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I can't imagine this is a new question, but I was unable to find what I was looking for. I have seen it stated that if $X$ is a topological space, and $(A_{k})_{k \in \mathbb{N}}$ is a non-increasing sequence of non-empty compact sets, then $\cap_{k = 1}^{\infty} A_{k}$ is non-empty and compact. Is it true that for a general decreasing sequence of sets, i.e. if we drop the compactness assumption, that $\cap_{k = 1}^{\infty} A_{k} \neq \emptyset$? That is, if I'm not looking for the intersection to be compact, can I still get that it's non-empty? Thanks.

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No, e.g. $\bigcap [n,\infty[$ (which is empty), so that even closedness is not enough.

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  • $\begingroup$ IS there a limitation where it would work, e.g. I'm dealing with something "bounded" (I use quotations because I'm not sure if I am necessarily referring to bounded in the sense of metrics)? $\endgroup$ – AJY Feb 1 '15 at 4:58
  • $\begingroup$ Are you looking for a sequence of non-compact sets with nonempty limit? In that case simply take $\bigcap(\{0\}\cup [n,\infty[)$. $\endgroup$ – Alp Uzman Feb 1 '15 at 5:01
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Well, take $X=\mathbb R$ and $A_k = \left]0,k^{-1}\right[$ the sequence of open intervals ranging from $0$ to $k^{-1}$. The intersection of the $A_k$'s is empty.

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