Margin between a compact subspace and an including open subspace

olivierd

Aug 1, 2017 2 min read

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 $\forall x \in C,\ \dot B(x, r) \subseteq O$ . $\blacksquare$