Contents

# Contents

## Idea

The Lebesgue number lemma makes a statement about properties of open covers of sequentially compact metric spaces. It serves as ingredient of the proof that sequentially compact metric spaces are equivalently compact metric spaces.

## Statement

###### Proposition

Assuming excluded middle and countable choice then:

If $(X,d)$ is a metric space which is sequentially compact, then for every open cover $\{U_i \to X\}_{i \in I}$ there exists a positive real number $\delta \in \mathbb{R}$, $\delta \gt 0$, called a Lebesgue number for $X$, such that for every point $x \in X$ there exists an $i \in I$ such that the open ball around $x$ of radius $\delta$ is contained in $U_i$:

$\left( (X,d) \, \text{sequentially compact} \right) \;\Rightarrow\; \underset{\delta \gt 0}{\exists} \left( \underset{x \in X}{\forall} \left( \underset{i \in I}{\exists} \left( B^\circ_x(\delta) \subset U_i \right) \right) \right)$
###### Proof

Assume that the statement were not true. This would mean that for every $n \in \mathbb{N}$ there exists a point $x_n \in X$ such that for all $i \in I$ the open ball $B^\circ_{x_n}(1/(n+1))$ is not contained in $U_i$. These points $(x_n)_{n \in \mathbb{N}}$ would constitute a sequence and so by assumption on $X$ there would exist a sub-sequence $(x_{n_k})_{k \in \mathbb{N}}$ which converges to some $x_\infty \in X$. Hence then there would be some $i_\infty \in I$ with $x_\infty \in U_{i_\infty}$, and this, since $U_{i_\infty}$ is open, also a positive real number $\epsilon \gt 0$ with $B^\circ_{x_\infty}(\epsilon) \subset U_{i_\infty}$. By convergence of the sub-sequence $(x_{n_k})_k$ we could now choose a $k \in \mathbb{N}$ such that

$\frac{1}{n_k + 1} \lt \frac{\epsilon}{2} \phantom{AAA} \text{and} \phantom{AAA} d(x_{n_k}, x_{\infinity}) \lt \frac{\epsilon}{2} \,.$

This would imply that

$B^\circ_{x_{n_k}}(1/n_k) \subset B^\circ_{x_\infty}(\epsilon) \subset U_\infty \,.$

This contradicts the assumption that none of the $U_i$ contains the open ball $B^\circ_{x_{n_k}}(1/n_k)$, and hence we have a proof by contradiction.

## References

Named after Henri Lebesgue.

Last revised on January 3, 2019 at 10:41:50. See the history of this page for a list of all contributions to it.