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 $$(X, d)$$ is compact and an open cover of $$X$$ is given, then there exists a number $$\delta > 0$$ such that every subset of $$X$$ having diameter less than $$\delta$$ is contained in some member of the cover.

Such a number $$\delta$$ is called a Delta number of this cover. The notion of a Delta number itself is useful in other applications as well.

Direct Proof
Let $$\mathcal U$$ be an open cover of $$X$$. Since $$X$$ is compact we can extract a finite subcover $$\{A_1, \dots, A_n\} \subseteq \mathcal U$$. If any one of the $$A_i$$'s equals $$X$$ then any $$ \delta > 0 $$ will serve as a Delta number. Otherwise for each $$i \in \{1, \dots, n\}$$, let $$C_i := X \smallsetminus A_i$$, note that $$C_i$$ is not empty, and define a function $$f : X \rightarrow \mathbb R$$ by


 * $$f(x) := \frac{1}{n} \sum_{i=1}^n d(x,C_i). $$

Since $$f$$ is continuous on a compact set, it attains a minimum $$\delta$$. The key observation is that, since every $$x$$ is contained in some $$A_i$$, the extreme value theorem shows $$\delta > 0$$. Now we can verify that this $$\delta$$ is the desired Delta number. If $$Y$$ is a subset of $$X$$ of diameter less than $$\delta$$, choose $$x_0$$ as any point in $$Y$$, then by definition of diameter, $$Y\subseteq B_\delta(x_0)$$, where $$B_\delta(x_0)$$ denotes the ball of radius $$\delta$$ centered at $$x_0$$. Since $$f(x_0)\geq \delta$$ there must exist at least one $$i$$ such that $$d(x_0,C_i)\geq \delta$$. But this means that $$B_\delta(x_0)\subseteq A_i$$ and so, in particular, $$Y\subseteq A_i$$.

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

This enables us to perform the following construction:

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

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

$$\vdots$$

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

$$\vdots$$

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

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

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

Finally, define $$q \in \mathbb{Z}_{> 0}$$ such that $$n_{q} \geq M$$ and $$q \geq L$$. For all $$x' \in A_{n_{q}}$$, notice that:
 * $$ d(x_{n_{q}},x') \leq \operatorname{diam} (A_{n_{q}})<\frac{r}{2}$$, because $$n_{q} \geq M$$.
 * $$d(x_{n_{q}},x_{0})<\frac{r}{2}$$, because $$q \geq L$$ entails $$x_{n_{q}} \in B_{r/2}\left(x_{0}\right)$$.

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