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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.


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One thought on “The Lebesgue Number Lemma and uniform continuity

  1. 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?

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