Lebesgue's number lemma

Short description: Given a cover of a compact metric space, all small subsets are subset of some cover set

In topology, Lebesgue's number lemma, named after Henri Lebesgue, 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 Lebesgue number of this cover. The notion of a Lebesgue number itself is useful in other applications as well.

Proof

Direct Proof

Let ${𝒰}$ be an open cover of $X$. Since $X$ is compact we can extract a finite subcover $\{A_1,\dots ,A_n\}\subseteq {𝒰}$. If any one of the $A_i$'s equals $X$ then any $\delta >0$ will serve as a Lebesgue number. Otherwise for each $i\in \{1,\dots ,n\}$, let $C_i:=X\setminus A_i$, note that $C_i$ is not empty, and define a function $f:X\to \mathbb{ℝ}$ 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 Lebesgue 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 $f(x_0)\ge \delta$ there must exist at least one $i$ such that $d(x_0,C_i)\ge \delta$. 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, ${𝒜}=\{U_{\alpha }|\alpha \in J\}$ is an open covering of $X$ and the Lebesgue number $\delta$ does not exist. So, $\mathrm{\forall }\delta >0$, $\mathrm{\exists }A\subset X$ with $diam(A)<\delta$ such that $\mathrm{\neg }\mathrm{\exists }\beta \in J$ where $A\subset U_{\beta }$.

This allows us to make the following construction:

${\delta }_1=1$, $\mathrm{\exists }{A}_1\subset X$ where $(diam(A_1)<{\delta }_1)$ and $\mathrm{\neg }\mathrm{\exists }\beta (A_1\subset U_{\beta })$

${\delta }_2=\frac{1}{2}$, $\mathrm{\exists }{A}_2\subset X$ where $(diam(A_2)<{\delta }_2)$ and $\mathrm{\neg }\mathrm{\exists }\beta (A_2\subset U_{\beta })$

⋮

${\delta }_k=\frac{1}{k}$, $\mathrm{\exists }{A}_k\subset X$ where $(diam(A_k)<{\delta }_k)$ and $\mathrm{\neg }\mathrm{\exists }\beta (A_k\subset U_{\beta })$

⋮

For all $n\in {\mathbb{\mathbb{Z}}}^{+}$, $A_n\ne \mathrm{\varnothing }$ since $A_n\subset ̸U_{\beta }$.

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 $\{x_{n_k}\},k\in {\mathbb{\mathbb{Z}}}^{+}$ that converges to $x_0\in X$.

Using the fact that ${𝒜}$ is an open covering, $\mathrm{\exists }{\alpha }_0\in J$ where $x_0\in U_{{\alpha }_0}$. As $U_{{\alpha }_0}$ is open, $\mathrm{\exists }r>0$ such that $B_d(x_0,r)\subset U_{{\alpha }_0}$. By definition of convergence, $\mathrm{\exists }L\in {\mathbb{\mathbb{Z}}}^{+}$ such that $x_{n_p}\in B_d(x_0,\frac{r}{2})$ for all $p\ge L$.

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

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

• $d(x_{n_q},x^{\prime })\le diam(A_{n_q})<\frac{r}{2}$ because $n_q\ge M$.
• $d(x_{n_q},x_0)<\frac{r}{2}$ because $q\ge L$ which means $x_{n_q}\in B_d(x_0,\frac{r}{2})$.

By the triangle inequality, $d(x_0,x^{\prime })<r$, 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, https://archive.org/details/topologyfirstcou00munk_0/page/179