(Lemma) Lebesgue numbers exist for compact metric spaces

Lemma: Lebesgue number

If $\mathcal{C}$ is an open cover of a compact metric space, then there exists $\delta >0$ such that whenever $\text{diam}(A)<\delta$, we have $A\subseteq U$ entirely contained for some $U\in \mathcal{C}$. The quantity $\delta$ is known as the Lebesgue number of the cover.

Proof from Topology.

It suffices to show this for all balls $B_{\delta }(x)$, since if $\text{diam}(A)<\delta$ then $A\in B_{\delta }(x)$, where $x\in A$ is any point. Note that $B_{\delta }(x)\subseteq U$ if and only if $d(x,X\backslash U)\ge \delta$.

🔺 Exercise. Show that we do have $B_{\delta }(x)\subseteq U$ if and only if $d(x,X\backslash U)\ge \delta$. This trick is central to this proof!

Assume $X\notin \mathcal{C}$, since otherwise any $\delta >0$ works. Since $X$ is compact, assume $U_1,\dots ,U_n\in \mathcal{C}$ is a finite subcover of $X$. Define $Z_i=X\backslash U_i$, and let $f_i:X\to \mathbb{R}$ be the map $x\mapsto d(x,Z_i)$ for each $1\le i\le n$. Further, define $f:X\to \mathbb{R}$ be defined by

$$f(x)=\max \{f_i(x)|i=1,\dots ,n\}.$$

🔺 Exercise. Show that the maximum of finitely many continuous functions is continuous. (Hint: use induction).

Our goal is to prove that for all $x\in X$, we have $f(x)>0$. If we know this, then since $f$ continuous implies $f(X)$ is compact, there exists $\delta >0$ such that for all $x\in X$, we have

$$f(x)>\delta >0.$$

Then we may conclude that for all $x\in X$, there exists $i\in \{1,\dots ,n\}$ such that

$$f(x)=f_i(x)=d(x,X\backslash U_i)>\delta ⟺B_{\delta }(x)\subseteq U_i.$$

wip because the infimum is also strictly greater than 0? why is $f(x)$ strictly greater? Note that in general strictly greater than 0 because X compact.

To prove that $f(x)$ is always positive, let $x\in X$ so that $x\in U_i$ for some $i$. Then there exists $ϵ>0$ such that

$$x\in B_{ϵ}(x)\subseteq U_i⟹d(x,X\backslash U_i)\ge ϵ.$$

Since the distance is precisely how we defined $f_i(x)$, we conclude that $f_i(x)>0$.