What a Compact-Open Constraint Says
A function-space topology can be described by specifying what it means for a map to satisfy a local restriction. In the compact-open topology, the restriction asks that the image of an entire compact subset of the domain lie inside one open subset of the codomain. The domain matters: compact subsets of the real line can be infinite, while compact subsets of a discrete space must be finite.
Let \(X\) and \(Y\) be topological spaces, and let \(C(X,Y)\) denote the set of continuous maps from \(X\) to \(Y\). When \(K\subseteq X\) is compact and \(U\subseteq Y\) is open, write \([K,U]\) for the following subset of \(C(X,Y)\).
The containment \(f(K)\subseteq U\) means that every point of \(K\) is sent into \(U\). It does not require \(f(K)=U\), and it is not a condition on points outside \(K\). If \(K=\varnothing\), then \(f(K)=\varnothing\subseteq U\) for every \(f\) and every \(U\), so \([\varnothing,U]=C(X,Y)\), including when \(U=\varnothing\). If \(K\ne\varnothing\) and \(U=\varnothing\), then \([K,\varnothing]=\varnothing\).
By the definition of a topology generated by a subbasis, finite intersections of compact-open subbasic sets form a basis. Thus a typical basic set has the form \([K_1,U_1]\cap\cdots\cap[K_m,U_m]\), where every \(K_i\) is compact in \(X\) and every \(U_i\) is open in \(Y\). A map \(f\) belongs to this intersection exactly when \(f(K_i)\subseteq U_i\) for every index \(i\). This describes basic restrictions; an arbitrary open set is a union of such basic sets.
Proof. Membership in an intersection means membership in every set being intersected. By the definition of \([K_i,U_i]\), membership in its \(i\)-th set is equivalent to \(f(K_i)\subseteq U_i\). Applying this condition for each \(i\) proves the first assertion. Applying the same membership test to \(f_0\) proves the second. Finite intersections of subbasic sets form a basis for the generated topology by the definition of a generated topology, so these are exactly its basic sets. \(\square\)
Compact Subsets of the Real Line
For \(\mathbb{R}\) we use the standard topology. The Heine–Borel theorem says that a subset of \(\mathbb{R}\) is compact if and only if it is closed and bounded. Consequently, a compact set used in a compact-open constraint on \(C(\mathbb{R},Y)\) may be an interval, a finite set, or an infinite closed bounded set. The condition is imposed on every point of that set, including its endpoints.
Worked Example: A Constraint on an Interval in the Real Line
Take \(X=Y=\mathbb{R}\) with their standard topologies, \(K=[-1,1]\), and \(U=(0,2)\). The interval \(K\) is closed and bounded, hence compact by Heine–Borel; \(U\) is open. The subbasic set is \([K,U]=\{f\in C(\mathbb{R},\mathbb{R}): f(x)\in(0,2)\text{ for every }x\in[-1,1]\}\).
For \(f(x)=1+x/2\), the endpoint values are \(f(-1)=1-1/2=1/2\) and \(f(1)=1+1/2=3/2\). Since \(f\) is increasing, its values on \([-1,1]\) lie between \(1/2\) and \(3/2\), and \([1/2,3/2]\subset(0,2)\). Thus \(f\in[K,U]\).
For \(g(x)=1+x\), the endpoint values are \(g(-1)=0\) and \(g(1)=2\). Both endpoints are excluded from \(U=(0,2)\), so \(g\notin[K,U]\). In particular, replacing the strict containment \(g(K)\subseteq U\) by a condition that allows the endpoints of \(U\) would give the wrong answer.
Check: \(f(K)=[1/2,3/2]\subset(0,2)\), whereas \(g(K)=[0,2]\) is not a subset of \((0,2)\). These image calculations confirm the two membership decisions.
A compact set in \(\mathbb{R}\) need not be an interval. For example, \(\{0\}\cup[1/2,1]\) is closed and bounded, so it is compact. A constraint on this set controls values at \(0\) and throughout \([1/2,1]\), but places no direct requirement on values in the gap \((0,1/2)\). Continuity can relate values at nearby points, but the definition of the subbasic set itself asks only for containment on \(K\).
Worked Example: A Compact Set with a Gap
Again take \(X=Y=\mathbb{R}\) with the standard topologies. Let \(K=\{0\}\cup[1/2,1]\), which is closed and bounded and therefore compact, and let \(U=(-1,1)\). Then \([K,U]=\{f\in C(\mathbb{R},\mathbb{R}): f(0)\in(-1,1)\text{ and }f(x)\in(-1,1)\text{ for all }x\in[1/2,1]\}\).
For \(f(x)=x/2\), the value at the isolated listed point \(0\) is \(f(0)=0\). On \([1/2,1]\), the function is increasing, with endpoint values \(f(1/2)=(1/2)/2=1/4\) and \(f(1)=1/2\). Hence \(f(K)=\{0\}\cup[1/4,1/2]\), which is contained in \((-1,1)\), and \(f\in[K,U]\).
For \(g(x)=x\), we have \(g(0)=0\), and \(g([1/2,1])=[1/2,1]\). Since \(1\notin(-1,1)\), \(g(K)\) is not contained in \(U\), so \(g\notin[K,U]\). The failure occurs at \(x=1\in K\); the fact that \(g(x)\in U\) for points of the gap does not repair it.
Check: \(f(K)=\{0\}\cup[1/4,1/2]\subset(-1,1)\), while \(g(1)=1\notin(-1,1)\). These verify the conclusions directly from the defining containment.
Compact Subsets of a Compact Interval
Let \(I=[a,b]\) carry the subspace topology inherited from \(\mathbb{R}\), where \(a\le b\). The interval is compact by Heine–Borel. A subset \(K\subseteq I\) is compact precisely when it is closed in \(I\), that is, closed for the subspace topology. This includes \(K=\varnothing\) and, when \(a=b\), both possible subsets of the one-point interval.
Here are the compactness facts behind this characterization. A closed subset of a compact space is compact: given an open cover of that subset, add its open complement to obtain an open cover of the whole space, take a finite subcover, and then discard the complement if it was selected. Conversely, compact subsets of a Hausdorff space are closed. For this direction, separate each point outside the compact set from each point inside it by disjoint open neighborhoods; compactness reduces the neighborhoods covering the compact set to finitely many, and the intersection of the corresponding finitely many neighborhoods of the outside point misses the compact set. The interval is Hausdorff as a subspace of \(\mathbb{R}\), so both facts apply.
Thus on \(C(I,Y)\), a compact-open condition can use any relatively closed \(K\subseteq I\), not only subintervals. The condition still applies to all of \(K\) at once. In particular, a proper closed subset may omit a whole interval or have several separated pieces.
Worked Example: Restricting Maps on the Unit Interval
Let \(I=[0,1]\) with its subspace topology from \(\mathbb{R}\), let \(Y=\mathbb{R}\) with its standard topology, and take \(K=\{0\}\cup[1/2,1]\) and \(U=(-1,1)\). The set \(K\) is closed in \(I\), hence compact. The subbasic set requires \(f(0)\in(-1,1)\) and \(f(t)\in(-1,1)\) for every \(t\in[1/2,1]\).
For \(f(t)=t/2\), \(f(0)=0\), \(f(1/2)=1/4\), and \(f(1)=1/2\). Since \(f\) is increasing on \([1/2,1]\), its values there lie in \([1/4,1/2]\). Therefore \(f(K)=\{0\}\cup[1/4,1/2]\subset(-1,1)\), so \(f\in[K,U]\).
For \(g(t)=2t-1\), we have \(g(0)=-1\), \(g(1/2)=0\), and \(g(1)=1\). The values \(-1\) and \(1\) are not in the open set \(U\), and both corresponding inputs belong to \(K\). Thus \(g(K)\not\subseteq U\) and \(g\notin[K,U]\).
Check: The endpoints \(0\) and \(1\) belong to \(K\), and the membership condition uses the open interval \((-1,1)\), which excludes both \(-1\) and \(1\). This confirms why \(g\) fails even though its value at the midpoint \(1/2\) is in \(U\).
Compact Subsets of a Discrete Space
A space \(X\) is discrete when every subset of \(X\) is open. For any discrete space, its compact subsets are exactly its finite subsets. Every finite subset is compact because from any open cover one can select one covering set for each of its finitely many points. If a subset \(K\) is infinite, the open cover consisting of the singletons \(\{\{x\}:x\in K\}\) has no finite subcover: finitely many singletons cover only finitely many points. Therefore \(K\) is not compact.
Proof. If \(K\) is finite, let an open cover of \(K\) be given. For each point of \(K\), choose one member of the cover containing it. The chosen family has at most \(|K|\) members and covers \(K\); if \(K=\varnothing\), the empty subfamily covers it. Hence \(K\) is compact. If \(K\) is infinite, its singleton subsets are open in the discrete topology and cover \(K\). No finite collection of these singletons covers an infinite set, so this open cover has no finite subcover. Thus \(K\) is not compact. \(\square\)
Every map from a discrete space is continuous: the preimage of any open set is a subset of the discrete domain and hence is open. For a discrete domain, therefore, compact-open subbasic sets impose conditions on finitely many input points. The compact set must be finite, but the same open target set \(U\) must contain the images of all the points in that finite set.
Worked Example: A Finite Constraint on a Discrete Domain
Give \(X=\mathbb{N}=\{1,2,\ldots\}\) the discrete topology and \(Y=\mathbb{R}\) the standard topology. Choose \(K=\{1,3,5\}\), a finite and therefore compact subset of \(X\), and \(U=(0,2)\). The subbasic set consists of all maps \(f:\mathbb{N}\to\mathbb{R}\) satisfying \(f(1),f(3),f(5)\in(0,2)\); every such map is continuous.
For \(f(n)=1/n\), the three constrained values are \(f(1)=1\), \(f(3)=1/3\), and \(f(5)=1/5\). Each is strictly greater than \(0\) and strictly less than \(2\), so \(f\in[K,U]\). For \(g\) defined by \(g(1)=1\), \(g(3)=2\), \(g(5)=1/5\), and \(g(n)=0\) for all other \(n\), the value \(g(3)=2\) is not in the open interval \(U\). Thus \(g\notin[K,U]\); values at inputs outside \(K\), such as \(g(2)=0\), do not affect this membership test.
Check: The constrained values of \(f\) are \(1\), \(1/3\), and \(1/5\), all in \((0,2)\). The constrained value \(g(3)=2\) is not in \((0,2)\). This checks both decisions using exactly the points of \(K\).
A Common Mistake: Using a Noncompact Set
The compact-open subbasis only permits compact \(K\). For example, \((0,1)\subseteq\mathbb{R}\) is bounded but not compact: the open sets \((1/n,1)\), for integers \(n\ge2\), cover \((0,1)\), since for every \(x\in(0,1)\) there is an integer \(n\ge2\) with \(1/n<x\). A finite collection of these sets has a largest index \(N\), and their union is \((1/N,1)\), which misses points in \((0,1/N]\). So this cover has no finite subcover.
Common pitfall. Writing \([(0,1),U]\) as a compact-open subbasic set is invalid: the first entry is not compact. The correction is to choose a compact set, such as \([0,1]\), or a closed subinterval such as \([1/4,3/4]\), and then test the image of that entire chosen set. Boundedness alone is not enough on \(\mathbb{R}\); Heine–Borel requires both closedness and boundedness.
| Domain | Which subsets are compact? | What the subbasic condition tests |
|---|---|---|
| \(\mathbb{R}\) with the standard topology | Exactly the closed bounded subsets | Every value \(f(x)\) for \(x\in K\) |
| \([a,b]\) with the subspace topology | Exactly the relatively closed subsets | Every value on the chosen relatively closed set |
| A discrete space \(X\) | Exactly the finite subsets | Values at finitely many specified inputs |
Check Your Understanding
For each question, identify the compact set used and apply the containment \(f(K)\subseteq U\), with attention to whether the endpoints of \(U\) are included.
- In \(C(\mathbb{R},\mathbb{R})\), with standard topologies, does \(f(x)=x^2\) belong to \([[-1,1],(-1,2)]\)?
- In \(C([0,1],\mathbb{R})\), which subsets \(K\subseteq[0,1]\) are eligible in a compact-open subbasic set?
- Why is \(\{1/n:n\in\mathbb{N}\}\) not compact in the standard topology on \(\mathbb{R}\)?
- For a discrete domain \(X\), what does \([K,U]\) require when \(K=\{x_1,x_2,x_3\}\)?
- Explain why \([X,U]\) is not a compact-open subbasic set when \(X\) is an infinite discrete space.