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Let $(M,d)$ be a compact metric space. Suppose that $(F_n)$ is a decreasing sequence of nonempty closed sets in $M$, and that $\bigcap_{n=1}^\infty F_n$ is contained in some open set $G$. Then $F_n \subset G$ for all but finitely many $n$.

I know that $\bigcap_{n=1}^\infty F_n \neq \emptyset$, but I'm having trouble proving the above statement.

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Suppose not. Then there is an infinite sequence of $F_n$ such that all the $F_n$ aren't contained inside $G$. Since $G$ is open, each $F_n \setminus G$ is also closed; each is also nonempty by assumption. Consider the intersection of those.

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  • $\begingroup$ I know the intersection of those $F_n \setminus G$ is closed, but I'm not sure how to proceed. $\endgroup$ – user286129 Nov 2 '15 at 7:53
  • $\begingroup$ Is it true that $\bigcap_{n=1}^\infty (F_n\setminus G) = (\bigcap_{n=1}^\infty F_n) \setminus G$? $\endgroup$ – user286129 Nov 2 '15 at 8:20
  • $\begingroup$ @user286129 . YES...Let $ G^c$ be the complement of $G$.Both sides are equal to $\cap_n(F_n\cap G^c)$. $\endgroup$ – DanielWainfleet Nov 2 '15 at 8:32
  • $\begingroup$ Note that the closed sets are compact, nested intersections of non-empty compact sets are non-empy.. $\endgroup$ – Henno Brandsma Nov 2 '15 at 8:46

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