Number Smaller than Lebesgue Number is also Lebesgue Number

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $\epsilon \in \R_{>0}$ be a Lebesgue number for $M$.

Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$.

Then $\epsilon'$ is also a Lebesgue number for $M$.

Proof
By hypothesis, let $\epsilon \in \R_{>0}$ be a Lebesgue number for $M$.

Then by definition:
 * $\forall x \in A: \exists \map U x \in \UU: \map {B_\epsilon} x \subseteq \map U x$

where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.

Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$.

Let $y \in \map {B_{\epsilon'} } x$.

That is:
 * $\map {B_{\epsilon'} } x \subseteq \map U x$

The result follows by definition of Lebesgue number.