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If $(X,d)$ be a compact metric space then, arbitrary intersection of compact subsets is compact.

Is it true if $(X,d)$ is a metric space, but not compact?

There is similar type of questions already have been asked before, but those are from real-analysis where any closed and bounded set is compact. But in metric space closed and bounded set may not be compact. Hence this question is different from previously asked questions.

Tamas Kanti Garai
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  • You may find answer here: https://math.stackexchange.com/questions/1447506/arbitrary-intersection-of-compact-sets-is-compact – Aniruddha Deshmukh Nov 25 '18 at 06:35
  • If you are taking arbitrary intersections of compact sets, that will be a closed set (because each of the compact sets are closed sets and arbitrary intersection of closed sets is closed). As the intersection is a subset of a compact set, it's a compact set. – John_Wick Nov 25 '18 at 06:37

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Let $(X,d)$ be a metric space and $\{C_i\}_{i\in I}$ be some collection of compact subsets of $(X,d)$ . Let $\alpha\in I$ be fixed and consider the collection $\{C_i\cap C_{\alpha}\}_{i\in I}$ of subsets of $C_{\alpha}$ . Note that for each $i\in I$ we have $C_i\cap C_{\alpha}$ is a closed subset of the compact set $C_{\alpha}$.Therefore $\bigcap_{i\in I} C_i=\bigcap_{i\in I} (C_i\cap C_{\alpha})$ is also a closed subset of $C_{\alpha}$ being arbitrary intersection of closed subsets. But closed subset of compact set is compact , hence $\bigcap_{i\in I} C_i$ is compact subset of $(X,d)$ . Note that compactness is not relative property.

Sumanta
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