# Metric space compactness implies sequential compactness - unclear details

I am following through some proofs related to metric spaces - complete, compact and what not.

The book this proposition is taken from is from Royden and Fitzpatrick's Real Analysis.

Here is a proposition:

Proposition: If $X$ is a compact metric space, then it is sequentially compact.

Proof: Let $\{{x_n}\}$ be a sequence in $X$. For each $n$ (integer) let ${F_n}$ be the closure of the set $\{x_k |\ k \ge n\}$. Then $F_n$ is a descending (that is every set is a subset of the ones with smaller indices) sequence of non-empty closed sets. According to the Cantor Intersection Theorem, $\bigcap_{n=0}^{\infty}{F_n}$ is not empty.

The proof goes on from there, but this was the part that was bothering me.

The Cantor Intersection Theorem: Let $X$ be a metric space. Then $X$ is complete iff whenever ${F_n}$ is a contracting sequence of nonempty closed subsets of X, then $\bigcap_{n=0}^{\infty}{F_n} = \{x\}$ for some $x \in X$. Here a contracting sequence is such that $F_n \subseteq F_m$ if $m \leq n$, and also $\lim_{n\to0}{diam(F_n)} = 0$.

Cantor's Intersection Theorem is stated for complete spaces. Yet we use it here on a compact space without having proven that compactness implies completeness and totally boundedness ( in fact this proposition is part of a circular proposition of equivalent statements: that completeness + totally boundednes <=> compactness <=> sequential compactness) and so far it was proven that completeness + totally boundedness => compactness. Now this proposition will be used to prove compactness => sequential compactness. As you see we use an argument which I deem lacks a clear explanation.

Here is my take on trying to prove that this intersection is nonempty. We have shown that a space is compact iff every collection of closed sets with the finite intersection property has a nonempty intersection. From here we directly can show that if we take a finite subcollection of the indexed $F_n$, then the intersection will be $F_{n_{max}}$, where $n_{max}$ is the largest index in this finite subcollection, and hence will be nonempty. And since this finite subcollection was arbitrary, then for all finite subcollections this holds and then the whole intersection must be nonempty.

As I am writing this I realize that I am pretty sure of the correctness of my argument, but still would like to know if I am missing something in regards to the books argument with the Cantor Intersection Theorem.

If anyone has an idea if their argument is valid, can you please elaborate?

They mean another Cantor intersection theorem: if $F_n$ is a decreasing sequence of non-empty closed sets in a compact space $X$, $\cap_n F_n \neq \emptyset$. (If the intersection were empty, the complements would form an open cover of $X$ without a finite subcover).