Compact-Open and Uniform Topologies on Compact Domains

Two Topologies on a Space of Functions

There are two natural ways to describe closeness between continuous functions. The compact-open topology specifies that a function must send a chosen compact subset into a chosen open subset of the codomain. The supremum metric instead bounds the distance between two functions at every point of the domain, using one common bound. When the domain is compact Hausdorff and the codomain is metric, these descriptions generate the same topology. The compactness and Hausdorff assumptions enter at different stages of the proof.

Let \(C(X,Y)\) denote the set of continuous maps from \(X\) to \(Y\). For a compact subset \(K\subseteq X\) and an open subset \(V\subseteq Y\), write

The compact-open topology is the topology generated by these sets as a subbasis. Thus its basic neighborhoods are finite intersections of sets \([K,V]\). If \(Y\) has metric \(d\), define

For compact \(X\), this supremum is finite whenever \(f,g\in C(X,Y)\): each image is compact and hence bounded in the metric space \(Y\). The metric \(d_\infty\) is the supremum metric, and its open balls generate the uniform topology. If \(X=\varnothing\), there is just one map from \(X\) to \(Y\); taking the empty supremum to be \(0\), both topologies on this one-point function space agree. If \(X\ne\varnothing\) but \(Y=\varnothing\), the function space is empty and the topologies also agree. We may therefore prove the result for nonempty \(X\) and \(Y\).

Compact-Open Neighborhoods Contain Uniform Balls

First, we show that each compact-open subbasic condition is open in the uniform topology. Fix \(f\in[K,V]\), where \(K\) is compact and \(V\) is open. If \(K=\varnothing\) or \(V=Y\), then \([K,V]=C(X,Y)\), so there is no restriction to check. Otherwise \(K\ne\varnothing\) and \(Y\setminus V\ne\varnothing\). The set \(f(K)\) is compact, since it is the continuous image of a compact space, and \(f(K)\subseteq V\).

For each \(y\in f(K)\), choose \(r_y>0\) such that \(B_d(y,r_y)\subseteq V\), using openness of \(V\). The balls \(B_d(y,r_y/2)\) cover \(f(K)\), so compactness gives a finite subcover \(B_d(y_i,r_i/2)\), \(1\le i\le m\), where \(r_i=r_{y_i}\). Set \(\delta=\min_{1\le i\le m}r_i/4>0\). If \(d_\infty(f,g)<\delta\) and \(x\in K\), choose \(i\) with \(d(f(x),y_i)<r_i/2\). Then

Thus \(g(x)\in B_d(y_i,r_i)\subseteq V\) for every \(x\in K\), so \(g\in[K,V]\). This gives a uniform-metric ball around each \(f\in[K,V]\) contained in \([K,V]\). Finite intersections of such subbasic sets are therefore uniform-open as well.

The compactness used here is compactness of \(K\): it lets us replace the point-dependent radii \(r_y\) by one positive bound that works across \(f(K)\). Since compact-open basic neighborhoods are finite intersections, it follows that every compact-open open set is uniform-open.

Uniform Balls Contain Compact-Open Neighborhoods

For the reverse direction, we need a consequence of compact Hausdorffness. If \(O\) is open in a compact Hausdorff space \(X\) and \(x\in O\), then there is an open set \(U\) such that \(x\in U\) and \(\overline{U}\subseteq O\). Here \(\overline{U}\) denotes the closure of \(U\).

Proof. Put \(F=X\setminus O\), a closed subset of the compact space \(X\), hence compact. If \(F=\varnothing\), take \(U=X\). Otherwise, for each \(y\in F\), the Hausdorff property gives disjoint open sets \(U_y\) and \(W_y\) containing \(x\) and \(y\), respectively. The sets \(W_y\) cover \(F\); choose a finite subcover \(W_{y_1},\ldots,W_{y_n}\). Let \(U=U_{y_1}\cap\cdots\cap U_{y_n}\). Then \(U\) is open, contains \(x\), and is disjoint from \(W_{y_1}\cup\cdots\cup W_{y_n}\). This union is open, so its complement is closed and contains \(U\); it therefore contains \(\overline{U}\). Since the union covers \(F\), its complement is contained in \(O\). Hence \(\overline{U}\subseteq O\). \(\square\)

Now fix \(f\in C(X,Y)\) and \(\varepsilon>0\). For each \(x\in X\), the set \(O_x=f^{-1}(B_d(f(x),\varepsilon/3))\) is open by continuity and contains \(x\). By the preceding fact, choose an open set \(U_x\) with \(x\in U_x\) and \(\overline{U_x}\subseteq O_x\). The sets \(U_x\) cover \(X\); compactness gives a finite subcover \(U_{x_1},\ldots,U_{x_n}\). Define \(K_i=\overline{U_{x_i}}\). Each \(K_i\) is compact because it is closed in compact \(X\), and the \(K_i\) cover \(X\). Also \(K_i\subseteq O_{x_i}\), so \(f(K_i)\subseteq B_d(f(x_i),\varepsilon/3)\).

Consider the compact-open neighborhood of \(f\)

It contains \(f\) by the inclusions just established. If \(g\in N\) and \(x\in X\), choose \(i\) with \(x\in K_i\). Then both \(g(x)\) and \(f(x)\) lie within distance \(\varepsilon/3\) of \(f(x_i)\), and so

This bound holds for every \(x\), hence \(d_\infty(f,g)\le2\varepsilon/3<\varepsilon\). Thus \(N\) lies inside the uniform ball of radius \(\varepsilon\) around \(f\). Every uniform-open set is therefore compact-open, proving the other inclusion and the theorem.

A Common Pitfall

A finite collection of pointwise conditions does not generally control a function on an entire compact set. For a concrete incorrect argument, take \(X=[0,1]\), \(f(t)=0\), and the compact-open subbasic set \([X,(-1,1)]\). A mistaken calculation might check \(g\) at finitely many sample points and conclude \(g(X)\subset(-1,1)\). To see why this does not follow, let \(S\subset[0,1]\) be any finite set. Choose \(c\in(0,1)\setminus S\), and choose \(r>0\) small enough that \([c-r,c+r]\subset(0,1)\setminus S\). The continuous function

vanishes at every point of \(S\), yet \(g(c)=2\), so \(g\notin[X,(-1,1)]\). The correction is to use compact-open conditions on compact sets that cover the domain, not merely finitely many sampled points. In the proof, compact Hausdorffness supplies compact closures contained in suitable open sets, and compactness supplies a finite cover by those closures.

The two directions of the theorem use the assumptions in distinguishable ways. Compactness makes the point-dependent controls uniform over compact sets and yields finite covers. Hausdorffness is used to shrink open neighborhoods so their closures stay within prescribed open sets. The conclusion is an equality of topologies, not just an agreement about which sequences converge: the proof constructs neighborhoods in both directions.