Compact-Open Convergence and Uniform Convergence on Compacts

Two Ways to Control Continuous Maps

Let \(C(X,Y)\) denote the set of continuous maps from \(X\) to \(Y\). A neighborhood of a map can impose conditions on the image of a whole compact subset of the domain, or it can bound the distance between two maps at every point of that subset. The compact-open topology uses the first kind of condition. Uniform convergence on compact subsets uses the second. We will show that, when \(X\) is locally compact Hausdorff and \(Y\) is metric, these give the same topology on \(C(X,Y)\).

The supremum defining \(p_K\) is finite: \(x\mapsto d(f(x),g(x))\) is continuous, and its image on compact \(K\) is compact, hence bounded in \(\mathbb{R}\). A basic neighborhood in the uniform-on-compacts topology is a finite intersection of the displayed sets, potentially using different compact sets and tolerances. In particular, it does not require one bound to hold over all of \(X\).

We use one fact about compact Hausdorff spaces to turn varying pointwise constraints into finitely many constraints on compact pieces.

Proof. First fix \(x\in K\) and an open set \(O\subseteq K\) containing \(x\). If \(K\setminus O\) is empty, take \(W=K\). Otherwise, for each \(y\in K\setminus O\), the Hausdorff property gives disjoint open neighborhoods \(A_y\) of \(x\) and \(B_y\) of \(y\). The sets \(B_y\) cover the compact set \(K\setminus O\), so finitely many, say \(B_{y_1},\ldots,B_{y_n}\), cover it. Put \(W=\bigcap_{j=1}^n A_{y_j}\). This is an open neighborhood of \(x\), disjoint from \(\bigcup_{j=1}^n B_{y_j}\). Since that union is open and disjoint from \(W\), it is also disjoint from the closure \(\overline{W}\) in \(K\). Consequently \(\overline{W}\subseteq O\). Now apply this construction to each member of the given cover at each point it contains. The resulting open neighborhoods cover \(K\), so finitely many cover \(K\). Their closures are compact, still cover \(K\), and each is contained in its corresponding cover member. \(\square\)

The Two Neighborhood Translations

Proof. We prove both inclusions of topologies. First consider a compact-open subbasic set \([K,U]\), and take \(f\in[K,U]\). If \(K=\varnothing\), then \([K,U]=C(X,Y)\). If \(Y\setminus U=\varnothing\), the same conclusion holds. Otherwise, the function \(y\mapsto d(y,Y\setminus U)\), the distance from \(y\) to the nonempty closed set \(Y\setminus U\), is continuous and positive on the compact set \(f(K)\). It therefore has a positive minimum \(\delta\) there. If \(p_K(f,g)<\delta\), then for every \(x\in K\), \(d(g(x),f(x))<\delta\leq d(f(x),Y\setminus U)\), so \(g(x)\in U\). Thus the uniform-on-compacts neighborhood defined by \(K\) and \(\delta\) is contained in \([K,U]\). Every compact-open subbasic set is therefore open in the uniform-on-compacts topology.

For the other inclusion, fix \(f\in C(X,Y)\), a compact \(K\subseteq X\), and \(\varepsilon>0\). We show that every \(g\) with \(p_K(f,g)<\varepsilon\) has a compact-open neighborhood contained in \(\{h:p_K(f,h)<\varepsilon\}\). There is nothing to prove if \(K=\varnothing\). Otherwise let \(r=p_K(f,g)\) and \(m=\varepsilon-r>0\), and put \(\alpha=m/4\). For each \(x\in K\), continuity of \(f\) and \(g\) gives a relatively open neighborhood \(O_x\) of \(x\) in \(K\) such that, for every \(y\in O_x\), \(d(f(y),f(x))<\alpha\) and \(d(g(y),g(x))<\alpha\). The set \(K\) is compact Hausdorff, so the lemma gives a finite compact cover \(C_1,\ldots,C_q\) of \(K\), with each \(C_i\subseteq O_{x_i}\) for some \(x_i\in K\). Each \(C_i\) is also compact as a subset of \(X\). Let \(U_i\) be the open ball in \(Y\) with center \(g(x_i)\) and radius \(2\alpha\). The bounds on \(g\) show \(g(C_i)\subseteq U_i\), so \(g\) belongs to the compact-open neighborhood \(\bigcap_{i=1}^q[C_i,U_i]\).

If \(h\) belongs to this neighborhood and \(y\in C_i\), then \(d(h(y),g(x_i))<2\alpha\). The triangle inequality gives \(d(f(y),h(y))\leq d(f(y),f(x_i))+d(f(x_i),g(x_i))+d(g(x_i),h(y))<\alpha+r+2\alpha=r+3m/4<\varepsilon\). Since the \(C_i\) cover \(K\), this holds for every \(y\in K\), and hence \(p_K(f,h)<\varepsilon\). We have found a compact-open neighborhood inside the specified uniform neighborhood at each of its points, so that uniform neighborhood is compact-open open. This proves the reverse inclusion and the equality of the topologies. \(\square\)

The proof of the reverse inclusion uses that compact subsets of a Hausdorff space are compact Hausdorff; local compactness is part of the stated hypothesis but is not needed for this particular equality. The metric on \(Y\) supplies distances and open balls. No completeness assumption on \(Y\) is required.

Worked Neighborhood Conversions

What the Equivalence Does—and Does Not—Say

The equality is useful because it lets us choose the language that best fits a problem. To show a map lies in a compact-open neighborhood, it is often enough to check that its image on a compact set stays inside an open target set. To discuss closeness of maps, a bound on \(p_K\) is often more direct. The theorem guarantees that these descriptions give the same neighborhoods; it does not turn compact control into control over the entire domain when the domain is noncompact.

Common pitfall. A compact-open condition \([K,U]\) does not generally say that a map is uniformly close to a particular nonconstant map on \(K\). For a concrete check, take \(K=[-1,1]\), \(U=(-2,2)\), \(f(x)=x\), and \(h(x)=0\). Then \(h(K)=\{0\}\subseteq U\), so \(h\in[K,U]\), but \(p_K(f,h)=\sup_{-1\leq x\leq1}|x-0|=1\). In particular, this compact-open condition does not imply \(p_K(f,h)<\tfrac12\). The correction is to compare the target values with \(f\), using sufficiently small compact pieces as in the first worked example, or to use the uniform-on-compacts neighborhood directly.