Open Cover may not have Lebesgue Number

From ProofWiki
Jump to: navigation, search

Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Let $\mathcal C$ be an open cover of $M$.


Then it may not necessarily be the case that $\mathcal C$ has a Lebesgue number.


Proof

Let $M := \left({0 \,.\,.\, 1}\right) \subseteq \R$ under the Euclidean metric.

Let $\mathcal C := \left\{{\left({\dfrac 1 n \,.\,.\, 1}\right): n \ge 2}\right\}$.

For any $\epsilon \in \R_{>0}$, take $n > \dfrac 1 \epsilon$ and $x = \dfrac 1 n$.

There is no $\left({\dfrac 1 m \,.\,.\, 1}\right)$ such that $B_\epsilon \left({x}\right) \subseteq \left({\dfrac 1 m \,.\,.\, 1}\right)$, since $B_\epsilon \left({x}\right) = B_\epsilon \left({\dfrac 1 n}\right) = \left({0 \,.\,.\, \dfrac 1 n + \epsilon}\right)$.

$\blacksquare$


Sources