Sequence of compact sets

Question

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 $n$?

Answer 1

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.

Answer 2

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 $\bigcap (A\setminus A_n)=A\setminus\bigcup A_n=\varnothing$.