Proving that the infinite intersection of a sequence of closed subsets in a compact space is non-empty.

Question

Let $X$ be a compact space, and let $\{C_i\}_{i \in \mathbb{N}}$ be a collection of nonempty closed sets in $X$ satisfying $C_{i+1} \subset C_i $ for each $i \in \mathbb{N}$. Prove tthat $\bigcap_{i=0}^\infty C_i \neq \emptyset$.

I have tried to configure a proof of my own, however not been able to finish my question, so I maybe making a wrong assumption.

First I proved if $\bigcap_{i=0}^\infty C_i= \emptyset $ iff for any $\alpha \in C_i$ there exists $N$ such that for $ \alpha \notin C_n$ for all $n >N$. Secondly, if $\{X -C_i\}_{i \in \mathbb{N}}$ is an open covering, however not sure what to do from here.

Answer 1

You have taken the right path. By compactness $X$ is covered by a finite union of the set $X \setminus C_i$. This means $\cap_{i=1}^{n}C_i=\emptyset$. But $\cap_{i=1}^{n}C_i=C_n$ so $C_n$ is empty. This is a contradiction.

Answer 2

If $\bigcap_i C_i = \emptyset$, by de Morgan

$$X=X-(\bigcap_i C_i)=\bigcup_i (X-C_i)$$

so indeed $\{X-C_i, i \in \Bbb N\}$ is an open (!) cover of $X$.

Let $X-C_{i_1}, \ldots, X-C_{i_n}$ be a finite subcover.

The $C_i$ have the property that $\bigcap_{i=1}^N C_i = C_N$, because the $C_i$ decrease with increasing index, and the intersection is thus the one with the largest index. So if $N=\max\{i_1,\ldots, i_n\}$, any point $x \in C_N$ (which exists) lies in all $C_{i_1}, C_{i_2}, \ldots, C_{i_n}$ as well, so is not covered, contradiction. So the intersection cannot be empty.