The Lebesgue Number Lemma

Recall from the Bolzano Weierstrass Topological Spaces page that a topological space $X$ is said to be a Bolzano Weierstrass space if every infinite subset of $X$ has an accumulation point.

If we now consider a general metric space $(X, d)$ that is also a BW space then we can prove the following remarkable theorem which says that for any open cover $\mathcal F$ of $X$ there exists an $\epsilon > 0$ such that for all $x \in M$ there exists a set $U$ from the cover $\mathcal F$ for which the open ball centered at $x$ with radius $\epsilon$ is fully contained in $U$.

We now formally state and prove this remarkable result known as the Lebesgue Number Lemma.

Lemma (The Lebesgue Number Lemma): Let $(X, d)$ be a metric space that is also a BW space and let $\mathcal F$ be an open cover of $X$. Then there exists an $\epsilon > 0$ such that for all $x \in X$ there exists a $U \in \mathcal F$ such that $B(x, \epsilon) \subseteq U$.

• Proof: Let $X$ be a BW space and let $d : M \times M \to [0, \infty)$ be a metric defined on $X$. Let $\mathcal F$ be any open cover of $X$.

• Suppose instead that there exists an $x^* \in X$ such that for all $\epsilon > 0$ and for all $U \in \mathcal F$ we have that:

(1)

\begin{align} \quad B(x^*, \epsilon) \not \subseteq U \end{align}

• Then for all $n \in \mathbb{N}$ and for all $U \in \mathcal F$ there exists a point $x_n \in X$ such that

(2)

\begin{align} \quad B \left ( x_n, \frac{1}{n} \right ) \not \subseteq U \quad (*) \end{align}

• So, for every $n \in \mathbb{N}$ there exists a point, call it $x_n \in X$ such that $x_n \not \in U$ for all $U \in \mathcal F$. Let $A$ be the set of all of such points, i.e., $A = \{ x_1, x_2, ... \}$. We claim that $A$ is an infinite set.

• Suppose not, i.e., suppose that $A$ is a finite set, say $A = \{ x_1, x_2, ..., x_n \}$ for some $n \in \mathbb{N}$. Then we must have that $x_m = x_n$ for all $m \geq n$. But then for all $m \geq n$ we would have that $B \left (x_m, \frac{1}{m} \right )$ is not contained in any element $U$ of $\mathcal F$. Note that $B \left ( x_m, \frac{1}{m} \right )$ is a local basis of $x_m$ though, and by definition for all open neighbourhoods of $x_m$ in $X$ there should exist a local basis element that is fully contained in $U$. So we see that a contradiction has arisen. So indeed, $A = \{ x_1, x_2, ... \}$ is an infinite set.

• Since $X$ is a BW space, this infinite set $A$ has an accumulation point. Let $x \in X$ be this accumulation point. Since $\mathcal F$ covers $X$ we have that there exists a $U \in \mathcal F$ such that $x \in U$. Furthermore, there exists an $r > 0$ such that:

(3)

\begin{align} \quad x \in B(x, r) \subseteq U \end{align}

• Since $(X, d)$ is a metric space, $X$ is Hausdorff. So every open neighbourhood of the accumulation point $x$ of $A$ contains infinitely many points of $A$. In particular, the open neighbourhood $B \left ( x, \frac{r}{2} \right )$ of $x$ contains infinitely many points $A$. S

• So, there exists an $m \in \mathbb{N}$ with $\frac{1}{m} < \frac{r}{2}$ such that $x_m \in B \left ( x, \frac{r}{2} \right )$. If not, i.e., if for all $m \in \mathbb{N}$ with $\frac{1}{m} < \frac{r}{2}$ we have that $x_m \not \in B \left ( x, \frac{r}{2} \right )$ then $B \left ( x, \frac{r}{2} \right )$ contains at most finitely many points which is a contradiction.

• So $x_m$ is such that:

(4)

\begin{align} \quad x_m \in B \left ( x_m, \frac{1}{m} \right ) \subset B(x, r) \subset U \end{align}

• But this is a contradiction since from $(*)$, $x_m$ is such that $x_m \not \subseteq U$ for all $U \in \mathcal F$. Therefore the assumption that there exists a point such that $x^* \in X$ such that for all $\epsilon > 0$ and for all $U \in \mathcal F$, $B(x^*, \epsilon) \not \subseteq U$ is false.

• So, there exists an $\epsilon > 0$ such that for all $x \in X$ there exists a $U \in \mathcal F$ such that $B(x, \epsilon) \subseteq U$.

From the Lebesgue Number lemma, we know that such numbers $\epsilon$ are well-defined.

Definition: If $(X, d)$ is a metric space and BW space, and if $\mathcal F$ is an open cover of $X$ then a Lebesgue Number is a number $\epsilon > 0$ such that for all $x \in X$ there exists a $U \in \mathcal F$ such that $x \in B(x, \epsilon) \subseteq U$.