I am studying 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.

I found this proof.

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 $$ B(y, \delta) \subseteq B(y, r(x)) \stackrel{\color{Red}{(\triangle)}}{\subseteq} B(x, 2 r(x)) \subseteq A_x \in \mathcal A.$$

I'm not sure why the containment marked by $\color{Red}{(\triangle)}$ holds. Why is this true? How could I prove it?


As $y \in B(x, r(x))$, $d(x, y) < r(x)$.

Let $z \in B(y, r(x))$, then $d(y, z) < r(x)$.

By the triangle inequality,

$$d(x, z) \leq d(x, y) + d(y, z) < r(x) + r(x) = 2r(x),$$

so $z \in B(x, 2r(x))$.

Therefore, $B(y, r(x)) \subseteq B(x, 2r(x))$.


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