Separating Continuous Maps
A topological space of continuous maps can be studied by specifying which collections of maps count as open. In the compact-open topology, a basic test asks whether a map sends a chosen compact subset of the domain entirely into a chosen open subset of the target. To show that this function space is Hausdorff, we will apply that test to a particularly simple compact set: a singleton.
Let \((X,\mathcal{T})\) and \((Y,\mathcal{U})\) be topological spaces. Write \(C(X,Y)\) for the set of continuous maps from \(X\) to \(Y\). A subset \(K\subseteq X\) is compact if every cover of \(K\) by open sets has a finite subcover. A space is Hausdorff if each pair of distinct points has disjoint open neighborhoods.
In particular, open sets in this topology are unions of finite intersections of these subbasic sets. The definition places a compactness requirement on \(K\); not every subset of \(X\) is an eligible choice.
Proof. Let \(\mathcal{V}\) be any open cover of \(\{x\}\). Since it covers \(x\), at least one member \(V\in\mathcal{V}\) contains \(x\). Then the single set \(V\) covers \(\{x\}\), so \(\mathcal{V}\) has a finite subcover. Thus \(\{x\}\) is compact. \(\square\)
Evaluation at One Fixed Point
For each fixed \(x\in X\), define the point-evaluation map \(\operatorname{ev}_x:C(X,Y)\to Y\) by \(\operatorname{ev}_x(f)=f(x)\). This map reads off one value of a function. Its continuity follows directly from the compact-open subbasis and the compactness of \(\{x\}\).
Proof. Let \(U\) be any open subset of \(Y\). By the definition of point evaluation, \(\operatorname{ev}_x^{-1}(U)=\{f\in C(X,Y):f(x)\in U\}\). The condition \(f(x)\in U\) is equivalent to \(f(\{x\})\subseteq U\), so \(\operatorname{ev}_x^{-1}(U)=[\{x\},U]\). The singleton \(\{x\}\) is compact by the lemma, and therefore \([\{x\},U]\) is a compact-open subbasic set. It is open in \(C(X,Y)\). Since the preimage of every open \(U\subseteq Y\) is open, \(\operatorname{ev}_x\) is continuous. \(\square\)
This statement concerns evaluation at a fixed point of the domain. The separation argument below only needs this fixed-point property: two distinct functions differ at some point, and their values there can be separated in a Hausdorff target.
The Hausdorffness Theorem
Proof. Take distinct \(f,g\in C(X,Y)\). Since they are different functions, there is a point \(x\in X\) for which \(f(x)\ne g(x)\). Because \(Y\) is Hausdorff, there are disjoint open sets \(U,V\subseteq Y\) such that \(f(x)\in U\) and \(g(x)\in V\). The sets \([\{x\},U]\) and \([\{x\},V]\) are open in \(C(X,Y)\), since \(\{x\}\) is compact. Also, \(f\in[\{x\},U]\) and \(g\in[\{x\},V]\). These two open sets are disjoint: if \(h\) belonged to both, then \(h(x)\in U\) and \(h(x)\in V\), which would imply \(h(x)\in U\cap V=\varnothing\), an impossibility. Thus every distinct pair of points of \(C(X,Y)\) has disjoint open neighborhoods. If \(C(X,Y)\) has fewer than two elements, the Hausdorff condition holds vacuously; the same conclusion therefore covers those cases. \(\square\)
The hypotheses have distinct roles. The difference \(f(x)\ne g(x)\) supplies two target points to separate. Hausdorffness of \(Y\) supplies disjoint target neighborhoods. Compactness of the singleton ensures that the corresponding conditions on maps are open in the compact-open topology. No compactness or separation assumption on \(X\) is needed.
Worked Example: A Singleton Domain
Let \(X=\{p\}\) with its unique topology \(\{\varnothing,\{p\}\}\), and let \(Y=\mathbb{R}\) with its standard topology. Identify each map in \(C(X,\mathbb{R})\) with its single value at \(p\). Under this identification, the compact-open topology agrees with the standard topology on \(\mathbb{R}\).
The only subsets of \(X\) are \(\varnothing\) and \(\{p\}\), and both are compact. For any open \(U\subseteq\mathbb{R}\), \([\varnothing,U]\) is all of \(C(X,\mathbb{R})\), because the empty image is contained in every \(U\). The set \([\{p\},U]\) consists exactly of maps whose value at \(p\) belongs to \(U\). Thus the subbasic sets correspond to all standard open subsets \(U\) of \(\mathbb{R}\); they generate precisely the standard topology under the identification.
For a specific separation, take the maps \(f(p)=2\) and \(g(p)=5\). The open intervals \(U=(3/2,5/2)\) and \(V=(9/2,11/2)\) contain \(2\) and \(5\), respectively, because \(3/2<2<5/2\) and \(9/2<5<11/2\). They are disjoint since \(5/2<9/2\). The compact-open neighborhoods \([\{p\},U]\) and \([\{p\},V]\) therefore contain \(f\) and \(g\) and have no map in common.
Check: the map corresponding to a real number \(r\) belongs to \([\{p\},U]\) exactly when \(r\in U\). So the neighborhoods used for separation are exactly the ordinary open neighborhoods of \(2\) and \(5\), as expected.
Worked Example: A Noncompact Domain
Let \(X=\mathbb{R}\) with its standard topology, and let \(Y=\mathbb{R}\) with its standard topology. The domain is not compact: the open sets \((-n,n)\), for \(n\in\mathbb{N}\), cover \(\mathbb{R}\), while any finite subcollection is contained in \((-N,N)\) for its largest index \(N\), which does not cover \(\mathbb{R}\).
Consider the continuous maps \(f(t)=t\) and \(g(t)=t+1\). They differ at \(0\), where \(f(0)=0\) and \(g(0)=1\). Choose the open sets \(U=(-1/4,1/4)\) and \(V=(3/4,5/4)\). The values lie in the required sets because \(-1/4<0<1/4\) and \(3/4<1<5/4\). The sets are disjoint because \(1/4<3/4\).
The singleton \(\{0\}\) is compact, so \([\{0\},U]\) and \([\{0\},V]\) are open in \(C(\mathbb{R},\mathbb{R})\). The first contains \(f\), and the second contains \(g\). If a map \(h\) belonged to both, then \(h(0)\) would belong to \(U\cap V\), which is empty. These are therefore disjoint neighborhoods separating the maps, even though the domain is noncompact.
Check: the relevant values are \(f(0)=0\in U\) and \(g(0)=0+1=1\in V\). The construction requires only compactness of \(\{0\}\), not of all of \(\mathbb{R}\).
Why the Target Hypothesis Matters
The theorem gives a sufficient condition: a Hausdorff target guarantees a Hausdorff function space. It does not say that every non-Hausdorff target produces a non-Hausdorff function space. For example, if the domain is empty, there is exactly one map from it to any target, so its function space is Hausdorff vacuously. The next example shows that without a Hausdorff target, the conclusion can fail.
Worked Example: An Indiscrete Target
Let \(X=\{p\}\) have its unique topology, and let \(Y=\{a,b\}\) have the indiscrete topology \(\{\varnothing,Y\}\). The target is not Hausdorff: the only open neighborhood of either \(a\) or \(b\) is \(Y\), so the distinct points \(a,b\) have no disjoint open neighborhoods.
There are exactly two maps from \(X\) to \(Y\), one sending \(p\) to \(a\), and one sending \(p\) to \(b\). Both are continuous, since the preimage of either open set \(\varnothing\) or \(Y\) is open in \(X\). To calculate the compact-open topology, the compact subsets of \(X\) are \(\varnothing\) and \(\{p\}\), and the open subsets of \(Y\) are \(\varnothing\) and \(Y\). The four resulting subbasic sets are \([\varnothing,\varnothing]=C(X,Y)\), \([\varnothing,Y]=C(X,Y)\), \([\{p\},\varnothing]=\varnothing\), and \([\{p\},Y]=C(X,Y)\). Thus every subbasic set is either empty or the whole function space, and finite intersections and arbitrary unions of these sets give only those same two sets. The compact-open topology is indiscrete on the two maps.
The two distinct maps cannot be separated by disjoint open neighborhoods: their only open neighborhood in this topology is the whole function space. Hence \(C(X,Y)\) is not Hausdorff in this example. This confirms why the Hausdorff condition on \(Y\) is useful for the general theorem.
Check: the computation uses every possible compact \(K\) and every possible open \(U\). In particular, no subbasic set contains just one of the two maps.
A Common Pitfall
When separating \(f\) and \(g\), it is tempting to choose disjoint open neighborhoods of their values and then impose those conditions on the entire image of each function. That approach can fail, and it may not even produce a compact-open subbasic set: the chosen subset of the domain must be compact.
For the maps \(f(t)=t\) and \(g(t)=t+1\) on \(\mathbb{R}\), using \(K=\mathbb{R}\) is invalid because \(\mathbb{R}\) is not compact, as shown above. Moreover, neither map sends all of \(\mathbb{R}\) into the bounded intervals \(U=(-1/4,1/4)\) or \(V=(3/4,5/4)\). The correction is to use the compact singleton \(K=\{0\}\): then the subbasic conditions test only \(f(0)=0\) and \(g(0)=1\), and the two disjoint intervals give the required open neighborhoods. The proof works because it localizes the separation at a point where the functions differ.
Check Your Understanding
Use the compact-open subbasis and the definition of Hausdorffness to answer these questions.
- Why is \([\{x\},U]\) open in \(C(X,Y)\) whenever \(U\) is open in \(Y\)? State the fact about \(\{x\}\) that is needed.
- In the Hausdorffness proof, where is the Hausdorff hypothesis on \(Y\) used?
- Can the theorem’s conclusion fail when \(X\) is noncompact? Explain what the noncompact-domain example shows.
- For the two-point indiscrete target, list all possible compact-open subbasic sets and determine the resulting topology on \(C(X,Y)\).
- Why does the theorem also apply when \(C(X,Y)\) has fewer than two elements?