TITLE: Hausdorff metric and Vietoris topology
QUESTION [9 upvotes]: I am supposed to show that on a compact metric space, the Hausdorff metric and the Vietoris topology induce the same topology. Does anybody know how this can be done? I wanted to start by showing that they contain the same basis elements, but this was not successful.

REPLY [6 votes]: Suppose that $\langle X , d \rangle$ is a compact metric space with $d \leq 1$.
And set $2^X := \{ F \subseteq X : F\text{ is closed} \}$.

The Hausdorff metric on $2^X$ is defined by $$d^H ( F, E ) = \begin{cases}
0, &\text{if }F = E = \varnothing \\
1, &\text{if }F = \varnothing \neq E,\text{ or }E = \varnothing \neq F \\
\max \{ \max_{x \in F} d(x,E) , \max_{y \in E} d(y,F) \}, &\text{if }F \neq \varnothing \neq E 
\end{cases}$$
The Vietoris topology on $2^X$ is generated by the family of all sets of the form $$\langle U; V_1 , \ldots , V_n \rangle := \{ F \in 2^X : F \subseteq U, F \cap V_i \neq \varnothing\text{ for each }i \leq n\}$$ where $U, V_1 , \ldots , V_n$ ($n \geq 0$) are open subsets of $X$.

To show that the Vietoris topology is finer than the topology induced by the Hausdorff metric, pick $F \in 2^X$, and let $\varepsilon > 0$.

If $F = \varnothing$, note that either $B^H ( F , \varepsilon ) = \{ \varnothing \} = \langle \varnothing; \rangle$ or $B ( F , \varepsilon ) = 2^X = \langle X; \rangle$.
If $F \neq \varnothing$, then cover $F$ by open balls $V_1 , \ldots , V_n$ of radius $\frac{\varepsilon}{2}$ such that $F \cap V_i \neq \varnothing$ for each $i \leq n$.  Setting $U = \bigcup_{i \leq n} V_i$, show that $F \in \langle U ; V_1 , \ldots , V_n \rangle \subseteq B^H ( K , \varepsilon )$.

To show that the topology induced by the Hausdorff metric is finer than the Vietoris topology, let $U , V_1 , \ldots , V_n \subseteq X$ be open, and let $F \in \langle U ; V_1 ,\ldots , V_n \rangle$.

If $F = \varnothing$, it follows that $n = 0$ and $F \in  B^H ( \varnothing , \frac{1}{2} ) = \{ \varnothing \} = \langle U ; \rangle$.
If $F \neq \varnothing$, note that since $X \setminus U$ is a closed (compact) set disjoint from $F$, then $\delta = \min_{x \in X \setminus U} d(x,F) > 0$.  For each $i \leq n$ pick some $x_i \in F \cap V_i$, and let $\delta_i > 0$ be such that $B ( x_i , \delta_i ) \subseteq V_i$. Setting $\epsilon = \min \{ \delta , \delta_1 , \ldots , \delta_n \}$, show that $F \in B^H ( F , \varepsilon ) \subseteq \langle U;V_1 ,\ldots , V_n \rangle$.