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

 $\displaystyle d \left({y, x}\right)$ $<$ $\displaystyle \epsilon'$ Definition of Open Ball of Metric Space $\displaystyle \implies \ \$ $\displaystyle d \left({y, x}\right)$ $<$ $\displaystyle \epsilon$ as $\epsilon' < \epsilon$ $\displaystyle \implies \ \$ $\displaystyle y$ $\in$ $\displaystyle B_\epsilon \left({x}\right)$ Definition of Open Ball of Metric Space $\displaystyle \implies \ \$ $\displaystyle y$ $\in$ $\displaystyle U \left({x}\right)$ Definition of Subset: $B_\epsilon \left({x}\right) \subseteq U \left({x}\right)$

That is:

$B_{\epsilon'} \left({x}\right) \subseteq U \left({x}\right)$

The result follows by definition of Lebesgue number.

$\blacksquare$