nLab Lebesgue number lemma

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.

• Wikipedia, Lebesgue’s number lemma