Number Smaller than Lebesgue Number is also Lebesgue Number

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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.

$\blacksquare$


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