Number Smaller than Lebesgue Number is also Lebesgue Number
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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 $y \in \map {B_{\epsilon'} } x$.
\(\ds \map d {y, x}\) | \(<\) | \(\ds \epsilon'\) | Definition of Open Ball of Metric Space | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map d {y, x}\) | \(<\) | \(\ds \epsilon\) | as $\epsilon' < \epsilon$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds \map {B_\epsilon} x\) | Definition of Open Ball of Metric Space | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds \map U x\) | Definition of Subset: $\map {B_\epsilon} x \subseteq \map U x$ |
That is:
- $\map {B_{\epsilon'} } x \subseteq \map U x$
The result follows by definition of Lebesgue number.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $7.2$: Sequential compactness