0
$\begingroup$

Let $\mathcal{V}$ be a finite open cover of a compact metric space $(X, d)$ and $\delta$ is Lebesgue number for open cover $\mathcal{V}$. Let $A\subseteq X$ be a non-empty set.

Is there $0<\epsilon<\delta$ such that if $d(a, b)<\epsilon$,for all $a,b\in A$, then $A\subseteq U$ for some $U\in \mathcal{V}$?

Please help me to know it.

Thanks a lot

$\endgroup$
1
$\begingroup$

Yes, by definition of the Lebesgue number your statement holds for all such $\varepsilon$ with $0< \varepsilon < \delta$:

If $A$ is such that $d(a,b) < \varepsilon$ for all $a,b \in A$ then

$$\operatorname{diam}(A) = \sup\{d(a,b): a,b \in A\} \le \varepsilon$$

as $\varepsilon$ is an upperbound, and the $\sup$ is the least upperbound.

So $\operatorname{diam}(A) < \delta$ and by said definition, $A \subseteq U$ for some $U \in \mathcal{U}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.