Lebesgue number lemma

Statement

Every open cover, [ilmath]\mathcal{U} [/ilmath], of a compact metric space has a Lebesgue number.

Proof

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

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

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

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