Is there anything wrong with my proof of Cantor's intersection theorem?

Question

Let a compact topological space $(X, \mathcal{T})$ contain a shrinking sequence of closed, nonempty sets $\{F_n\}_{n=1}^\infty$. Prove that $\cap_{n=1}^\infty F_n$ is nonempty.

For each $F_n$, let $\{U_k\}_n$ be the most refined open covering of the sequence of refinements) over $F_n$. Because $F_n$ is shrinking, we have the chain $\{U_k\}_1 > \{U_k\}_2 > ... $. Now each open covering of a closed set must contain the closed set and also the neighborhoods around each of the limit points (in order to be open), making them strictly greater than the open set in question. Therefore $\{V_k\} = \lim_{n\to\infty} \{U_k\}_n$ is non-empty and is an open cover of $X = \cap_{n=1}^\infty F_n$. Because $X$ is closed (infinite intersection of closed sets) and therefore compact, $X$ has a finite subcover, which would be impossible if $X = \{\}$

Answer 1

It looks to me like you are trying to do something like this:

Suppose $F_n\neq \emptyset$ but $\bigcap_nF_n=\emptyset$.

Then, $\mathscr A=\{F_n^c\}_n$ is an open cover of $X$:

let $x\in X$. If $x\notin F_n$ for any integer $n$, then $x\in F^c_1.$ If $x\in F_n$ for some integer $n$, let $j$ be the greatest integer such that $x\in F_j$. Thus is possible since $\bigcap_nF_n=\emptyset$. Then, $x\in F^c_{j+1}$.

But $\mathscr A$ has no finite subcover.

Answer 2

Why use all those excessively complex limit things?
Assume $\cap_n F_n$ empty.
Thus X = $\cup_n(X - F_n)$
Take a finite subcover C.
Thus exists n with $F_n$ = $\cap${ X - K : K in C }.
So oh my, $F_n$ is empty.

Exercise.
Give a real quick example why compact is necessary.

Answer 3

Let $\{\overline{F_n}\}_{n=0}^\infty$ be the open cover, and let $C$ be a finite subcover of it. Look carefully at the requirement for C: $\forall n\in \mathbb{N}\,\exists p \in X$ such that $p \in F_n$ but $p \not\in F_{n+1}$. Therefore $\forall n\in \mathbb{N}(\overline{F_n} \in C)$, contradicting finiteness. Therefore there exists an open cover without a finite subcover, a contradiction.