Lebesgue's number lemma

Template:Short description 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 $\delta >0$ such that every subset of $X$ having diameter less than $\delta$ 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 $(X,d)$ is compact and an open cover of $X$ is given, then the cover admits some Lebesgue's number $\delta >0$.

The notion of Lebesgue's numbers 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's 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 $f:X\to ℝ$ 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's 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 $f(x_0)\ge \delta$ 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

Suppose for contradiction 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 $\mathrm{diam}(A)<\delta$ such that there does not exist $\beta \in J$ with $A\subset U_{\beta }$.

This enables us to perform the following construction:

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

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

$⋮$

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

$⋮$

Note that $A_n\ne \mathrm{\varnothing }$ for all $n\in {\mathbb{Z}}^{+}$, since $A_n\subset ̸U_{\beta }$. 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 ${\delta }_M=\frac{1}{M}<\frac{r}{2}$. Hence for all $z\in {\mathbb{Z}}_{>0}$, we have $M\le z$ implies $\mathrm{diam}(A_M)<\frac{r}{2}$.

Finally, define $q\in {\mathbb{Z}}_{>0}$ such that $n_q\ge M$ and $q\ge L$. For all $x^{\prime }\in A_{n_q}$, notice that:

• $d(x_{n_q},x^{\prime })\le \mathrm{diam}(A_{n_q})<\frac{r}{2}$, because $n_q\ge M$.
• $d(x_{n_q},x_0)<\frac{r}{2}$, because $q\ge L$ entails $x_{n_q}\in B_{r/2}\left(x_0\right)$.

Hence $d(x_0,x^{\prime })<r$ by the triangle inequality, which implies that $A_{n_q}\subset U_{{\alpha }_0}$. This yields the desired contradiction.

References

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