This is the Wikipedia proof of Cantor's intersection theorem:
$C_0\supseteq C_1\supseteq C_2…C_k\supseteq C_{k+1}$
so that this true $\bigcap_\limits{k}^{}C_k\neq\emptyset$
Assume, by way of contradiction, that $\bigcap_\limits{n}^{}C_k\neq\emptyset$.For each $n$, let $U_n=C_0\setminus C_n$ Since ${\displaystyle \bigcup U_{n}=C_{0}\setminus \bigcap C_{n}} $ and ${\displaystyle \bigcap C_{n}=\emptyset }$ $\bigcap C_{n}=\emptyset$ , thus ${\displaystyle \bigcup U_{n}=C_{0}} {\displaystyle \bigcup U_{n}=C_{0}}$.
Since $ C_{0}\subset S $ is compact and ${\displaystyle (U_{n})} $ is an open cover of it, we can extract a finite cover. Let ${\displaystyle U_{k}}$ be the largest set of this cover; then ${\displaystyle C_{0}}\subset {\displaystyle U_{k}} $. But then ${\displaystyle C_{k}=C_{0}\setminus U_{k}=\emptyset }$ , a contradiction.$\blacksquare$
I want to know how $U_k$ happens to be a cover of $C_0$ how is ${\displaystyle C_{0}}\subset {\displaystyle U_{k}} $ instead of ${\displaystyle C_{0}}= {\displaystyle U_{k}} $ Thanks for reading!
Best Answer
I'll give a more detailed version.
Suppose that $C_0 \supseteq C_1 \supseteq C_2 \supseteq \ldots C_{k} \ldots \supseteq C_{k+1} \ldots$, where all $C_k$ are compact non-empty (and thus closed, as we are in the reals).
Suppose for a contradiction that $\bigcap_n C_n = \emptyset$. The idea is to use that $C_0$ is compact, so we define an open cover of $C_0$ by setting $U_k = C_0 \setminus C_k$ for $k \ge 1$. Note that these are open in $C_0$ as $C_0 \setminus C_k = C_0 \cap (X \setminus C_k)$ is a relatively open subset of $C_0$ (using that all $C_k$ are closed so have open complement).
Also $U_1 \subseteq U_2 \subseteq U_3 \ldots U_k \subseteq U_{k+1} \ldots$, as the $C_k$ are decreasing.
Take $x \in C_0$. Then there is some $C_k$ such that $x \notin C_k$ (or else $x \in \bigcap_n C_n = \emptyset$), and so this $x \in U_k$ for that $k$.
This shows that the $U_n$ form an open cover of $C_0$, so finitely many $U_k$, say $U_{k_1}, U_{k_2},\ldots, U_{k_m}, k_1 < k_2 \ldots k_m$ cover $C_0$, so using the increasingness, we see hat $C_0 \subseteq U_{k_m}$. But take any $p \in C_{k_m}$ (by non-emptiness), then $p \in C_0$ and $p \notin U_{k_m}$, contradiction. So $\bigcap_n C_n \neq \emptyset$.