Lebesgue's number lemma

In topology, the Delta number , is a useful tool in the study of compact metric spaces. It states:

If the metric space ${\displaystyle (X,d)}$ is compact and an open cover of ${\displaystyle X}$ is given, then there exists 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.

Such a number ${\displaystyle \delta }$ is called a Delta number of this cover. The notion of a Delta number 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 Delta 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 Delta 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, ${\displaystyle Y\subseteq A_{i}}$ .

Proof by Contradiction

Assume ${\displaystyle X}$  is sequentially compact, ${\displaystyle {\mathcal {A}}=\{U_{\alpha }|\alpha \in J\}}$  is an open covering of ${\displaystyle X}$  and the Lebesgue number ${\displaystyle \delta }$  does not exist. So, ${\displaystyle \forall \delta >0}$ , ${\displaystyle \exists A\subset X}$  with ${\displaystyle diam(A)<\delta }$  such that ${\displaystyle \neg \exists \beta \in J}$  where ${\displaystyle A\subset U_{\beta }}$ .

This allows us to make the following construction:

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

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

⋮

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

⋮

For all ${\displaystyle n\in \mathbb {Z} ^{+}}$ , ${\displaystyle A_{n}\neq \emptyset }$  since ${\displaystyle A_{n}\not \subset U_{\beta }}$ .

It is therefore possible to generate a sequence ${\displaystyle \{x_{n}\}}$  where ${\displaystyle x_{n}\in A_{n}}$  by axiom of choice. By sequential compactness, there exists a subsequence ${\displaystyle \{x_{n_{k}}\},k\in \mathbb {Z} ^{+}}$  that converges to ${\displaystyle x_{0}\in X}$ .

Using the fact that ${\displaystyle {\mathcal {A}}}$  is an open covering, ${\displaystyle \exists \alpha _{0}\in J}$  where ${\displaystyle x_{0}\in U_{\alpha _{0}}}$ . As ${\displaystyle U_{\alpha _{0}}}$  is open, ${\displaystyle \exists r>0}$  such that ${\displaystyle B_{d}(x_{0},r)\subset U_{\alpha _{0}}}$ . By definition of convergence, ${\displaystyle \exists L\in \mathbb {Z} ^{+}}$  such that ${\displaystyle x_{n_{p}}\in B_{d}\left(x_{0},{\frac {r}{2}}\right)}$  for all ${\displaystyle p\geq L}$ .

Furthermore, ${\displaystyle \exists M\in \mathbb {Z} ^{+}}$  where ${\displaystyle \delta _{M}={\frac {1}{K}}<{\frac {r}{2}}}$ . So, ${\displaystyle \forall z\in \mathbb {Z} ^{+}z\geq M\Rightarrow diam(A_{M})<{\frac {r}{2}}}$ .

Finally, let ${\displaystyle q\in \mathbb {Z} ^{+}}$  such that ${\displaystyle n_{q}\geq M}$  and ${\displaystyle q\geq L}$ . For all ${\displaystyle x'\in A_{n_{q}}}$ , notice that:

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

By the triangle inequality, ${\displaystyle d(x_{0},x')<r}$ , implying that ${\displaystyle A_{n_{q}}\subset U_{\alpha _{0}}}$  which is a contradiction.

References

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