Hausdorff metric and Vietoris topology 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.
 A: 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 $F \in B^H ( F , \varepsilon ) \subseteq \langle U;V_1 ,\ldots , V_n \rangle$.

A: You can use the following definitions:
$V(U) := \{A ⊆ X: A ⊆ U\}$ for $U ⊆ X$ open,
$W(U) := \{A ⊆ X: A ∩ U ≠ ∅\}$ for $U ⊆ X$ open,
$N(K, ε) := \{A ⊆ X: A ⊆ K_ε\}$ for $ε > 0$, $K ⊆ X$,
$M(K, ε) := \{A ⊆ X: K ⊆ A_ε\}$ for $ε > 0$, $K ⊆ X$,
where $A_ε = \bigcup_{x ∈ A}\{y ∈ X: d(x, y) < ε\}$.
Then the sets $V(U)$, $W(U)$ form a subbasis of Vietoris topology and open ball in Hausdorff metric $B_H(K, ε) = \bigcup_{0 < δ < ε} (N(K, δ) ∩ M(K, δ))$. So if you want to prove that the topologies are the same, it is enough to show that $(∀U) (∀K ∈ V(U)) (∃ε): N(K, ε) ∩ M(K, ε) ⊆ V(U)$ and the same for $W(U)$. And for the other direction $(∀K)(∀ε)(∃U_0, …, U_n): K ∈ V(U_0) ∩ W(U_1) ∩ … ∩ W(U_n) ⊆ N(K, ε)$ and the same for $M(K, ε)$. In some inclustions, you'll need that all subsets which are points of our space are by definitions compact.
Also note that the definitions work for $K = ∅$. $\{∅\}$ is clopen in both topologies.
