Let $(X, \tau)$ be a topological space and let $2^{X}:=\mathcal{C}_X$ the family of closed sets in $X$. Prove if $(X, \tau)$ is compact then, $(2^{X}, \tau_{2^{X}})$ is compact, where $\tau_{2^{X}}$ is some topology for $2^{X}$.
My attempt. Let $U_1, U_2, \cdots, U_n$ open sets in $X$ and let
$$\left< U_1, \cdots, U_n \right>=\{ A \in 2^{X}: A \cap U_i \neq \emptyset, \ \forall i=1, \cdots,n \}.$$
Let the collection
$$ \mathcal{B}=\{\left< U_1, \cdots, U_n \right>: n \in \mathbb{N}, \ U_i \in \tau, \ \forall i=1, \cdots, n\}.$$
I have already proved that the collection $\mathcal{B}$ is a base for a topology in $2^{X}$, so it is also a subbase. Now I want to use the Alexander Subbase Theorem. I've wanted to take a collection of elements from the subbase, but couldn't find the finite collection.
Best Answer
In order for $\mathcal B$ be a base for some topology on $2^X$, we need that every element of $2^X$ is in some element of $\mathcal B$; in particular, the empty set is in some element of $\mathcal B$, which is absurd. Because of that, I will assume that $2^X$ is the set of all non-empty closed subsets of $X$.
Now, since we have that $\langle U_1,\dots,U_n \rangle = \bigcap_{i=1}^n \langle U_i \rangle$ for each $n \in \Bbb N$ and $U_1,\dots,U_n \in \tau$, the family $\mathscr S := \{\langle U \rangle : U \in \tau \}$ is a subbase for $2^X$.
So, if we prove that every cover of $2^X$ by elements from $\mathscr S$ has a finite subcover, we’re done. That is, if $\mathscr U \subseteq \tau$ is such that $\bigcup_{U \in \mathscr U} \langle U \rangle = 2^X$, we need to prove that there exists a finite subset $\mathscr V$ of $\mathscr U $ such that $\bigcup_{U \in \mathscr V} \langle U \rangle = 2^X$.
Hint: Consider $A := X \setminus \bigcup \mathscr U$. Note that $A = \varnothing$; otherwise $A \in 2^X$, and then […]. Thus $X = \bigcup \mathscr U$, and by compactness it follows that […].