Theorem: Lebesgue's Number Lemma/Compact Space

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:
:$\map d {x, a} < \dfrac \epsilon 2$
{{begin-eqn}}
{{eqn | q = \forall y \in X \setminus A_i
      | l = \map d {x, y}
      | o = \le
      | r = \map d {x, a} + \map d {a, y}
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \map d {a, y}
      | c = Premise
}}
{{eqn | ll = \leadsto
      | l = \map {f_i} x
      | o = \le
      | r = \frac \epsilon 2 + \map {f_i} a
}}
{{eqn | o = <
      | r = \epsilon + \map {f_i} a
}}
{{eqn | ll = \leadsto
      | l = \map {f_i} x - \map {f_i} a
      | o = <
      | r = \epsilon
}}
{{eqn | q = \forall y \in X \setminus A_i
      | l = \map d {a, y}
      | o = \le
      | r = \map d {a, x} + \map d {x, y}
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | r = \map d {x, a} + \map d {x, y}
      | c = {{Metric-space-axiom|3}}
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \map d {x, y}
}}
{{eqn | ll = \leadsto
      | l = \map {f_i} a
      | o = \le
      | r = \frac \epsilon 2 + \map {f_i} x
}}
{{eqn | o = <
      | r = \epsilon + \map {f_i} x
}}
{{eqn | ll = \leadsto
      | l = -\epsilon
      | o = <
      | r = \map {f_i} x - \map {f_i} a
}}
{{end-eqn}}
Combining the above, we have:
{{begin-eqn}}
{{eqn | l = -\epsilon
      | o = <
      | m = \map {f_i} x - \map {f_i} a
      | mo = <
      | r = \epsilon
}}
{{eqn | ll = \leadsto
      | o =
      | m = \size {\map {f_i} x - \map {f_i} a}
      | mo = <
      | r = \epsilon
}}
{{eqn | ll = \leadsto
      | o =
      | m = \map {d_\R} {\map {f_i} x, \map {f_i} a}
      | mo = <
      | r = \epsilon
}}
{{end-eqn}}
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:
:$f \sqbrk X$ 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:
{{begin-eqn}}
{{eqn | q = \forall y \in X \setminus A_i
      | l = y
      | o = \notin
      | r = \map {B_\epsilon} {x_0}
}}
{{eqn | ll = \leadsto
      | l = \map d {x_0, y}
      | o = \ge
      | r = \epsilon
}}
{{eqn | ll = \leadsto
      | l = \map {f_i} {x_0}
      | o = \ge
      | r = \epsilon
}}
{{eqn | ll = \leadsto
      | l = \map f {x_0}
      | o = \ge
      | r = \frac 1 n \map {f_i} {x_0}
}}
{{eqn | o = \ge
      | r = \frac \epsilon n
}}
{{eqn | o = >
      | r = 0
}}
{{end-eqn}}
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:
:$\map {B_{\epsilon_0}} x \subseteq A_i$
{{qed}}
{{Namedfor|Henri Léon Lebesgue|cat = Lebesgue}}
Category:Lebesgue's Number Lemma
Category:Compact Topological Spaces
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