Why Evaluation Needs a Compact Neighborhood
A function space is a space whose points are themselves functions. The evaluation map takes a function and a point in its domain and returns the value of that function at that point. Even when every function in the space is continuous, it does not follow automatically that evaluation is jointly continuous: the function and its input may vary at the same time. The compact-open topology is designed to control a function on compact subsets of its domain. Local compactness supplies just the compact set needed to control nearby inputs as well as nearby functions.
Let \(X\) and \(Y\) be topological spaces, and let \(C(X,Y)\) denote the set of continuous maps from \(X\) to \(Y\). A subset \(K\subseteq X\) is compact if every cover of \(K\) by sets open in \(X\) has a finite subcover. For compact \(K\subseteq X\) and open \(V\subseteq Y\), write \([K,V]=\{g\in C(X,Y):g(K)\subseteq V\}\).
A subbasis generates a topology by taking arbitrary unions of finite intersections of its members. Thus a basic open neighborhood in \(C(X,Y)\) is a finite intersection of sets \([K_i,V_i]\). A basic open neighborhood in \(C(X,Y)\times X\) has the form \(N\times U\), where \(N\) is open in the compact-open topology and \(U\) is open in \(X\).
We use the following form of local compactness: \(X\) is locally compact if every \(x\in X\) has a compact neighborhood, meaning a compact set \(K\subseteq X\) for which some open set \(U\) satisfies \(x\in U\subseteq K\). A space is Hausdorff if distinct points have disjoint open neighborhoods. The theorem below assumes \(X\) is locally compact Hausdorff. Its proof uses the compact-neighborhood property in the stated form; the Hausdorff hypothesis is part of the usual setting for this result.
The Fixed-Point Evaluation Map
Proof. Let \(V\subseteq Y\) be open. The preimage under \(\operatorname{ev}_x\) is \(\operatorname{ev}_x^{-1}(V)=\{g\in C(X,Y):g(x)\in V\}=[\{x\},V]\). The singleton \(\{x\}\) is compact: any open cover of it has a member containing \(x\), and that one member is a finite subcover. Therefore \([\{x\},V]\) is a subbasic open set in the compact-open topology. The preimage of every open set is open, so \(\operatorname{ev}_x\) is continuous. \(\square\)
This proposition controls variation of the function while the input remains fixed. Joint continuity must also control what happens when the input moves to a nearby point. For that, we need an open neighborhood of \(x\) contained in a compact set on which the function is controlled.
Joint Continuity of Evaluation
Proof. We prove continuity at an arbitrary \((f,x)\in C(X,Y)\times X\). Let \(V\) be any open neighborhood of \(f(x)\) in \(Y\). Since \(f\) is continuous, \(O=f^{-1}(V)\) is open in \(X\), and \(x\in O\). By local compactness, choose a compact set \(K\subseteq X\) and an open set \(U\subseteq X\) such that \(x\in U\subseteq K\). We may ensure \(K\subseteq O\): apply local compactness within the open neighborhood \(O\), equivalently choose a compact neighborhood contained in \(O\). Hence \(K\subseteq O=f^{-1}(V)\). It follows that \(f(K)\subseteq V\), so \(f\in[K,V]\). The set \([K,V]\times U\) is an open neighborhood of \((f,x)\) in the product topology. If \((g,y)\in[K,V]\times U\), then \(y\in U\subseteq K\), and \(g\in[K,V]\) means \(g(K)\subseteq V\). Thus \(g(y)\in V\), so \(\operatorname{ev}([K,V]\times U)\subseteq V\). We have found a product neighborhood of \((f,x)\) mapped into each prescribed open neighborhood \(V\) of \(f(x)\). This proves continuity at \((f,x)\); since \((f,x)\) was arbitrary, evaluation is continuous. \(\square\)
The compact containment used here is often expressed as follows: in a locally compact Hausdorff space, if \(x\in O\) and \(O\) is open, there is a compact neighborhood \(K\) of \(x\) with \(K\subseteq O\). This is the precise local property the proof needs. The key inclusion is \(K\subseteq f^{-1}(V)\); applying \(f\) gives \(f(K)\subseteq V\), exactly the condition that puts \(f\) in the compact-open neighborhood \([K,V]\).
Worked Example: A Moving Input on the Real Line
Let \(X=Y=\mathbb{R}\), both with their standard topologies, and let \(f(t)=t^2\). We exhibit a neighborhood of \((f,1)\) that evaluation maps into \(V=(0,3)\).
Take \(K=[1/2,3/2]\). It is compact by the Heine–Borel theorem, and its interior \(U=(1/2,3/2)\) is an open neighborhood of \(1\) contained in \(K\). On \(K\), the function \(t\mapsto t^2\) is increasing because \(K\) consists of positive numbers. Therefore \(f(K)=[(1/2)^2,(3/2)^2]=[1/4,9/4]\subseteq(0,3)\). Thus \(f\in[K,(0,3)]\).
For every \(g\in[K,(0,3)]\) and \(y\in U\), we have \(y\in K\) and \(g(K)\subseteq(0,3)\), so \(g(y)\in(0,3)\). Consequently \([K,(0,3)]\times U\) is a neighborhood of \((f,1)\) mapped into the requested target neighborhood.
Check: at the original pair, \(\operatorname{ev}(f,1)=f(1)=1^2=1\in(0,3)\). At the endpoints of \(K\), \(f(1/2)=1/4\) and \(f(3/2)=9/4\), confirming both bounds used to contain \(f(K)\) in \((0,3)\).
Worked Example: A Finite Discrete Domain
Let \(X=\{a,b\}\) with the discrete topology, so every subset is open. This space is compact: an open cover has at most two points to cover, and choosing one cover member for each point gives a finite subcover. Let \(Y=\mathbb{R}\) with its standard topology, and choose \(f\in C(X,Y)\) with \(f(a)=2\) and \(f(b)=-1\). We verify the neighborhood condition at \((f,a)\) for \(V=(1,3)\).
The singleton \(K=\{a\}\) is compact, and \(U=\{a\}\) is an open neighborhood of \(a\) contained in \(K\). Since \(f(K)=\{2\}\subseteq(1,3)\), we have \(f\in[K,(1,3)]\). For every \(g\in[K,(1,3)]\) and \(y\in U\), the only possibility is \(y=a\), and the definition of \([K,(1,3)]\) gives \(g(a)\in(1,3)\). Hence \([K,(1,3)]\times\{a\}\) is an open neighborhood mapped into \(V\).
Check: the value at the chosen pair is \(f(a)=2\), and \(1<2<3\). The open input neighborhood is contained in the compact set, and the function value on that compact set lies in the specified target open set.
Why Local Compactness Cannot Simply Be Dropped
Here is an explicit failure when the domain is not locally compact. Let \(X=\mathbb{Q}\) have the subspace topology inherited from the standard topology on \(\mathbb{R}\). This space is Hausdorff but is not locally compact. Let \(Y=S=\{0,1\}\) have topology \(\{\varnothing,\{1\},S\}\). For each open set \(A\subseteq\mathbb{Q}\), define \(f_A:\mathbb{Q}\to S\) by \(f_A(q)=1\) when \(q\in A\), and \(f_A(q)=0\) otherwise. This map is continuous: the preimage of \(\{1\}\) is \(A\), while the preimages of \(\varnothing\) and \(S\) are \(\varnothing\) and \(\mathbb{Q}\). In particular the constant-one map \(c=f_{\mathbb{Q}}\) is continuous.
Worked Example: Evaluation Fails on the Rational Line
We show that evaluation is not continuous at \((c,0)\). Its value there is \(c(0)=1\), and \(\{1\}\) is open in \(S\). Suppose \(N\times U\) is any basic product neighborhood of \((c,0)\), where \(N\) is open in \(C(\mathbb{Q},S)\) and \(U\) is open in \(\mathbb{Q}\) with \(0\in U\). By the definition of the compact-open topology, \(N\) contains a finite intersection of subbasic neighborhoods of \(c\).
A subbasic set \([K,V]\) containing \(c\) has \(V\) open in \(S\) and \(c(K)\subseteq V\). Since \(c\) is constantly \(1\), the only possibilities for \(V\) are \(\{1\}\) and \(S\). A constraint with \(V=S\) imposes no condition. Each remaining constraint has the form \([K_i,\{1\}]\), with \(K_i\) compact in \(\mathbb{Q}\). Let \(K\) be the union of the finitely many \(K_i\), taking \(K=\varnothing\) if there are no such constraints. A finite union of compact sets is compact: restrict any open cover to each compact set, take a finite subcover for each, and combine those finitely many finite subcovers.
Every compact subset \(K\) of \(\mathbb{Q}\) has empty interior in \(\mathbb{Q}\). To see this, the inclusion \(\mathbb{Q}\to\mathbb{R}\) is continuous, so \(K\), viewed as a subset of \(\mathbb{R}\), is compact. It is closed in \(\mathbb{R}\): for a point \(r\notin K\), separate \(r\) from each \(k\in K\) by disjoint open intervals; finitely many of the intervals around points \(k\) cover \(K\), and the intersection of the corresponding finitely many intervals around \(r\) is a neighborhood disjoint from \(K\). If \(K\) had nonempty interior in \(\mathbb{Q}\), it would contain \(\mathbb{Q}\cap(a,b)\) for some real \(a<b\). The rationals are dense in \(\mathbb{R}\), so closedness would force \(K\) to contain \([a,b]\), including irrational points, a contradiction. Thus \(K\) has empty interior.
Since \(U\) is a nonempty open set and \(K\) has empty interior, choose \(x\in U\setminus K\). The singleton \(\{x\}\) is closed in \(\mathbb{Q}\), so \(A=\mathbb{Q}\setminus\{x\}\) is open. Hence \(f_A\in C(\mathbb{Q},S)\). For each constraint \([K_i,\{1\}]\), we have \(K_i\subseteq K\subseteq A\), and therefore \(f_A(K_i)=\{1\}\). The constraints with target \(S\) also hold automatically. Thus \(f_A\in N\), but \(x\in U\) and \(f_A(x)=0\). Therefore \((f_A,x)\in N\times U\) while \(\operatorname{ev}(f_A,x)\notin\{1\}\).
Every basic product neighborhood of \((c,0)\) contains a point whose evaluation is not in \(\{1\}\). Thus the preimage of the open set \(\{1\}\) is not a neighborhood of \((c,0)\), and evaluation is not continuous there. This also verifies that \(\mathbb{Q}\) is not locally compact: a compact neighborhood would contain a nonempty open subset of \(\mathbb{Q}\), contradicting the empty-interior property just proved.
A Common Pitfall: Keeping the Inclusions in the Right Spaces
In the theorem’s proof, the relevant sets live in different spaces: \(O=f^{-1}(V)\) is a subset of \(X\), while \(V\) is a subset of \(Y\). The needed statement is \(K\subseteq O=f^{-1}(V)\), which implies \(f(K)\subseteq V\). One must not replace that reasoning with a claim that \(f(K)\subseteq O\); that compares an image in \(Y\) with a set in \(X\), and it is not the condition defining \([K,V]\).
For a concrete check, take \(X=Y=\mathbb{R}\), \(f(t)=t+10\), \(V=(10,11)\), and \(K=[1/4,3/4]\). Then \(O=f^{-1}(V)=(0,1)\), so \(K\subseteq O\). Direct calculation gives \(f(K)=[10+1/4,10+3/4]=[41/4,43/4]\), which is contained in \(V=(10,11)\), as required. But \(f(K)\) is not contained in \(O=(0,1)\). The correct inclusion follows by applying \(f\) to \(K\subseteq f^{-1}(V)\).
The theorem matters whenever a space of continuous maps is used as a domain for another construction: it guarantees that a continuously varying function evaluated at a continuously varying point still varies continuously. The proof identifies the precise mechanism. A compact-open neighborhood controls the function on \(K\), and an open neighborhood of the input lies inside \(K\). Without a compact neighborhood, as on \(\mathbb{Q}\), those two controls need not fit together.
Check Your Understanding
For each question, identify which compact set and which open input neighborhood are being used, or explain why they cannot be chosen.
- For the standard topology on \(\mathbb{R}\), let \(f(t)=t^2\) and consider \((f,2)\). Given \(V=(3,6)\), find a compact neighborhood \(K\) of \(2\) with \(f(K)\subseteq V\).
- Why is \(\operatorname{ev}_x\) continuous for each fixed \(x\), even when \(X\) is not locally compact?
- In the rational-line example, why does \(f_A\) remain in every subbasic constraint \([K_i,\{1\}]\) after choosing \(x\notin\bigcup_i K_i\)?
- Which inclusion ensures \(f\in[K,V]\) in the joint-continuity proof: \(K\subseteq f^{-1}(V)\) or \(f(K)\subseteq K\)? Explain the spaces involved.
- For a compact neighborhood \(K\) of \(x\), why must the input neighborhood be chosen inside \(K\), rather than merely to intersect \(K\)?