly smaller) base open set. Let C be the closure of O. Since X is locally compact, C is compact. Find such an O_a and C_a for each point a in K.

Now K is covered by open sets O_a, and a finite subcover will do. Hereinafter, consider the finite family of open sets O_a and their compact closures C_a.

Let S_a be the set of continuous functions that map C_a into the open ball of radius γ centered at g(a). Let S be the intersection over all S_a. This is an open set in the compact open topology. We need to prove every function in S is within ε of f.

Let b be an arbitrary point in K. Find a such that O_a contains b. Now S(b) is within γ of g(a). Also, g(b) is within γ of g(a). By the triangular inequality, S(b) and g(b) differ by no more than 2γ. This keeps S(b) within ε of f(b). The open set lies entirely within our ball. The open ball is indeed open relative to the compact open topology.

Let's look at the converse. Let S be the intersection over B(K_i,U_i). Let f be a function belonging to S. Focus in on a particular K and U. The image of K is a compact set, contained in the open set U. Place each point of the image in an open ball, inside U. Finitely many balls cover the image. Let r be the smallest radius. Any function within r of f still maps K into U. Of course we must deal with K₁ through K_n, mapping into U₁ through U_n, but this is handled by setting r to the smallest radius across the board. Thus an open ball of functions, of radius r, centered at f, across the domains K₁ through K_n, (the finite union of compact sets is compact), is contained in S. Thus S is covered by open balls, and is open under compact convergence.

When X is locally compact and Y is a metric space, the two topologies, compact open and compact convergence, are equivalent. The latter is often more convenient, since the open balls form a base, rather than a subbase. It is more intuitive to say, "The functions close to f on K", rather than "The finite intersection of functions that map various compact sets K into various open sets U." Of course X and Y could be exotic spaces, whence the compact open topology cannot be expressed in a simpler, equivalent form.