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On a discrete domain the compact-open topology is the topology of pointwise convergence

Statement

Let $X$ be a discrete topological space (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) and $Y$ a topological space. Every compact subset of $X$ is finite (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), and on $C (X, Y)$ the compact-open topology (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces) coincides with the topology of pointwise convergence (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ ); hence on every $F \subseteq C (X, Y)$ the compact-open subspace topology is the subspace topology inherited from the product $Y^{X}$ . In particular finite intersections of sets ${f : f (x) \in V}$ , with $x \in X$ and $V \subseteq Y$ open, form a basis; these are open-coordinate constraints, not requirements that a coordinate equal a prescribed value.

Facts & Assumptions

[F1]

In the discrete topology on $X$ every subset is open, and a subset $K \subseteq X$ is compact exactly when every open cover of $K$ by open sets of $X$ (equivalently of the subspace $K$ ) has a finite subcover; the empty space and every finite space are compact. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it)

[F2]

The compact-open topology on $C (X, Y)$ is generated by the subbasis of sets $S (K, V) = {f \in C (X, Y) : f [K] \subseteq V}$ with $K \subseteq X$ compact and $V \subseteq Y$ open. (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces)

[F3]

The topology of pointwise convergence on $C (X, Y)$ is the subspace topology inherited from the product $Y^{X}$ ; its subbasis consists of the traces of the sets $π_{x}^{- 1} [V] = {f \in Y^{X} : f (x) \in V}$ , $x \in X$ , $V \subseteq Y$ open, and its basic open sets impose open-set constraints at finitely many points of $X$ . (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ , Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace)

[F4]

In the product topology on $Y^{X}$ the basic open sets are the boxes that restrict only finitely many coordinates. (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space)

Proof

Given: A discrete space $X$ , a topological space $Y$ , and the two topologies on $C (X, Y)$ .

1.1F1

Every compact subset $K \subseteq X$ is finite: the family of singleton subsets ${x}$ , $x \in K$ , is an open cover of the subspace $K$ by [F1]. Compactness of $K$ gives a finite subcover, exhibiting $K$ as the union of finitely many singletons; for $K = \emptyset$ , the empty family suffices.

1.2F2F3

For every finite $K \subseteq X$ and every open $V \subseteq Y$ the subbasic compact-open set is the finite intersection $S (K, V) = ⋂_{x \in K} {f \in C (X, Y) : f (x) \in V}$ .

2.1step 1.1step 1.2F1F2F3

Each compact-open subbasic set is a finite intersection of pointwise subbasic sets by steps 1.1 and 1.2, and each pointwise subbasic set ${f \in C (X, Y) : f (x) \in V}$ equals $S ({x}, V)$ , which is compact-open subbasic because the singleton ${x}$ is compact by [F1]. A topology containing a family contains the topology generated by it, so the two topologies on $C (X, Y)$ each contain the other, hence are equal; by [F3] this common topology is the subspace topology inherited from $Y^{X}$ .

3.1step 2.1F3F4∎

Restricting an equality of topologies to a subset preserves it: for $F \subseteq C (X, Y)$ the traces on $F$ of the two topologies coincide. The pointwise topology is the subspace topology from $Y^{X}$ by [F3], and its basic open sets impose open-set constraints on finitely many coordinates by [F3, F4]; hence on $C (X, Y)$ , and on every $F \subseteq C (X, Y)$ , those sets form a basis.

Depends on

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Sources

Dikran D. Dikranjan, Introduction to Topological Groups (author lecture notes, Universita di Udine / Universidad Complutense de Madrid, 2007) (standard reference, not scraped)
Manfred Einsiedler and Thomas Ward, Ergodic Theory with a View Towards Number Theory, Appendix C (course-hosted full text) (standard reference, not scraped)