From Compact Pieces to Sets of Functions
A topology on a function space can be specified by asking what a collection of functions must do on selected parts of its domain. The compact-open topology uses compact subsets of the domain and open subsets of the codomain: a basic requirement is that every point of a chosen compact set be sent into a chosen open set. The requirement is a single condition on the image of the whole compact set, not merely a condition at one point.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) 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 subsets of \(X\) has a finite subcover. Compactness here is relative to the open sets of \(X\); in particular, the empty set is compact.
A subbasis is a collection of sets used to generate a topology. Its members need not themselves be a basis: finite intersections of subbasic sets form a basis, and arbitrary unions of those finite intersections form the topology. The finite intersections record simultaneous requirements on possibly different compact subsets and open subsets.
What the Subbasis Generates
Proof. Let \(\mathcal{B}\) be the collection of all finite intersections of members of \(\mathcal{S}\), including \(Z\) as the intersection of no sets. Let \(\mathcal{T}_*\) consist of all unions of members of \(\mathcal{B}\). The empty union shows \(\varnothing\in\mathcal{T}_*\), and \(Z\in\mathcal{T}_*\) because \(Z\in\mathcal{B}\). An arbitrary union of members of \(\mathcal{T}_*\) is still a union of members of \(\mathcal{B}\), so it belongs to \(\mathcal{T}_*\). For two such unions, their intersection is the union of the pairwise intersections of the sets being unioned. The intersection of two finite intersections of members of \(\mathcal{S}\) is again a finite intersection of members of \(\mathcal{S}\), so this union also belongs to \(\mathcal{T}_*\). Thus \(\mathcal{T}_*\) is a topology. It contains every member of \(\mathcal{S}\), since each is a finite intersection with one member. Conversely, any topology containing \(\mathcal{S}\) contains its finite intersections and their arbitrary unions. Therefore \(\mathcal{T}_*\) is the smallest topology containing \(\mathcal{S}\), which proves the claim. \(\square\)
Applied to \(C(X,Y)\), this says that the compact-open topology is made from arbitrary unions of sets of the form \([K_1,U_1]\cap\cdots\cap[K_n,U_n]\), where each \(K_i\) is compact in \(X\) and each \(U_i\) is open in \(Y\). The empty-set conventions cause no difficulty: \([\varnothing,U]=C(X,Y)\), and \([K,\varnothing]=\varnothing\) when \(K\ne\varnothing\). These follow directly from the definition of image containment.
Two useful identities also follow directly from the definition. For compact \(K\) and open \(U,V\subseteq Y\), \([K,U]\cap[K,V]=[K,U\cap V]\). For compact \(K,L\subseteq X\) and open \(U\subseteq Y\), \([K\cup L,U]=[K,U]\cap[L,U]\). The second identity applies because a finite union of compact sets is compact: given an open cover of \(K\cup L\), take a finite subcover of \(K\) and a finite subcover of \(L\), then combine them. In both identities, a map belongs to the left-hand set exactly when it satisfies the requirements on the right-hand side.
Worked Calculations
Worked Example: A Two-Point Domain
Give \(X=\{a,b\}\) and \(Y=\{0,1\}\) their discrete topologies, in which every subset is open. Compute \([\{a\},\{1\}]\) and \([X,\{1\}]\).
Every map \(X\to Y\) is continuous: the preimage of any open subset of \(Y\) is a subset of \(X\), hence is open. There are four maps, which we denote by their ordered values \((f(a),f(b))\): \((0,0),(0,1),(1,0),(1,1)\). Every subset of \(X\) is compact, since it is finite.
For \([\{a\},\{1\}]\), the condition is \(f(\{a\})\subseteq\{1\}\), equivalently \(f(a)=1\). The second coordinate may be either \(0\) or \(1\). Hence \([\{a\},\{1\}]=\{(1,0),(1,1)\}\). For \([X,\{1\}]\), both \(f(a)\) and \(f(b)\) must equal \(1\), so \([X,\{1\}]=\{(1,1)\}\).
Check: The map \((1,0)\) sends \(a\) into \(\{1\}\), so it belongs to the first set, but it sends \(b\) to \(0\), so it does not belong to the second. The map \((1,1)\) satisfies both requirements. The listed sets therefore agree with the defining image-containment conditions.
In fact, this compact-open topology on the four-element function set is discrete. For any map \((i,j)\), the intersection \([\{a\},\{i\}]\cap[\{b\},\{j\}]\) contains exactly that map. Each such singleton is open in the generated topology, and unions of singletons give every subset of the function set.
Worked Example: A Finite Constraint in a Discrete Domain
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) have the discrete topology, and let \(Y=\{0,1\}\) also have the discrete topology. Compute \([\{2,5\},\{1\}]\).
Every map \(f:X\to Y\) is continuous because every subset of \(X\) is open, so every preimage of an open set is open. The set \(K=\{2,5\}\) is compact: from any open cover, choose one covering set for \(2\) and one covering set for \(5\). These at most two sets cover \(K\).
By definition, \(f\in[\{2,5\},\{1\}]\) exactly when \(f(\{2,5\})\subseteq\{1\}\). Since \(f(\{2,5\})=\{f(2),f(5)\}\), this is equivalent to the two equations \(f(2)=1\) and \(f(5)=1\). Thus the subbasic set consists of all functions \(\mathbb{N}\to\{0,1\}\) having value \(1\) at both \(2\) and \(5\), with no restrictions at the other positive integers.
Check: Define \(f(n)=1\) for every \(n\in\mathbb{N}\). Then \(f(2)=f(5)=1\), so \(f\) belongs to the set. Define \(g(2)=1\), \(g(5)=0\), and \(g(n)=0\) for all other \(n\). Then \(g(5)=0\notin\{1\}\), so \(g\) does not belong. These tests verify that both required points, and only those points, determine membership.
For contrast, no infinite subset \(A\subseteq X\) is compact. The collection of all singleton sets \(\{\{a\}:a\in A\}\) is an open cover of \(A\), and no finite subcollection covers an infinite \(A\). Thus compact subsets of this discrete \(X\) are exactly its finite subsets: finite subsets are compact by the finite-choice argument above, and infinite subsets fail the open-cover definition.
Worked Example: A Compact Interval in the Domain
Let \(X=I=[0,1]\) with the standard topology inherited from \(\mathbb{R}\), and let \(Y=\mathbb{R}\) with its standard topology. Take \(K=[0,1/2]\) and \(U=(1/4,1)\). The set \(K\) is compact by the Heine–Borel theorem, which states that closed bounded subsets of \(\mathbb{R}\) are compact. Describe \([K,U]\) and test three continuous functions.
In this case, \([K,U]=\{f\in C(I,\mathbb{R}):f(t)\in(1/4,1)\text{ for every }t\in[0,1/2]\}\). For \(f(t)=t/2+1/2\), the values on \(K\) run from \(f(0)=1/2\) to \(f(1/2)=3/4\). Since this function is increasing, \(f(K)=[1/2,3/4]\), which is contained in \((1/4,1)\). Therefore \(f\in[K,U]\).
For \(g(t)=t\), the image of \(K\) is \([0,1/2]\); it contains \(0\), which is not in \(U\). Thus \(g\notin[K,U]\). For the constant function \(h(t)=1/4\), the image is \(\{1/4\}\). Since \(1/4\) is excluded from the open interval \(U=(1/4,1)\), we have \(h\notin[K,U]\).
Check: For \(f\), every value on \(K\) lies between \(1/2\) and \(3/4\), and \(1/4<1/2\le f(t)\le3/4<1\). For \(g\), \(g(0)=0\notin U\). For \(h\), \(h(0)=1/4\notin U\). Each membership conclusion follows by substituting values into the original condition \(f(K)\subseteq U\).
A Common Pitfall: Replacing an Open Set by Its Closure
The condition \(f(K)\subseteq U\) is strict containment in the specified open set \(U\). It does not mean that \(f(K)\) may also contain boundary points of \(U\). For example, keep \(I\) with its standard topology and consider \([I,(0,1)]\) in \(C(I,\mathbb{R})\). A mistaken calculation might replace \((0,1)\) by its closure \([0,1]\) and conclude that \(f(t)=t/2\) belongs, because its range is \([0,1/2]\subseteq[0,1]\).
The correction is to test the actual open set: \(f(0)=0\), and \(0\notin(0,1)\). Therefore \(f(I)=[0,1/2]\) is not a subset of \((0,1)\), so \(f\notin[I,(0,1)]\). The endpoint matters even though \(f(t)\in(0,1)\) for every \(t\) strictly between \(0\) and \(1\).
The compact-open subbasis gives a direct way to specify conditions on a whole compact piece of the domain. To form the topology, combine those conditions by finite intersections and arbitrary unions. When calculating a particular subbasic set, the decisive test remains exactly the same: the image of every point of \(K\) must lie in \(U\).
Check Your Understanding
For each question, begin with the definition of \([K,U]\) and check any claimed membership by testing the image of \(K\).
- In the two-point example, compute \([\{b\},\{0\}]\) using the ordered-value notation.
- For a discrete domain \(X=\mathbb{N}\), explain why \([K,U]=C(X,Y)\) when \(K=\varnothing\), for any open \(U\subseteq Y\).
- In \(C([0,1],\mathbb{R})\), determine whether \(f(t)=1+t\) belongs to \([[0,1/2],(1,2)]\). Pay particular attention to the endpoint \(t=1/2\).
- Explain why the intersection \([K,U]\cap[K,V]\) equals \([K,U\cap V]\) when \(U,V\) are open in the same codomain.
- Why is an infinite subset of a discrete space not compact? Give an open cover that has no finite subcover.