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's 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's number. If ${\displaystyle Y}$  is a subset of ${\displaystyle X}$  of diameter less than ${\displaystyle \delta }$ , choose ${\displaystyle x_{0}}$  as any point in ${\displaystyle Y}$ , then by definition of diameter, ${\displaystyle Y\subseteq B_{\delta }(x_{0})}$ , where ${\displaystyle B_{\delta }(x_{0})}$  denotes the ball of radius ${\displaystyle \delta }$  centered at ${\displaystyle x_{0}}$ . 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, ${\displaystyle Y\subseteq A_{i}}$ .

Suppose for contradiction that ${\displaystyle X}$  is sequentially compact, ${\displaystyle \{U_{\alpha }\mid \alpha \in J\}}$  is an open cover of ${\displaystyle X}$ , and the Lebesgue number ${\displaystyle \delta }$  does not exist. That is: for all ${\displaystyle \delta >0}$ , there exists ${\displaystyle A\subset X}$  with ${\displaystyle \operatorname {diam} (A)<\delta }$  such that there does not exist ${\displaystyle \beta \in J}$  with ${\displaystyle A\subset U_{\beta }}$ .

This enables us to perform the following construction:

Note that ${\displaystyle A_{n}\neq \emptyset }$  for all ${\displaystyle n\in \mathbb {Z} ^{+}}$ , since ${\displaystyle A_{n}\not \subset U_{\beta }}$ . It is therefore possible by the axiom of choice to construct a sequence ${\displaystyle (x_{n})}$  in which ${\displaystyle x_{i}\in A_{i}}$  for each ${\displaystyle i}$ . Since ${\displaystyle X}$  is sequentially compact, there exists a subsequence ${\displaystyle \{x_{n_{k}}\}}$  (with ${\displaystyle k\in \mathbb {Z} _{>0}}$ ) that converges to ${\displaystyle x_{0}}$ .

Because ${\displaystyle \{U_{\alpha }\}}$  is an open cover, there exists some ${\displaystyle \alpha _{0}\in J}$  such that ${\displaystyle x_{0}\in U_{\alpha _{0}}}$ . As ${\displaystyle U_{\alpha _{0}}}$  is open, there exists ${\displaystyle r>0}$  with ${\displaystyle B_{d}(x_{0},r)\subset U_{\alpha _{0}}}$ . Now we invoke the convergence of the subsequence ${\displaystyle \{x_{n_{k}}\}}$ : there exists ${\displaystyle L\in \mathbb {Z} ^{+}}$  such that ${\displaystyle L\leq k}$  implies ${\displaystyle x_{n_{k}}\in B_{r/2}(x_{0})}$ .

Furthermore, there exists ${\displaystyle M\in \mathbb {Z} _{>0}}$  such that ${\displaystyle \delta _{M}={\tfrac {1}{M}}<{\tfrac {r}{2}}}$ . Hence for all ${\displaystyle z\in \mathbb {Z} _{>0}}$ , we have ${\displaystyle M\leq z}$  implies ${\displaystyle \operatorname {diam} (A_{M})<{\tfrac {r}{2}}}$ .

Finally, define ${\displaystyle q\in \mathbb {Z} _{>0}}$  such that ${\displaystyle n_{q}\geq M}$  and ${\displaystyle q\geq L}$ . For all ${\displaystyle x'\in A_{n_{q}}}$ , notice that:

• ${\displaystyle d(x_{n_{q}},x')\leq \operatorname {diam} (A_{n_{q}})<{\frac {r}{2}}}$ , because ${\displaystyle n_{q}\geq M}$ .
• ${\displaystyle d(x_{n_{q}},x_{0})<{\frac {r}{2}}}$ , because ${\displaystyle q\geq L}$  entails ${\displaystyle x_{n_{q}}\in B_{r/2}\left(x_{0}\right)}$ .

Hence ${\displaystyle d(x_{0},x')<r}$  by the triangle inequality, which implies that ${\displaystyle A_{n_{q}}\subset U_{\alpha _{0}}}$ . This yields the desired contradiction.

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