Frechet Intersection Theorem

Question

This is a question from Royden Real Analysis 4th Editioin.

Let {$F_n$}$_{n=1}^\infty$ be a descending countable collection of nonempty closed subsets of a compact metric space $X$. Show that $\bigcap _{n=1}^{\infty}F_n \neq \phi$.

My attempt: $\\$ $X$ compact $\rightarrow X$ complete and {$F_n$}$_{n=1}^\infty$ is a descending sequence of nonempty cvlosed sets so by Cantor INtersection Thm there is $x_0 \in X$ such that $x_0 \in$ {$F_n$}$_{n=1}^\infty$.

I don't think this is right because then this theorem would just be a weaker version of Cantor's, but I can't find my error. Could someone please explain

Answer 1

By taking a connected subsequence, we can suppose WLOG that $F_n$'s are connected. Let $\text{diam}(F_n)=\sup\{d(x,y)\mid x,y\in F_n\}\geq 0$. Since $F_n$ are decreasing, then so is $\text{diam}(F_n)$. Therefore it converge. If $$\lim_{n\to \infty }\text{diam}(F_n)>0,$$ then the result is obvious. If $$\lim_{n\to \infty }\text{diam}(F_n)=0,$$ then it's Cantor Intersection Theorem, and the claim follow.

Answer 2

Take $x_n \in F_n$. Since $X$ is compact there is a convergent subsequence. By renumbering we can just assume that $x_n \to x$.

Since the $F_n$ are closed and nested, we see that $x_m \in F_n$ for all $m \ge n$, hence $x \in F_n$ for all $n$.

Hence $x \in \cap_n F_n$.