Home » Topology » The Lebesgue Number Lemma and uniform continuity

# The Lebesgue Number Lemma and uniform continuity

In this post, I’ll prove the Lebesgue Number Lemma and use it to prove that a continous function on a compact metric space is uniformly continuous.

Lemma 1 (Lebesgue Number Lemma). Let ${(X,d)}$ be a compact metric space and let ${\{V_i\}_{i \in I}}$ be an open cover of ${X}$. Then there exists a real number ${\delta > 0}$ such that every open ball of radius ${\delta}$ is contained in some ${V_i}$.

Proof: First, remark that if any refinement of the cover ${\{V_i\}_{i \in I}}$ satisfies this property, then ${\{V_i\}}$ also satisfies this property; thus, since ${X}$ is compact, we may replace the cover ${\{V_i\}}$ by a finite cover by open balls ${\{B(x_i, r_i)\}_{i=1}^n}$.

Define ${f_i}$ on ${X}$ by

$\displaystyle f_i(x) = \min(0, r_i - d(x, x_i)).$

Then ${f_i}$ is continous and its support is ${B(x_i, r_i)}$. Let ${f= \min(f_1, \dots, f_n)}$. Then ${f}$ is continous, and ${f>0}$ because ${\{B(x_i, r_i)\}}$ covers ${X}$. Since ${X}$ is compact, ${f}$ attains its minimum, which is ${>0}$; we call it ${\delta}$. Now, if ${x \in X}$, the statement ${f(x)\leq \delta}$ means precisely that for some ${i}$, ${d(x, x_i) \leq r_i - \delta}$, so the ball of radius ${\delta}$ around ${x}$ is contained in ${B(x_i, r_i)}$. $\Box$

Theorem 2 Let ${f}$ be a continous function on the compact metric space ${(X, d)}$. Then ${f}$ is uniformly continous.

Proof: Let ${\epsilon>0}$. For each ${w \in X}$, let

$\displaystyle V_w = \{y \in X : |f(w)-f(y)|<\epsilon/2\}.$

Then ${\{V_w\}_{w \in X}}$ is an open cover of ${X}$. Let ${\delta}$ be a Lebesgue Number for the cover ${\{V_w\}}$. Then, if ${x, y\in X}$ are such that ${d(x, y)< \delta}$, there exists a ${w \in X}$ such that ${B(x, \delta) \subseteq V_w}$; therefore, since ${x, y \in V_w}$, we have

$\displaystyle |f(x)-f(y)|\leq |f(x)-f(w)| + |f(w)-f(y)| < \epsilon.$

$\Box$

## One thought on “The Lebesgue Number Lemma and uniform continuity”

1. Patrick Da Silva says:

In Lemma $1$, $f_i = \min(0, r_i – d(x,x_i)$ should be $f_i = \max(0,r_i – d(x,x_i)$, $f = \min (f_1, \dots, f_n)$ should be $f = \max (f_1, \dots, f_n)$, and the statement $f(x) \le \delta$ should be replaced by $f(x) \ge \delta$, otherwise it doesn’t make sense. Do you agree?