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Let $(X,d)$ be a metric space and consider an increasing sequence $A_n$ of its subsets such that $A = \bigcup_n A_n$ is compact. Can it happen that $A\setminus A_n$ is compact for all finite $n$?

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No. Even if we don't require that the $A_n$'s are increasing.

Assume otherwise. Since $X$ is a metric space, compact sets are closed. Note that $\{A\setminus A_n\mid n\in\Bbb N\}$ has the finite intersection property. Therefore the intersection of all of them must be non-empty. But this is impossible since $\bigcap (A\setminus A_n)=A\setminus\bigcup A_n=\varnothing$.

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You forget the case that $A = A_1 \cup \cdots \cup A_n$ for some $n$. Note that the empty set is compact. – Omnomnomnom Jul 7 '14 at 16:17
This seems to miss the intention of the question. I expect Ilya to be able and handle the case when the sequence is eventually constant. – Asaf Karagila Jul 7 '14 at 16:24
@AsafKaragila: thanks, though I expected myself to be able to handle also the case in the question, granted that I've applied finite intersection property a couple of times before facing that question. Too much work... – Ilya Jul 7 '14 at 20:02

If $A-A_n$ is compact then it is closed and so $A_n$ is a relative open subset of the space $A$. Now we have that $A_n$ is an open cover of $A$ (considered in the space $A$) if this is compact we have $A=A_n$ for some $n$ since the sets are increasing. If the sets are non increasing we still get that $A$ is a finite sum.

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