I want to prove the Lebesgue number lemma:
Let $(X, d)$ be a compact metric space. Then given an open cover $\mathcal{A}$ of $X$, there exists $\delta \gt 0$ such that for each subset of $X$ having diameter less than $\delta$, there is an element of $\mathcal{A}$ containing it.
How can I prove this?
We give an approach using the extreme value theorem. The following definition is a key ingredient in the proof: $\newcommand{\diam}{\operatorname{diam}}$
For each $x \in X$, define $h(x) \in \mathbb R^{\geqslant 0}$ to be the infimum of $\diam S$ over all $S \subseteq X$ satisfying the following conditions:
- $x \in S$, and
- $S \nsubseteq U$ for any $U \in \mathcal A$.
We now study the map $h: X \to \mathbb R^{\geqslant 0}$ as defined above. Note first that $h(x) > 0$ for every $x \in X$. [Proof is left as exercise.]
Lipschitzness. The main technical idea is to show that $h$ is $1$-Lipschitz. Fix any $x, y \in X$; we want to show that $h(y) \geqslant h(x) - d(x,y)$. Further fix an arbitrary $\varepsilon > 0$. By the definition of $h(y)$, there exists $T \subseteq X$ such that
Now, consider $S = T \cup \{ x \}$. Clearly,
Therefore, by definition of $h(x)$, we can see that $h(x) \leqslant h(y) + d(x,y) + \varepsilon$. Since this is true for all $\varepsilon > 0$, it follows that $h(x) \leqslant h(y) + d(x,y)$.
Wrap-up of the proof. Since $h$ is Lipschitz, it is also continuous on $X$. Furthermore, being a continuous function over a compact set, $h$ is guaranteed (by the extreme value theorem) to attains its minimum over $X$, and this minimum is strictly positive. Let $\delta > 0$ be any number that is strictly smaller than $h(x)$ for all $x \in X$. It only remains to check that such a $\delta$ satisfies the requirements of the problem. I leave that as a simple exercise.
This is known as the Lebesgue number lemma. The $\delta > 0$ promised by it is called a Lebesgue number of the cover $\mathcal A$. Considering the importance of this result, I give two proofs for it. This answer describes an open cover based proof; another one based on my favorite extreme value theorem is in another answer.
Since $\mathcal A$ is an open cover, for each $x \in X$, there is a member $A_x \in \mathcal A$ such that $x \in A_x$. Since $A_x$ is open, there exists $r(x) > 0$ such that $B(x, 2 r(x)) \subseteq A_x \in \mathcal A$. (Notice that the radius of the ball is $2 r(x)$, not $r(x)$.) Now $\left\{ B(x, r(x)) \right\}_{x \in X}$ is an open cover of $X$; hence by compactness, there exists a finite set $S \subseteq X$ such that $X = \bigcup _{x \in S} \ B(x, r(x)) $. Finally, we claim that $\delta = \min \{ r(x) \, \colon \, x \in S \}$ works:
First, $\delta > 0$ since we are minimising over a finite set of strictly positive numbers.
Given $y \in X$, there exists $x \in S$ such that $y \in B(x, r(x))$. Then $$ B(y, \delta) \subseteq B(y, r(x)) \stackrel{\color{Red}{(\triangle)}}{\subseteq} B(x, 2 r(x)) \subseteq A_x \in \mathcal A. $$ [Exercise: Explain the inclusion marked with $\color{Red}{(\triangle)}$.] $\qquad \square$
Except for some notational changes, this proof is the same as the one gary's comment points to. Check this blog page.
From the open cover $\mathcal{A}= (U_i)_{i \in I}$ we get another cover with open sets $(U_{i,n})$ where
$$U_{i,n} = \{ x \in X \ | d(x, X \backslash U_i)> 1/n\}$$
Take a finite subcover
$$\{ U_{i_1, n_1}, \ldots, U_{i_k, n_k} \}$$
Any $\delta < \min \{ 1/n_1, \ldots, 1/n_k\}$ will work.
Seeing as this proof is not here yet, I will add it.
Suppose for the sake of contradiction that there is no such $\delta$, then for each $n$ we can find $x_n\in X$ such that
$$ B(x_n,1/n) \not\subseteq A$$
For any $A\in\mathcal{A}$
By sequential compactness of $X$ the sequence $(x_n)$ has a convergent subsequence, $(x_{n_k})$ converging to $x_0$, but now we obtain a contraindication since there is some $\varepsilon>0$ such that
$$B(x_0,\varepsilon)\subseteq A$$
For some $A\in \mathcal{A}$, and there is some $k\in\mathbb{N}$ such that $\frac{1}{n_k}<\frac{\varepsilon}{2}$ and such that $d(x_0,x_{n_k})<\frac{\varepsilon}{2}$, which means
$$ B(x_{n_k},1/n_k)\subseteq B(x_0,\varepsilon) \subseteq A\in\mathcal{A} $$
The desired contradiction.
I offer a proof I came up with a while ago that is not too dissimilar from Srivatsan's answer but is, I think, easier to follow and also easier to think about oneself.
Let $\mathscr{U}$ be an open cover of the compact metric space $(X;\varrho)$. Define $\psi:X\to(0,\infty)$ by: $$x\mapsto\sup\{0<t<\mathrm{diam}\,X:\exists U\in\mathscr{U},\,B(x,t)\subseteq U\}$$Since $\mathscr{U}$ is an open cover, we are taking a supremum over a nonempty set so $\psi(x)$ really is a positive real for each $x$, and also since $X$ is compact we are taking a supremum over a bounded set so $\psi(x)$ really is a real number for each $x$; this is well-defined. Clearly the purpose of $\psi$ is to measure "the most we can get away with", in terms of sizes of balls, and we want to look at extrema of $\psi$.
Fix $x,y\in X$. Observe that if $t>\varrho(x,y)$ and, for any $U\in\mathscr{U}$, $B(x,t)\subseteq U$ then $B(y,t-\varrho(x,y))\subseteq U$ and visa versa; thus $\psi(x)-\varrho(x,y)\le\psi(y)$ and $\psi(y)-\varrho(x,y)\le\psi(x)$; we see $\psi$ is Lipschitz with constant $1$ (this is recovered even in the case both $\psi(x),\psi(y)$ are lesser than $\varrho(x,y))$. Hence, $\psi$ is a continuous positive function on $X$ and attains a minimum positive value. Very directly we see that any $\delta$ in range $0<\delta<\min_X\psi$ will be a Lebesgue number; in fact, $\min_X\psi$ is the supremum over all Lebesgue numbers for the cover $\mathscr{U}$.
