Number Smaller than Lebesgue Number is also Lebesgue Number

Theorem
Let $M = \left({A, d}\right)$ 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:
 * $\forall x \in A: \exists U \left({x}\right) \in \mathcal U: B_\epsilon \left({x}\right) \subseteq U \left({x}\right)$

where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$ in $M$.

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

Let $y \in B_{\epsilon'} \left({x}\right)$.

That is:
 * $B_{\epsilon'} \left({x}\right) \subseteq U \left({x}\right)$

The result follows by definition of Lebesgue number.