nLab
Lebesgue number lemma

Context

Analysis

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

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)(X,d) is a metric space which is sequentially compact, then for every open cover {U iX} iI\{U_i \to X\}_{i \in I} there exists a positive real number δ\delta \in \mathbb{R}, δ>0\delta \gt 0, called a Lebesgue number for XX, such that for every point xXx \in X there exists an iIi \in I such that the open ball around xx of radius δ\delta is contained in U iU_i:

((X,d)sequentially compact)δ>0(xX(iI(B x (δ)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 nn \in \mathbb{N} there exists a point x nXx_n \in X such that for all iIi \in I the open ball B x n (1/(n+1))B^\circ_{x_n}(1/(n+1)) is not contained in U iU_i. These points (x n) n(x_n)_{n \in \mathbb{N}} would constitute a sequence and so by assumption on XX there would exist a sub-sequence (x n k) k(x_{n_k})_{k \in \mathbb{N}} which converges to some x Xx_\infty \in X. Hence then there would be some i Ii_\infty \in I with x U i x_\infty \in U_{i_\infty}, and this, since U i 0U_{i_0} is open, also a positive real number ϵ>0\epsilon \gt 0 with B x (ϵ)U i B^\circ_{x_\infty}(\epsilon) \subset U_{i_\infty}. By convergence of the sub-sequence (x n k) k(x_{n_k})_k we could now choose a kk \in \mathbb{N} such that

1n k+1<ϵ2AAAandAAAd(x n k,w)<ϵ2. \frac{1}{n_k + 1} \lt \frac{\epsilon}{2} \phantom{AAA} \text{and} \phantom{AAA} d(x_{n_k}, w) \lt \frac{\epsilon}{2} \,.

This would imply that

B x n k (1/n k)B x (ϵ)U . 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 iU_i contains the open ball B x n k (1/n k)B^\circ_{x_{n_k}}(1/n_k), and hence we have a proof by contradiction.

References

Named after Henri Lebesgue.

Revised on May 21, 2017 05:11:22 by Urs Schreiber (92.218.150.85)