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.


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

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