Neighborhood Calculations In The Compact-Open Topology

What a Compact-Open Neighborhood Requires

The compact-open topology turns conditions on whole compact subsets of a domain into open sets in a space of maps. A typical condition says that every point of a specified compact set must be sent into a specified open set. Finite intersections let us impose several such requirements at once. The practical calculation is therefore to identify the compact sets, identify their target open sets, and check each containment separately.

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 open sets has a finite subcover. In the definition below, compactness is with respect to the topology on \(X\), and openness is with respect to the topology on \(Y\).

Thus \(g\in[K,U]\) means exactly that \(g(x)\in U\) for every \(x\in K\). It is a containment condition on the entire image \(g(K)\), not merely a condition on one point of \(K\). If \(K=\varnothing\), the condition holds for every \(g\), since \(g(\varnothing)=\varnothing\). If \(U=\varnothing\), then \([K,U]\) is empty when \(K\ne\varnothing\), and is all of \(C(X,Y)\) when \(K=\varnothing\).

Finite Intersections and the Neighborhood Basis

Proof. By the definition of a topology generated by a subbasis, its open sets are arbitrary unions of finite intersections of subbasic sets. The whole space is the empty intersection. To check the membership statement, a map \(f\) belongs to an intersection precisely when it belongs to each factor. By the definition of \([K_j,U_j]\), this is equivalent to \(f(K_j)\subseteq U_j\) for every index \(j\). These finite intersections therefore form a basis: they cover the space, and the intersection of two of them is again a finite intersection of subbasic sets. \(\square\)

For a particular \(f\in C(X,Y)\), the basis elements that contain \(f\) give a neighborhood basis at \(f\). More explicitly, if \(O\) is any open set in the compact-open topology with \(f\in O\), then the definition of the generated topology expresses \(O\) as a union of finite intersections. At least one of those intersections contains \(f\), and that intersection is contained in \(O\). Consequently, to describe a neighborhood of \(f\), it is enough to specify finitely many conditions \(f(K_j)\subseteq U_j\). The conditions need not involve disjoint compact sets or disjoint open sets.

Compact Images and Finite-Cover Refinements

Compactness supplies a useful way to reduce a possibly large open cover of the image of a compact set to finitely many target open sets. The resulting finite union gives a compact-open neighborhood. This does not say that the image lies in one member of the cover; the finite union is important.

Proof. First, \(f(K)\) is compact. Indeed, if open sets in \(Y\) cover \(f(K)\), their preimages under the continuous map \(f\) are open in \(X\) and cover \(K\). Compactness of \(K\) gives finitely many of those preimages covering \(K\), and the corresponding finitely many open sets cover \(f(K)\). Apply compactness of \(f(K)\) to the given cover \(\{U_i:i\in J\}\). A finite subfamily \(U_{i_1},\ldots,U_{i_m}\) covers \(f(K)\), so \(f(K)\subseteq U_{i_1}\cup\cdots\cup U_{i_m}\). This is exactly the assertion that \(f\) belongs to the indicated subbasic open set. Since the finite union is contained in \(\bigcup_{i\in J}U_i\), every map whose image of \(K\) lies in the finite union also has its image in the full union. This proves the stated inclusion. If \(K=\varnothing\), the image is empty and the conclusion holds with the empty finite subfamily. \(\square\)

There is another useful refinement when a compact set has already been covered by compact pieces. Suppose \(K_1,\ldots,K_m\) are compact subsets of \(X\) whose union is \(K\), and \(U_1,\ldots,U_m\) are open in \(Y\). Then \(\bigcap_{j=1}^{m}[K_j,U_j]\subseteq[K,\bigcup_{j=1}^{m}U_j]\). For any map in the intersection, each point \(x\in K\) lies in at least one \(K_j\); its image therefore lies in the corresponding \(U_j\), and hence in the union. This argument uses the cover of \(K\) by the \(K_j\); it does not require those pieces to be disjoint.

A Common Mistake: Union Is Not Intersection

A frequent error is to replace a condition that the image lies in a union by the requirement that it lie in every member of the union. These are different conditions. If \(U_1,U_2\) are open, then \([K,U_1]\cap[K,U_2]=[K,U_1\cap U_2]\), because membership requires the image to lie in both sets. In contrast, \([K,U_1\cup U_2]\) only requires each image point to lie in at least one of them.

For a concrete check, let \(K=[0,1]\), \(f(t)=t\), \(U_1=(-1,3/5)\), and \(U_2=(2/5,2)\). We have \(f(K)=[0,1]\subseteq(-1,2)=U_1\cup U_2\), so \(f\in[K,U_1\cup U_2]\). But \(1\notin U_1\), so \(f(K)\nsubseteq U_1\); and \(0\notin U_2\), so \(f(K)\nsubseteq U_2\). Hence \(f\notin[K,U_1]\cap[K,U_2]\). The incorrect calculation would assert that membership in \([K,U_1\cup U_2]\) implies membership in the intersection. The correction is to check containment in the union, or to impose the genuinely stronger separate conditions only when those are intended.

Finite intersections are useful precisely because each factor records one independent containment requirement. When compactness gives a finite subcover of an image, it justifies using a finite union of target open sets. When a domain set is covered by compact pieces, separate conditions on those pieces can give a smaller neighborhood. In either calculation, keeping track of whether the target condition is a union or an intersection prevents the most common membership error.