Lebesgue's number lemma

Let ${\displaystyle {\mathcal {U}}}$ be an open cover of ${\displaystyle X}$. Since ${\displaystyle X}$ is compact we can extract a finite subcover ${\displaystyle \{A_{1},\dots ,A_{n}\}\subseteq {\mathcal {U}}}$. If any one of the ${\displaystyle A_{i}}$'s equals ${\displaystyle X}$ then any ${\displaystyle \delta >0}$ will serve as a Lebesgue number. Otherwise for each ${\displaystyle i\in \{1,\dots ,n\}}$, let ${\displaystyle C_{i}:=X\smallsetminus A_{i}}$, note that ${\displaystyle C_{i}}$ is not empty, and define a function ${\displaystyle f:X\rightarrow \mathbb {R} }$ by

${\displaystyle f(x):={\frac {1}{n}}\sum _{i=1}^{n}d(x,C_{i}).}$

Since ${\displaystyle f}$ is continuous on a compact set, it attains a minimum ${\displaystyle \delta }$. The key observation is that, since every ${\displaystyle x}$ is contained in some ${\displaystyle A_{i}}$, the extreme value theorem shows ${\displaystyle \delta >0}$. Now we can verify that this ${\displaystyle \delta }$ is the desired Lebesgue number. If ${\displaystyle Y}$ is a subset of ${\displaystyle X}$ of diameter less than ${\displaystyle \delta }$, then there exists ${\displaystyle x_{0}\in X}$ such that ${\displaystyle Y\subseteq B_{\delta }(x_{0})}$, where ${\displaystyle B_{\delta }(x_{0})}$ denotes the ball of radius ${\displaystyle \delta }$ centered at ${\displaystyle x_{0}}$ (namely, one can choose ${\displaystyle x_{0}}$ as any point in ${\displaystyle Y}$). Since ${\displaystyle f(x_{0})\geq \delta }$ there must exist at least one ${\displaystyle i}$ such that ${\displaystyle d(x_{0},C_{i})\geq \delta }$. But this means that ${\displaystyle B_{\delta }(x_{0})\subseteq A_{i}}$ and so, in particular, ${\displaystyle Y\subseteq A_{i}}$.

• Munkres, James R. (1974), Topology: A first course, p. 179, ISBN 978-0-13-925495-6