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Let $(X,\varrho)$ be a metric space and $K\subset X$ compact.

Then, for every $\,\varepsilon > 0$, $\,K$ can be covered with a finite number of balls of radius $\varepsilon$. Show that the reciprocal is not true.

My approach. Since $K$ is compact, every open cover of $K$ contains a finite sub-cover. But every one of the open sets of the finite sub-cover can written as a union of open balls with radius $\varepsilon$.

Jeybe
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1 Answers1

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Let $K$ be compact, then $$ \{B(x,\varepsilon) : x\in K\} $$ is an open cover of $K$, and hence it has a finite subcover: $$ K\subset B(x_1,\varepsilon)\cup\cdots\cup B(x_n,\varepsilon). $$ Note however, that every non-compact subset of this compact set $K$ has the same property, as it can also be covered by finitely many such balls.

In particular, every bounded (and hence not necessarily compact) subset of $\mathbb R$ can be covered by finitely many such balls.