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 be a compact metric space and let be an open cover of . Then there exists a real number such that every open ball of radius is contained in some . *

*Proof:* First, remark that if any refinement of the cover satisfies this property, then also satisfies this property; thus, since is compact, we may replace the cover by a finite cover by open balls .

Define on by

Then is continous and its support is . Let . Then is continous, and because covers . Since is compact, attains its minimum, which is ; we call it . Now, if , the statement means precisely that for some , , so the ball of radius around is contained in .

**Theorem 2** * Let be a continous function on the compact metric space . Then is uniformly continous. *

*Proof:* Let . For each , let

Then is an open cover of . Let be a Lebesgue Number for the cover . Then, if are such that , there exists a such that ; therefore, since , we have

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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?