# Lebesgue's Number Lemma

## Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $M$ be sequentially compact.

Then there exists a Lebesgue number for every open cover of $M$.

## Proof

Aiming for a contradiction, suppose $\UU$ is an open cover of $M$ for which no Lebesgue number exists.

Then for any $n \in \N$, there exists some $x_n \in M$ such that $\map {B_{1 / n} } {x_n} \subseteq U$ is false for each $U \in \UU$. (Otherwise $1/n$ would be a Lebesgue number for $\UU$.)

As $M$ is sequentially compact, $\sequence {x_n}$ has a subsequence, say $\sequence {x_{\map n r} }$ which converges to some $x \in M$.

Since $\UU$ covers $M$, $x \in U_0$ for some $U_0 \in \UU$.

Since $U_0$ is open, $\exists m \in \N: \map {B_{2 / m} } x \subseteq U_0$.

Now $\map {B_{1 / m} } x$ contains $x_{\map n r}$ for all $r \ge R$, say.

Choose $r \ge R$ such that $\map n r \ge m$ and write $s = \map n r$.

Then $\map {B_{1 / s} } {x_s} \subseteq \map {B_{2 / m} } x$ since:

 $\ds \map d {x_s, y}$ $<$ $\ds \frac 1 s$ $\ds \leadsto \ \$ $\ds \map d {x, y}$ $\le$ $\ds \map d {x, x_s} + \map d {x_s, y}$ $\ds$ $<$ $\ds \frac 1 m + \frac 1 s$ $\ds$ $\le$ $\ds \frac 2 m$

So $\map {B_{1 / s} } {x_s} \subseteq U_0$ which contradicts the choice of $x_s$.

So, by Proof by Contradiction, there has to be a Lebesgue number for $\UU$.

$\blacksquare$

## Source of Name

This entry was named for Henri Léon Lebesgue.