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Question: Let $\{O_{\lambda}:\lambda\in\Lambda\}$ be an open cover of the metric space $(X,d).$ If $X',$ the set of all accumulation points of $X$ is compact then show that $\exists~\delta>0$ such that for $x\in X'$ there exists $\lambda_x\in\Lambda$ such that $B(x,\delta)\subset O_{\lambda_x}.$

Unfortunately I could not solve it.

Please help.

My try: I tried to think of a solution using the following result: If $A$ is a compact subset of a metric space $(X,d)$ then for any open set $U$ containing $A$, there exists a $\delta>0$ such that $B(A,\delta) \subset U.$

Jave
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  • Idea: Try proving it by contradiction. Namely suppose the space is compact but it doesn't satisfy that property. Then, for any $n$, there is $x$ such that the ball of radius $1/n$ around $x$ is not contained in any set in the cover. You use this to come up with a sequence, which by compactness should have a convergent subsequence. Then the limit point must be contained in some set in the cover. Try to derive a contradiction from this. – Anguepa Nov 04 '18 at 17:41
  • It is otherwise clear but I cannot reach to any contradiction. Please help – Jave Nov 04 '18 at 18:05
  • Consider the limit Ive described $x$. It must belong in some set in the cover $O$. In fact since $O$ is open we have that $B(x, \varepsilon) \subseteq O$ for some positive $\varepsilon$. Fix $n$ such that $1/n < \varepsilon/2$. Show using the triangular inequality that, for any $y$ with $d(x,y)<\varepsilon/2$, $B(y, \varepsilon/2) \subseteq B(x, \varepsilon)$. In particular $B(y,1/n)\subseteq B(x, \varepsilon) \subseteq U$. Use this to derive a contradiction. – Anguepa Nov 04 '18 at 18:14

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