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


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 $U \in \mathcal{U}$.


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