Why Evaluation Requires Nearby-Point Control
A function space has two kinds of input: a function and a point in its domain. The evaluation map sends them to the value of the function at that point. To prove continuity, it is not enough to control a function at one fixed point: the point in the domain may also move. The compact-open topology provides the needed control on compact sets, and local compactness lets us choose such a set around each point.
Throughout, \(X\) and \(Y\) are topological spaces, and \(C(X,Y)\) denotes the set of continuous maps from \(X\) to \(Y\). A subset \(K\subseteq X\) is compact if every open cover of \(K\) has a finite subcover. We give \(C(X,Y)\) the compact-open topology.
We use the convention that \(X\) is locally compact if every point has a compact neighborhood: for each \(x\in X\), some compact \(C\subseteq X\) contains an open set containing \(x\). The Hausdorff condition means that any two distinct points have disjoint open neighborhoods. The key geometric fact is that, in a locally compact Hausdorff space, a compact set can be chosen around a point while staying inside any prescribed open set.
Shrinking a Neighborhood Inside a Compact Set
Proof. Choose a compact neighborhood \(C\) of \(x\), and an open set \(G\) with \(x\in G\subseteq C\). A compact subset of a Hausdorff space is closed: for any \(z\notin C\), separate \(z\) from each point \(c\in C\) by disjoint open neighborhoods; finitely many of the neighborhoods of points \(c\) cover \(C\), and the intersection of the corresponding finitely many neighborhoods of \(z\) is disjoint from \(C\). Thus \(C\) is closed in \(X\).
We also need a shrinking property inside the compact Hausdorff space \(C\). Let \(G_0=C\cap G\cap O\), a relatively open subset of \(C\) containing \(x\), and put \(F=C\setminus G_0\). The set \(F\) is compact, because it is closed in the compact space \(C\). For each \(y\in F\), the Hausdorff property gives disjoint relatively open sets \(P_y,Q_y\subseteq C\) with \(x\in P_y\) and \(y\in Q_y\). Finitely many sets \(Q_{y_1},\ldots,Q_{y_m}\) cover \(F\). Set \(P=\bigcap_{i=1}^m P_{y_i}\) and \(Q=\bigcup_{i=1}^m Q_{y_i}\). Then \(P\) is relatively open and contains \(x\), while \(Q\) is relatively open and contains \(F\). Since \(P\cap Q=\varnothing\), the relative closure \(\overline{P}^{\,C}\) is disjoint from \(Q\), so \(\overline{P}^{\,C}\subseteq C\setminus Q\subseteq G_0\). If \(F=\varnothing\), take \(P=C\), for which the same containment holds.
By the definition of the subspace topology, write \(P=C\cap H\) for some open \(H\subseteq X\). Define \(U=H\cap G\), which is open in \(X\) and contains \(x\). Since \(G\subseteq C\), we have \(U\subseteq P\). Let \(K=\overline{U}\), the closure in \(X\). Because \(C\) is closed and \(U\subseteq C\), we have \(K\subseteq C\). Also, \(K\subseteq\overline{P}^{\,C}\), since \(P\subseteq C\) and \(C\) is closed. Hence \(K\subseteq G_0\subseteq O\). Finally, \(K\) is closed in the compact set \(C\), so it is compact. Thus \(x\in U\subseteq K\subseteq O\). \(\square\)
The lemma is exactly the bridge between continuity of a single function and an open set in the function space: it supplies a compact \(K\) on which nearby functions can be constrained, while ensuring that nearby domain points lie in \(K\).
The Evaluation Theorem
Proof. We verify continuity at an arbitrary \((f,x)\in C(X,Y)\times X\). Let \(V\subseteq Y\) be open with \(f(x)\in V\). Since \(f\) is continuous, \(O=f^{-1}(V)\) is open in \(X\) and contains \(x\). By the lemma, choose an open \(U\) and compact \(K\) with \(x\in U\subseteq K\subseteq O\). Because \(K\subseteq f^{-1}(V)\), we have \(f\in[K,V]\). Thus \([K,V]\times U\) 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 \(g(K)\subseteq V\). Consequently \(g(y)\in V\). We have found a neighborhood of \((f,x)\) mapped into \(V\), which proves continuity at \((f,x)\). Since the point and the open set were arbitrary, evaluation is continuous. \(\square\)
The proof does not require \(Y\) to be Hausdorff, locally compact, or metrizable. Its only use of the target topology is that \(V\) is open and \(f^{-1}(V)\) is open by continuity.
Using the Neighborhood Construction
Worked Example: A Finite Discrete Domain
Let \(X=\{a,b\}\) have the discrete topology, let \(Y=\mathbb{R}\) have its standard topology, and define \(f(a)=2\), \(f(b)=-1\). At the point \((f,a)\), take \(V=(1,3)\). The value is \(f(a)=2\in(1,3)\). The set \(K=\{a\}\) is compact, and \(U=\{a\}\) is open with \(a\in U\subseteq K\subseteq f^{-1}(V)\), since \(f^{-1}(V)=\{a\}\).
Therefore \([\{a\},(1,3)]\times\{a\}\) is an open neighborhood of \((f,a)\). For any \((g,y)\) in it, \(y=a\) and \(g(a)\in(1,3)\), so \(\operatorname{ev}(g,y)\in(1,3)\). This directly checks the neighborhood construction. In fact, every subset of a finite space is compact, and the discrete topology makes every subset open.
Worked Example: Evaluation on the Real Line
Take \(X=Y=\mathbb{R}\) with their standard topologies, \(f(t)=t\), \(x=2\), and \(V=(1,3)\). Then \(f(x)=2\in V\), and \(f^{-1}(V)=(1,3)\). Choose \(U=(3/2,5/2)\) and \(K=[3/2,5/2]\). The set \(U\) is open, \(K\) is compact, and \(2\in U\subseteq K\subseteq(1,3)\); the last inclusion follows because \(3/2>1\) and \(5/2<3\).
Thus \([K,V]\times U\) is an open neighborhood of \((f,2)\). For any \((g,y)\) in this neighborhood, \(y\in U\subseteq K\) and \(g(K)\subseteq(1,3)\), so \(g(y)\in(1,3)\). The strict endpoints matter: both endpoints of \(K\) lie inside the target interval.
Worked Example: A Compact Interval as Domain
Let \(X=[0,1]\) with its subspace topology from \(\mathbb{R}\), let \(Y=\mathbb{R}\), and set \(f(t)=t^2\). At \(x=1/2\), the value is \(f(1/2)=1/4\). Choose \(V=(0,1)\), \(U=(3/8,5/8)\) as a relatively open subset of \([0,1]\), and \(K=[1/4,3/4]\subseteq[0,1]\). We have \(1/2\in U\subseteq K\). For \(t\in K\), \(1/4\le t\le3/4\), so \(1/16\le t^2\le9/16\). Since \(1/16>0\) and \(9/16<1\), \(f(K)\subseteq(0,1)\).
Hence \(f\in[K,V]\), and \([K,V]\times U\) is a neighborhood of \((f,1/2)\) whose evaluations all lie in \((0,1)\). The calculation checks both inclusions required by the theorem: the moving point lies in the constrained compact set, and the entire image of that set lies in the target open set.
Common Pitfall and the Role of the Hypotheses
Common pitfall. A compact-open condition on the singleton \(\{x\}\) controls \(g(x)\), but it does not control \(g(y)\) when \(y\) moves away from \(x\). For a concrete failure of that attempted argument, take \(X=Y=\mathbb{R}\), \(f(t)=0\), \(x=0\), and \(V=(-1,1)\). The set \([\{0\},V]\times(-1,1)\) is a neighborhood of \((f,0)\). But \(g(t)=4t\) belongs to \([\{0\},V]\), because \(g(0)=0\in(-1,1)\), while at \(y=1/2\) we have \(y\in(-1,1)\) and \(g(y)=2\notin(-1,1)\). The correction is to constrain \(g\) on a compact set \(K\) that contains an open neighborhood \(U\) of \(x\).
The Hausdorff and local compactness assumptions cannot simply be discarded. The following example verifies that evaluation can fail when the domain is not locally compact Hausdorff.
Worked Example: Evaluation Fails on the Rationals
Give \(X=\mathbb{Q}\) the subspace topology from \(\mathbb{R}\), and let \(Y=\mathbb{R}\) have its standard topology. First, \(X\) is not locally compact. If a compact set \(C\subseteq\mathbb{Q}\) were a neighborhood of some \(x\in\mathbb{Q}\), it would contain a relative open neighborhood of \(x\), hence contain \((x-r,x+r)\cap\mathbb{Q}\) for some \(r>0\). The inclusion \(\mathbb{Q}\to\mathbb{R}\) is continuous, so \(C\) is compact as a subset of \(\mathbb{R}\), and therefore closed in \(\mathbb{R}\). Its containing \((x-r,x+r)\cap\mathbb{Q}\) would force it to contain the whole interval \((x-r,x+r)\), since that set of rationals is dense in the interval. This is impossible because \(C\subseteq\mathbb{Q}\) and the interval contains irrationals.
Now let \(f:\mathbb{Q}\to\mathbb{R}\) be the zero function and consider \((f,0)\). If evaluation were continuous there, the inverse image of \(V=(-1,1)\) would contain a product neighborhood \(N\times U\) of \((f,0)\), where \(N\) is open in \(C(\mathbb{Q},\mathbb{R})\) and \(U\) is an open neighborhood of \(0\) in \(\mathbb{Q}\). Since finite intersections of compact-open subbasic sets form a basis, \(N\) contains a neighborhood of \(f\) of the form \(\bigcap_{i=1}^m[K_i,V_i]\), with each \(K_i\subseteq\mathbb{Q}\) compact and each \(V_i\subseteq\mathbb{R}\) open. As \(f\in[K_i,V_i]\), each \(V_i\) contains \(0\).
Let \(K=\bigcup_{i=1}^m K_i\), a compact subset of \(\mathbb{Q}\). Each \(K_i\), and thus \(K\), is compact and closed in \(\mathbb{R}\). It has empty interior in \(\mathbb{Q}\): otherwise it would contain \((q-r,q+r)\cap\mathbb{Q}\) for some rational \(q\) and \(r>0\), and closedness in \(\mathbb{R}\) would force it to contain irrationals in that interval. Since \(U\) is a nonempty relatively open set, choose \(q\in U\setminus K\). If \(K\ne\varnothing\), define \(g(t)=2d(t,K)/(d(t,K)+|t-q|)\) for \(t\in\mathbb{Q}\), where \(d(t,K)=\inf\{|t-s|:s\in K\}\). The distance function is continuous, \(d(q,K)>0\) because \(K\) is closed in \(\mathbb{R}\) and \(q\notin K\), and the denominator is nonzero for every \(t\). Thus \(g\) is continuous, \(g(t)=0\) on \(K\), and \(g(q)=2\). If \(K=\varnothing\), instead take the constant function \(g(t)=2\).
In either case, \(g\in[K_i,V_i]\) for every \(i\), because \(g(K_i)=\{0\}\subseteq V_i\) when \(K_i\ne\varnothing\), and the condition is vacuous when \(K_i=\varnothing\). Hence \(g\in N\), while \(q\in U\) and \(g(q)=2\notin(-1,1)\). This contradicts \(\operatorname{ev}(N\times U)\subseteq V\). Therefore evaluation is not continuous at \((f,0)\). \(\square\)
The proof of continuity on a locally compact Hausdorff domain has a precise division of labor: continuity of \(f\) supplies the open set \(f^{-1}(V)\), and the shrinking lemma supplies \(U\subseteq K\subseteq f^{-1}(V)\). The compact-open neighborhood controls every function on \(K\), while the product neighborhood keeps the evaluation point inside \(K\). Omitting either inclusion breaks the argument.
Check Your Understanding
In each question, distinguish the compact set constraining a function from the open neighborhood in which the evaluation point moves.
- For \(X=\mathbb{R}\), \(f(t)=t\), \(x=0\), and \(V=(-2,2)\), choose \(U\) and compact \(K\) with \(0\in U\subseteq K\subseteq f^{-1}(V)\).
- Explain why \(K\subseteq f^{-1}(V)\) implies \(f\in[K,V]\).
- In the evaluation theorem, identify exactly where local compactness is used and exactly where the Hausdorff condition is used.
- Why does \([\{x\},V]\times U\) generally fail to control evaluation when \(U\) contains points other than \(x\)?
- In the rationals example, why is the function \(g\) constructed to vanish on \(K\) but not at \(q\)?