We define a continuum $X$ as a compact, connected and nonempty metric space and we denote the closed hyperspace as $2^{X} = \{A \subset X: A\neq \emptyset \text{ is closed }\}$.
If $X$ is a continuum, the Hausdorff metric is defined in terms of the $X$ metric, so two metrics of X generate two metrics for $2^{X}$.
I have to show that all Hausdorff metrics on a continuum generate the same topology (provided that the metrics taken for $X$ generate the same topology on $X$)
For example I have the following Hausdorff metrics: Let $A,B \in 2^{X}$
$H(A,B)= \inf\{ \epsilon >0: A \subset N(\epsilon,B) \land B \subset N(\epsilon,A)\}$
$H'(A,B)= \max \{ \sup \{d(a,B):a \in A\}, \sup\{d(b,A):b \in B\} \}$
and are equivalent metrics, $H(A,B)= H'(A,B)$.
I know the follow theorem: Let $U_{1},..,U_{n}$ open subsets of X. Consider the family of subsets of $2^{X}$ given by
$\mathcal{B}:=\{\langle U_{1},\ldots,U_{n} \rangle: n \in \mathbf{N} \text{ and } U_{1},\ldots,U_{n}\text{ are open in }X\}$
where
$\langle U_{1},\ldots,U_{n}\rangle =\{ A \in 2^{X}: A \subseteq \bigcup_{i=1}^{n} U_{i}\text{ and } \forall i \in \{1,\ldots,n\}: A \cap U_{i} \neq \emptyset \}$
then $\mathcal{B}$ is a base for the so-called Vietoris topology. And the Vietoris topology is the same topology given by the Hausdorff metric.
I don't know if I can use this theorem to prove what I need or is there another way to prove it. Someone knows?
I thank you in advance for the help.