Lebesgue's number lemma

In topology, the Lebesgue covering lemma is a useful tool in the study of compact metric spaces.

Given an open cover of a compact metric space, a Lebesgue's number of the cover is a number ${\displaystyle \delta >0}$ such that every subset of ${\displaystyle X}$ having diameter less than ${\displaystyle \delta }$ is contained in some member of the cover.

The existence of Lebesgue's numbers for compact metric spaces is given by the Lebesgue's covering lemma:

If the metric space ${\displaystyle (X,d)}$ is compact and an open cover of ${\displaystyle X}$ is given, then the cover admits some Lebesgue's number ${\displaystyle \delta >0}$.

The notion of Lebesgue's numbers itself is useful in other applications as well.

Proof

Direct proof

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, ${\displaystyle Y\subseteq A_{i}}$.

Proof by contradiction

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:

${\displaystyle \delta _{1}=1,\quad \exists A_{1}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{1})<\delta _{1}\quad {\text{and}}\quad \neg \exists \beta (A_{1}\subset U_{\beta })}$

${\displaystyle \delta _{2}={\frac {1}{2}},\quad \exists A_{2}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{2})<\delta _{2}\quad {\text{and}}\quad \neg \exists \beta (A_{2}\subset U_{\beta })}$

${\displaystyle \vdots }$

${\displaystyle \delta _{k}={\frac {1}{k}},\quad \exists A_{k}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{k})<\delta _{k}\quad {\text{and}}\quad \neg \exists \beta (A_{k}\subset U_{\beta })}$

${\displaystyle \vdots }$

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 ${\displaystyle B_{r}(x_{0})\subset U_{\alpha _{0}}}$. 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.

References

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