Lebesgue number lemma

Statement

Every open cover, $\mathcal{U}$, of a compact metric space has a Lebesgue number.

Proof

For every $x\in X$ there is a $U\in\mathcal{U}$ such that $x\in U$, and that $U$ is an open set, thus:

• $\forall x\in X\ \exists U\in\mathcal{U}\ \exists\epsilon_x>0[B_{\epsilon_x}(x)\subseteq U]$ (where $B_r(x)$ denotes the open ball of radius $r$ centred at $x$)
• Note that $B_{\frac{1}{2}\epsilon_x}(x)\subseteq B_{\epsilon_x}(x)$ so:
  • $\forall x\in X\ \exists U\in\mathcal{U}\ \exists\epsilon_x>0[B_{\frac{1}{2}\epsilon_x}(x)\subseteq B_{\epsilon_x}(x)\subseteq U]$
• The set $\{B_{\frac{1}{2}\epsilon_x}(x)\ \vert\ x\in X\}$ is trivially an open cover of $X$. By the compactness property of $X$
  • $\exists$ a finite subcover of $\{B_{\frac{1}{2}\epsilon_x}(x)\ \vert\ x\in X\}$, call this:
    • $\left\{B_{\frac{1}{2}\epsilon_{x_i} }(x_i)\right\}_{i=1}^n$ for some $x_i\in X$, is a covering of $X$.
• Define $\delta:=\text{min}\left\{\right\}$
• Now we must show that for any set of diameter $<\delta$, $A$, that there $\exists U\in\mathcal{U}[A\subseteq U]$.
  • Let $A$ be given. Where $A$ has diameter $<\delta$
    • Let $\alpha\in A$ be any point in $A$.

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TODO: Discussion of this proof, finish it, although I've done the work, all you need to do is show $d(\alpha,x_i)<\frac{1}{2}\epsilon_{x_i}$ then $d(\alpha,\beta)<\deta$ for some $\beta\in A$. Thus $d(x_i,\beta)<\frac{1}{2}\epsilon_{x_i}+\delta$ and note $\delta<\frac{1}{2}\epsilon_{x_i}$ always, thus $d(x_i,\beta)<\epsilon_{x_i}$

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