Two Topologies on a Space of Maps
There are different ways to describe how maps vary. The pointwise topology controls function values at finitely many specified points. The compact-open topology controls the image of a compact set all at once. The first control is always available in the second topology, because a single point is compact. If the domain is discrete, compact sets are finite, so these two kinds of control generate the same topology.
Let \(X\) and \(Y\) be topological spaces, and write \(C(X,Y)\) for the set of continuous maps from \(X\) to \(Y\). The product topology on \(Y^X=\prod_{x\in X}Y\) has basic open sets that restrict the coordinates at finitely many points. Give \(C(X,Y)\) the subspace topology inherited from this product; this is the pointwise topology. Thus a typical basic neighborhood of \(f\in C(X,Y)\) specifies open sets around \(f(x)\) for finitely many \(x\in X\).
Compactness here means that every open cover has a finite subcover. The notation \([K,U]\) therefore describes a simultaneous condition on every point of \(K\), not a bound on how far apart the values \(f(x)\) are. We compare the topologies by examining their subbasic and basic neighborhoods.
Pointwise Open Sets Are Compact-Open
Proof. A subbasic open set in the pointwise topology restricts a single coordinate \(x\in X\) to an open set \(U\subseteq Y\); it is \(\{f\in C(X,Y):f(x)\in U\}\). The singleton \(\{x\}\) is compact: from any open cover of it, one member covering \(x\) is already a finite subcover. Also, \(f(x)\in U\) is equivalent to \(f(\{x\})\subseteq U\). Thus this pointwise subbasic set is exactly \([\{x\},U]\), which is compact-open subbasic. Since every pointwise open set is a union of finite intersections of pointwise subbasic sets, it is compact-open. If \(X\) is empty, there are no coordinate restrictions and the pointwise topology is the indiscrete topology on \(C(X,Y)\); the compact-open topology contains it as well. \(\square\)
The direction of containment matters: a compact-open condition can constrain many function values at once, whereas a pointwise basic condition constrains only finitely many. The theorem says every finite-coordinate condition is compact-open; it does not say that every compact-open condition can be reduced to finitely many coordinates.
Worked Example: A Neighborhood on a Finite Discrete Domain
Let \(X=\{a,b,c\}\) have the discrete topology and let \(Y=\mathbb{R}\) with its standard topology. Consider the continuous map \(f\) given by \(f(a)=1\), \(f(b)=0\), and \(f(c)=3\). Take \(K=\{a,c\}\), \(U=(0,4)\), \(L=\{b\}\), and \(V=(-1,1)\). Both \(K\) and \(L\) are compact because they are finite. The compact-open neighborhood \([K,U]\cap[L,V]\) consists exactly of maps \(g\) satisfying \(g(a),g(c)\in(0,4)\) and \(g(b)\in(-1,1)\).
Written as pointwise conditions, this is \([a,U]\cap[c,U]\cap[b,V]\), where \([a,U]\) here means \(\{g:g(a)\in U\}\). Substitution checks that \(f\) belongs: \(1\in(0,4)\), \(3\in(0,4)\), and \(0\in(-1,1)\). Thus, in this example, the two compact-open restrictions combine into three finite pointwise restrictions.
Check: the coordinate conditions recover exactly the original two image conditions: \(g(K)\subseteq U\) means both \(g(a)\in U\) and \(g(c)\in U\), and \(g(L)\subseteq V\) means \(g(b)\in V\).
Why the Topologies Agree on Discrete Domains
Proof. First we identify the compact subsets of a discrete space. Every finite subset is compact: given an open cover, choose one cover member for each of its finitely many points. Conversely, if \(K\subseteq X\) is infinite, the family of open singletons \(\{\{x\}:x\in K\}\) covers \(K\), and no finite subfamily covers \(K\). Hence a subset of a discrete space is compact exactly when it is finite, including the empty set.
Now let \([K,U]\) be compact-open subbasic. The result just proved says that \(K\) is finite. If \(K\ne\varnothing\), then \([K,U]=\bigcap_{x\in K}\{f\in C(X,Y):f(x)\in U\}\), a finite intersection of pointwise subbasic open sets. If \(K=\varnothing\), then \([K,U]=C(X,Y)\), which is open in every topology. Thus each compact-open subbasic set is pointwise open, so the compact-open topology is contained in the pointwise topology. The preceding theorem gives the reverse containment, and therefore the topologies are equal. Finally, every map from a discrete \(X\) to \(Y\) is continuous: the preimage of any open subset of \(Y\) is a subset of \(X\), hence open. This observation is consistent with the argument and does not require restricting the available maps further. \(\square\)
Worked Example: A Compact-Open Condition on the Natural Numbers
Give \(X=\mathbb{N}=\{1,2,\dots\}\) the discrete topology, and give \(Y=\{0,1\}\) the discrete topology. Define \(f:\mathbb{N}\to Y\) by \(f(2)=f(5)=0\) and \(f(n)=1\) for all other \(n\). Take \(K=\{2,5\}\) and \(U=\{0\}\). The compact-open neighborhood \([K,U]\) consists of exactly those maps \(g\) for which \(g(2)=0\) and \(g(5)=0\).
Since \(K\) is finite, the same neighborhood is the pointwise open set \(\{g:g(2)\in\{0\}\}\cap\{g:g(5)\in\{0\}\}\). The given \(f\) belongs because \(f(2)=0\) and \(f(5)=0\). In contrast, the infinite set \(\mathbb{N}\) is not compact in this topology: the open cover by singletons \(\{\{n\}:n\in\mathbb{N}\}\) has no finite subcover.
Check: the equality of the two displayed neighborhoods imposes precisely two conditions, not a condition at every natural number. For example, \(g(2)=g(5)=0\) and \(g(7)=1\) still satisfies both conditions.
Containment Can Be Strict
The discrete-domain theorem does not say that the topologies must agree for every nondiscrete domain. Here is a concrete case where a single compact-open neighborhood cannot be a pointwise neighborhood. We use \(C([0,1],\mathbb{R})\), with the standard topologies on both spaces. The Heine–Borel theorem gives compactness of \([0,1]\).
Worked Example: A Compact-Open Neighborhood Not Pointwise Open
Let \(0\) denote the constant-zero function and consider \([[0,1],(-1,1)]\). This is compact-open subbasic, and it contains \(0\), since \(0([0,1])=\{0\}\subseteq(-1,1)\). We show it is not pointwise open.
Take any pointwise basic neighborhood of \(0\). It restricts values at some finite set \(F\subseteq[0,1]\): for each \(p\in F\), it requires \(g(p)\in V_p\), where \(V_p\subseteq\mathbb{R}\) is open and contains \(0\). Choose \(x_0\in(0,1)\setminus F\). This is possible because \((0,1)\) is infinite and \(F\) is finite. Let \(d\) be the minimum of \(x_0\), \(1-x_0\), and the numbers \(|x_0-p|\) for \(p\in F\); if \(F\) is empty, take \(d=\min(x_0,1-x_0)\). Every number in this finite minimum is positive, so \(d>0\). Set \(r=d/2\) and define \(g(x)=2\max(0,1-|x-x_0|/r)\) for \(x\in[0,1]\).
The absolute-value function, scalar operations, and the maximum of two continuous real-valued functions are continuous, so \(g\) is continuous. At every \(p\in F\), \(|p-x_0|\ge d>r\), giving \(g(p)=0\in V_p\). Thus \(g\) lies in the chosen pointwise basic neighborhood. But \(g(x_0)=2\), and \(2\notin(-1,1)\), so \(g\notin[[0,1],(-1,1)]\). Every pointwise basic neighborhood of \(0\) therefore contains a map outside the compact-open neighborhood. No pointwise open neighborhood of \(0\) is contained in it, so it is not pointwise open.
Check: the function used in the argument has value \(0\) at every constrained point and value \(2\) at \(x_0\). Those values verify, respectively, membership in the arbitrary pointwise neighborhood and failure of membership in \([[0,1],(-1,1)]\).
A Common Mistake and the Practical Distinction
Common pitfall. A tempting but incorrect argument is: “A compact-open condition restricts a whole set \(K\), so it cannot be pointwise open.” For \(K=\{x\}\), however, the condition \(f(K)\subseteq U\) is just \(f(x)\in U\). The singleton is compact, so this condition is pointwise subbasic and compact-open subbasic at the same time. The corrected comparison is that the compact-open topology always contains the pointwise topology, and the discrete-domain theorem identifies a setting where the reverse containment holds as well.
The difference is useful when interpreting neighborhoods. Pointwise control asks for values to fall in specified open sets at finitely many chosen inputs. Compact-open control asks that every value on a specified compact subset fall in one chosen open set. On a discrete domain the compact subsets are finite, so this second demand reduces to finitely many pointwise demands. On a nondiscrete domain, compact sets can contain infinitely many points, and the example above shows that finitely many pointwise tests need not control all of them.
Do not strengthen the conclusion to “the topologies are equal only when \(X\) is discrete.” The proof establishes equality for discrete domains, while the strict example establishes that equality can fail for a particular nondiscrete domain and target. It does not classify all cases of equality.
Check Your Understanding
For each question, distinguish a finite-coordinate condition from a condition on an entire compact set.
- For \(x\in X\) and open \(U\subseteq Y\), express the pointwise condition \(f(x)\in U\) using compact-open notation and explain why it is open.
- Prove directly from the open-cover definition that every finite subset of a topological space is compact.
- In an infinite discrete space \(X\), why does the singleton cover show that \(X\) is not compact?
- For a discrete domain \(X\), rewrite \([K,U]\) as a pointwise open set when \(K\) is finite. What is \([\varnothing,U]\)?
- In the nondiscrete example, why must the finite set \(F\) of constrained points leave a point \(x_0\) in \((0,1)\setminus F\)?