Proof of the Lebesgue number lemma

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?

Solution 1:

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

• $y \in T$;
• $T$ is not contained in any $U \in \mathcal A$;
• $\diam T \leqslant h(y) + \varepsilon$.

Now, consider $S = T \cup \{ x \}$. Clearly,

• $x \in S$;
• $S$ is not contained in any $U \in \mathcal A$ (why?);
• $\diam S \stackrel{\color{Red}{(!!)}}{\leqslant} \diam T + d(x,y) \leqslant h(y) + \varepsilon + d(x,y)$. [Exercise: Justify the inequality marked $\color{Red}{(!!)}$.]

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.

Solution 2:

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.

Solution 3:

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.