Compact and $T_2$ - space question Give $X$ is compact $T_2$-space and $(F_n)$ is the sequence of closed sets that $F_n \supset F_{n+1}$.
I have proved that $\displaystyle F=\bigcap_{n=1}^{+\infty}F_n \neq \varnothing$. So, i feeling that if $G$ is open contain $F$ then exists $n_0 \in \mathbb{N}$ that $F_{n_0} \subset G$. I have tried to use contracdiction to prove it but i failed. I wondering that my feeling is right or not?
 A: HINT: For each $n\in\Bbb Z^+$ let $H_n=F_n\setminus G$ and consider $\bigcap_{n\ge 1}H_n$.
A: From the hint above of Prof. Brian M. Scott, I have proved this problem as follows.
Suppose that $F_n \not\subset G, \forall n \in \mathbb{N}$.
Fix $H_n=F_n \setminus G \neq \varnothing$. We claim that $H_n$ is closed and $H_n \supset H_{n+1}$ for all $n\in \mathbb{N}$.
Since $H_n \supset H_{n+1}, \forall n \in \mathbb{N}$ then $\displaystyle \bigcap_{j \in J}H_j \neq \varnothing$ for all $J$ is finite in $\mathbb{N}$. Moreover, since $X$ is compact space then  $\displaystyle \bigcap_{n=1}^{+\infty}H_n \neq \varnothing$ (*).
Other side, $\displaystyle \bigcap_{n=1}^{+\infty}H_n= \bigcap_{n=1}^{+\infty}(F_n\setminus G)= F\setminus G = \varnothing$ (contradict with (*)).
So, there exists $n_0$ that $F_{n_0} \subset G$.
A: As a mild variation: $X\setminus G$ is closed and $X$ is compact, so $X\setminus G$ is compact. The sets $X\setminus F_n$ cover $X\setminus G$ (since $F\subset G$), and form an increasing sequence. Therefore, $X\setminus G \subset X\setminus F_{n_0}$ for some $n_0$. The result follows by taking complements.
