Compactess of the sum closed balls whose centers is a compact set.

Question

Let $(\mathcal{X},d) $ be a compact and metric space and let $A \subset \mathcal{X}$ be a compact subset and let $\alpha >0 $. Consider the set $$ C(A, \alpha) = \{ x \in \mathcal{X }\, | \, d(x,A) \leq \alpha\} $$ How to prove or disprove that $C(A, \alpha) $ is compact?

Thanks in advance.

José Carlos Santos · Accepted Answer · 2018-08-02T20:38:40.663

1

The set $C(A,\alpha)$ is closed because the function$$\begin{array}{rccc}\Delta\colon&\mathcal{X}&\longrightarrow&\mathbb R\\&a&\mapsto&d(a,A)\end{array}$$is continuous and $C(A,\alpha)=\Delta^{-1}\bigl([0,\alpha]\bigr)$.

Now, use the fact that a closed subset of a compact metric space is always compact.

edited Aug 02 '18 at 20:38

answered Aug 02 '18 at 16:41

José Carlos Santos

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Dear @José Carlos Santos ,Why $C(A,\alpha)$ is closed?. So far I know $C(A,\alpha)$ is an uncontable sum of closed balls. But any uncontable sum of closed balls not necessary it closed. – jaogye Aug 02 '18 at 17:39
@jaogye I've edited my answer. I hope that everything is clear now. – José Carlos Santos Aug 02 '18 at 20:39
Dear @José Carlos Santos many thanks for answer. – jaogye Aug 03 '18 at 06:58
@jaogye I'm glad I could help. – José Carlos Santos Aug 03 '18 at 07:30

Compactess of the sum closed balls whose centers is a compact set.

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