Lebesgue's Number Lemma/Compact Space
Theorem
Let $M = \struct {X, d}$ be a metric space.
Let $M$ be compact.
Then there exists a Lebesgue number for every open cover of $M$.
Proof
Let $\UU$ be an open cover of $M$.
By definition of compact, there exists a finite subcover $\set {A_i}_{1 \le i \le n} \subseteq \UU$.
First, suppose some $A_i = X$.
Then, let $\epsilon = 1$.
For any $x \in X$:
- $\map {B_\epsilon} x \subseteq X = A_i \in \UU$
Therefore, $\epsilon$ is a Lebesgue number for $\UU$.
Now, suppose every $A_i \subsetneq X$.
Then, for every $1 \le i \le n$:
- $X \setminus A_i$
is non-empty.
By Distance in Pseudometric is Non-Negative:
- $\map d {x, y} \ge 0$
for every $x, y \in X$
Therefore, for every $x \in X$ and $1 \le i \le n$:
- $\set {\map d {x, y} : y \in X \setminus A_i}$
is bounded below.
Therefore, by the Greatest Lower Bound Property:
- $\ds \inf_{y \mathop \in X \setminus A_i} \map d {x, y}$
is well-defined.
For each $1 \le i \le n$, define $f_i : X \to \R_{\ge 0}$ as:
- $\ds \map {f_i} x = \inf_{y \mathop \in X \setminus A_i} \map d {x, y}$
We must show that every $f_i$ is continuous.
Let $\epsilon > 0$ and $a \in X$ be arbitrary.
Let $x \in X$ be arbitrary, satisfying:
- $\map d {x, a} < \dfrac \epsilon 2$
| \(\ds \forall y \in X \setminus A_i: \, \) | \(\ds \map d {x, y}\) | \(\le\) | \(\ds \map d {x, a} + \map d {a, y}\) | Metric Space Axiom $(\text M 2)$: Triangle Inequality | ||||||||||
| \(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \map d {a, y}\) | Premise | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {f_i} x\) | \(\le\) | \(\ds \frac \epsilon 2 + \map {f_i} a\) | |||||||||||
| \(\ds \) | \(<\) | \(\ds \epsilon + \map {f_i} a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {f_i} x - \map {f_i} a\) | \(<\) | \(\ds \epsilon\) | |||||||||||
| \(\ds \forall y \in X \setminus A_i: \, \) | \(\ds \map d {a, y}\) | \(\le\) | \(\ds \map d {a, x} + \map d {x, y}\) | Metric Space Axiom $(\text M 2)$: Triangle Inequality | ||||||||||
| \(\ds \) | \(=\) | \(\ds \map d {x, a} + \map d {x, y}\) | Metric Space Axiom $(\text M 3)$ | |||||||||||
| \(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \map d {x, y}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {f_i} a\) | \(\le\) | \(\ds \frac \epsilon 2 + \map {f_i} x\) | |||||||||||
| \(\ds \) | \(<\) | \(\ds \epsilon + \map {f_i} x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -\epsilon\) | \(<\) | \(\ds \map {f_i} x - \map {f_i} a\) |
Combining the above, we have:
| \(\ds -\epsilon\) | \(<\) | \(\, \ds \map {f_i} x - \map {f_i} a \, \) | \(\, \ds < \, \) | \(\ds \epsilon\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds \size {\map {f_i} x - \map {f_i} a} \, \) | \(\, \ds < \, \) | \(\ds \epsilon\) | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds \map {d_\R} {\map {f_i} x, \map {f_i} a} \, \) | \(\, \ds < \, \) | \(\ds \epsilon\) |
As $x \in X$ and $\epsilon > 0$ were arbitrary, it follows that $f_i$ is continuous at $a$ by definition.
But as $a \in X$ was also arbitrary, it follows that $f_i$ is continuous on $M$.
Define $f : X \to \R$ as:
- $\ds \map f x = \frac 1 n \sum_{i = 1}^n \map {f_i} x$
By the Sum Rule and Multiple Rule:
- $f$ is continuous on $M$.
By Continuous Image of Compact Space is Compact:
- $f \sqbrk X$ is compact
By Compact Subspace of Real Numbers is Closed and Bounded:
By Infimum of Bounded Below Set of Reals is in Closure and Closed Set Equals its Closure:
- $\epsilon_0 = \ds \inf_{x \in X} \map f x \in f \sqbrk X$
from which it follows that there is some $x_0 \in X$ such that:
- $\map f {x_0} = \epsilon_0$
By definition of cover, there exists some $1 \le i \le n$ such that:
- $x_0 \in A_i$
By definition of open set, there exists some $\epsilon > 0$ such that:
- $\map {B_\epsilon} {x_0} \subseteq A_i$
Therefore:
| \(\ds \forall y \in X \setminus A_i: \, \) | \(\ds y\) | \(\notin\) | \(\ds \map {B_\epsilon} {x_0}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map d {x_0, y}\) | \(\ge\) | \(\ds \epsilon\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {f_i} {x_0}\) | \(\ge\) | \(\ds \epsilon\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map f {x_0}\) | \(\ge\) | \(\ds \frac 1 n \map {f_i} {x_0}\) | |||||||||||
| \(\ds \) | \(\ge\) | \(\ds \frac \epsilon n\) | ||||||||||||
| \(\ds \) | \(>\) | \(\ds 0\) |
Therefore:
- $\epsilon_0 > 0$
It remains to show that $\epsilon_0$ is a Lebesgue number for $\UU$.
Let $x \in X$ be arbitrary.
We want to show that:
- $\map {B_{\epsilon_0}} x \subseteq A_i$
for some $1 \le i \le n$.
By definition of infimum:
- $\map f x \ge \epsilon_0$
Therefore:
- $\ds \sum_{i \mathop = 1}^n \map {f_i} x \ge n \epsilon_0$
It follows that:
- $\map {f_i} x \ge \epsilon_0$
for some $1 \le i \le n$.
By definition of infimum again:
- $\forall y \in X \setminus A_i: \map d {x, y} \ge \epsilon_0$
That is:
- $\forall y \in X: \map d {x, y} < \epsilon_0 \implies y \in A_i$
which is precisely:
- $\map {B_{\epsilon_0}} x \subseteq A_i$
$\blacksquare$
Source of Name
This entry was named for Henri Léon Lebesgue.