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.

Proof

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 $\delta$, then there exists $x_{0}\in X$ such that $Y\subseteq B_{\delta }(x_{0})$, where $B_{\delta }(x_{0})$ denotes the ball of radius $\delta$ centered at $x_{0}$ (namely, one can choose $x_{0}$ as any point in $Y$). 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

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

This allows us to make the following construction:

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

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

⋮

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

⋮

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

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

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

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

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

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

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

References

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