Let $X$ be a Hausdorff space. Let $K(X)$ be the collection of all compact subsets of $X$. A topology on $K(X)$ is defined by a subbasis given by sets of the form $I_U=\{K\in K(X)\,|\,K\subset U\}$ and sets of the form $D_U=\{K\in K(X)\,|\,K\cap U\neq\emptyset\}$, for every $U$ open in $X$.
- Show that $K(X)$ is Hausdorff.
- Show that $K(X)$ is compact if and only if $X$ is compact.
I may use the result that a topological space given by a subbasis is compact if and only if every open covering consisting of subbasis elements contains a finite collection that also covers the space.
For 1., take two compact subsets $K,M\in K(X)$ and $K\neq M$. Take a point $x_0\in M$. Because $X$ is Hausdorff, we can take disjoint opens $U$ and $V$ with $K\subset U$ and $x_0\in V$. This means that $K\in I_U$ and $M\in D_V$. Assume $F\in I_U\cap D_V$, then $F\subset U$ and $F\cap V\neq\emptyset$, which contradicts $U\cap V=\emptyset$, so $I_U\cap D_V=\emptyset$. We conclude that $K(X)$ is Hausdorff.
For 2., I honestly have no idea how to start even with the given hint
All help is welcome, thank you!
Best Answer
I'll denote $I_U$ by the more common $\langle U \rangle$, and $D_U$ by $[U]$ (common in hyperspace theory). I assume $\emptyset \notin K(X)$ (it's usually excluded, because it's an isolated point anyway and it doesn't affect this compactness result anyway).
If $K(X)$ is compact, let $\mathcal{U}$ be an open cover of $X$. Then $\{[U]\mid U \in \mathcal{U}\}$ is an open cover of $K(X)$.
Let $\{[U]\mid U \in \mathcal{V}\}$ be a finite subcover (for some finite $\mathcal{V} \subseteq \mathcal{U}$) and then using the compact singletons it's easy to see that $\mathcal{V}$ is a finite subcover of $\mathcal{U}$.
For the converse assume $X$ is compact and that $$\{[U]\mid U \in \mathcal{U}_1\} \cup \{\langle U \rangle\mid U \in \mathcal{U}_2\}$$ is a cover of $K(X)$ by subbasic elements.
If $\bigcup \mathcal{U}_1 = X$, we note that $\mathcal{U}_1$ would have a finite subcover and the corresponding $[U]$ would form a finite subcover and we'd be done. So assume that $K=X \setminus \bigcup \mathcal{U}_1 \neq \emptyset$. This $K$ is in $K(X)$ (as it's a closed subset of the compact $X$), and so is covered by some $\langle U_2 \rangle$ for some $U_2 \in \mathcal{U}_2$ (it cannot be covered by the other subbasic cover elements). But then $K'=X\setminus U_2$ is compact too and covered by $\mathcal{U}_1$ and so by a finite subfamily $\mathcal{U}'_1$ of it.
But then $$\{[U]\mid U \in \mathcal{U}'_1\} \cup \{\langle U_2 \}$$ is a finite subcover of the subbasic cover we started with. QED.