Margin between a compact subspace and an including open subspace

In Chapter 31 (“Compact-Open Topology”), the fact that, on the set of the continuous mappings \R \to \R the second topology (the uniform convergence topology) is finer than the first (the compact-open topology) is said to be “clear” but actually needs some non-trivial proof. Part of this is the lemma that I state and prove below.

For any x \in E and real r > 0, let \dot B(x, r) be the open ball centered on x and with radius r. We know that an open ball is an open subset.

The lemma says:

Let (E, d) be a metric space, O an open subspace of E and C \subseteq O a compact subspace of E. Then there exists r > 0 such that for all x in C, \dot B(x, r) \subseteq O.

Proof:

For each x \in C, since x \in O, there exists an r_x > 0 such that \dot B(x, r_x) \subseteq O.

The collection of all \dot B(x, \frac 1 2 {r_x}) with x \in C is an open cover of C. Hence there exists a finite J \subseteq C such that \{\ \dot B(x, \frac 1 2 {r_x})\ \}_{x \in J} covers C.

Since J is finite, there exists a real r > 0 such that for all x \in J, r \le \frac 1 2 {r_x}.

Let x be an eventual point of C.

Since \{\ \dot B(z, \frac 1 2 {r_z})\ \}_{z \in J} covers C, there exists z \in C such that x \in \dot B(z, \frac 1 2 {r_z}), that is, d(z, x) < \frac 1 2 {r_z}.

Let y be an eventual point of \dot B(x, r).

We have d(z, x) < \frac 1 2 {r_z} and d(x, y) < r \le \frac 1 2 {r_z}, which leads to d(z, y) < r_z, that is, y \in \dot B(z, r_z). But z is in C and r_z was taken such that \dot B(z, r_z) \subseteq O. Hence y \in O.

Hence \dot B(x, r) \subseteq O.

We thus have r > 0 such that \forall x \in C,\ \dot B(x, r) \subseteq O. \blacksquare

Uniform convergence topology and the open set topology

Chapter 31, “The Compact-Open Topology”, describes the compact-open topology, and goes on to compare it with two other topologies on \Mor(X, Y) in the case where X = Y = \R. It asserts that “clearly” the second of these three is finer than the first, and the third finer than the second.

This may well be “clear” intuitively, but is not so easy to prove. Here I will prove that the third (that the author does not name, but that I call the open set topology) is finer than the second, which is the uniform convergence topology.

I will only need to specify that X is a topological space and Y a metric space with distance d.

I define E = \Mor(X, Y) the set of continuous mappings from X to Y.

I note \dot B(y, r) the open ball in Y centred on y \in Y and with radius r > 0. It is the set of all y' \in Y such that d(y, y') < r.

Furthermore, for any H subset of X \times Y, I note \kappa(H) the set of all continuous mappings \phi: X \to Y the graph of which is a subset of H; in other words, \kappa(H) = \{\ \phi \in E\ |\ \forall x \in X, (x, \phi(x)) \in H\ \}.

The uniform convergence topology on E is such that a subset A of E is open if and only if for any \phi \in A, there exists r > 0 such that for any \psi \in E, if \forall x \in X, \psi(x) \in \dot B(\phi(x), r), then \psi \in A. We can express this condition in terms of subsets of X \times Y that I have named open tubes: for any mapping (not necessarily continuous) \phi: X \to Y and for any r > 0, the open tube \dot T(\phi, r) is \{\ (x, y) \in X \times Y\ |\ d(\phi(x), y) < r\ \}. Then A \subseteq E is open for the uniform convergence topology if and only if for any \phi \in A, there exists r > 0 such that \kappa(\dot T(\phi, r)) \subseteq A.

The third topology ‑ the “open set” topology ‑ is the topology on E generated by all the \kappa(O) for all O open subsets of X \times Y. (The book says the topology is this collection itself, but I have not managed to prove that it directly forms a topology ‑ as is more explicitly requested in exercise 215.)

To show that the open set topology is finer than the uniform convergence one, we consider an open set for the uniform convergence topology and endeavour to show that it is open in the open set one.

Let A be an open set in the uniform convergence topology.

Let \phi be an eventual member of A.

There exists an r > 0 such that \kappa(\dot T(\phi, r)) \subseteq A.

I prove here that, \phi being continuous, \dot T(\phi, r) is an open subset of X \times Y. It includes the graph of \phi. Hence \kappa(\dot T(\phi, r)) contains \phi, and is by definition an open set for the open set topology on E. Since it is a subset of A, A is a neighbourhood of \phi for the open set topology.

Hence for any \phi \in A, A is a neighbourhood of \phi for the open set topology. This implies that A is itself an open set in the open set topology. \blacksquare