Compactness and Hausdorff metric

Question

I have this theorem "$X$ is compact $\leftrightarrow\exp X$ is compact", but i can not find source of it. It concerns Hausdorff metric.

Answer 1

The theorem that the continuous image of a compact set is compact will give the result as $exp$ and $log$ are both continuous functions. The theorem states:

If $f:X \rightarrow Y$ is a continuous map of topological spaces and $A \subset X$ is compact, then so is $f(A)$.

Answer 2

This is indeed true, for the Vietoris topology on the hyperspace $\operatorname{Exp}(X)$. For compact metric spaces this topology is metrisable and the Hausdorff metric is a compatible metric for it. This can be found in Nadler's book on hyperspaces, e.g., or spread over several exercises in Engelking's book "General Topology".

The Vietoris topology (on the set $\operatorname{Exp}(X)$ of closed non-empty subsets of a space $X$) has as a subbase all sets of the form $\langle U \rangle = \{ A \subset X: A \subset U \}$ and $[U] = \{A \subset X: A \cap U \neq \emptyset\}$, where $U$ ranges over the non-empty open subsets of $X$ (and all $A$ are closed non-empty).

Alexander's subbase lemma implies that if $X$ is compact, so is $\operatorname{Exp}(X)$. Proof: let $\{ \langle U_i \rangle: i \in I\} \cup \{ [V_j]: j \in J\}$ be a covering of $\operatorname{Exp}(X)$ by subbasic elements.

If the $V_j$ already cover $X$, this means that every point in the hyperspace intersects some $V_j$. But then same holds for a finite subcover of the $V_j$, and we are done. So assume $A = X \setminus \cup_j V_j$ is non-empty. This set is contained in some $U_{i_0}$. Then $X \setminus U_{i_0}$ is covered by the $V_j$ and as a closed subset of $X$ is compact, and thus covered by $V_{j_1},\ldots,V_{j_n}$. Then $\{[V_{j_1}],\ldots,[V_{j_n}],\langle U_{i_0} \rangle\}$ is a finite subcover.

There are also direct proofs from the metric, but I find these a bit more messy, although here the messy bits are in the proof of the equivalence of these topologies... Anyway, this is one way to go about this.