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Let $(X,T)$ be a compact metrizable space. For every metric $d$ on $X$ which is $T-$ compatible, the Hausdorff metric on $2^{X}$ gives a topology on $2^{X}$.

(Up to homeomorphism) is this topology independent of the metric $d$?

If the answer is yes, (up to homeomorphism) we actualy obtain a unique topology on the hyperspace. Now assume that $X$ and $Y$ are two compact metric space such that $Y$ is countable. Under what natural topology on $(2^{X})^{Y}=\{f:Y\to 2^{X} \mid \text{f is continuous} \}$ we have $$(2^{X})^{Y}\simeq 2^{X \times Y}$$

Where $\simeq$ means "homeomorphic to"?

We add the empty set to the hyper space as an isolated point, in order to avoid the obvious false for finite $X$ and $Y$.

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  • $\begingroup$ @Mirko I am not familiar with Vietoris topology. Could you please more explain on your comment? $\endgroup$ – Ali Taghavi Apr 25 '15 at 0:23
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    $\begingroup$ for compact metric spaces the Hausdorff topology on the hyperspace (generated by the Hausdorff metric) coincides with the Vietoris topology (defined purely in terms of the topology). See math.stackexchange.com/questions/403678/… also math.stackexchange.com/questions/791334/… and Infinite Dimensional Analysis: A Hitchhiker's Guide, Charalambos,Aliprantis, Border,Theorem 3.91,p.120 books.google.com/books?id=4hIq6ExH7NoC $\endgroup$ – Mirko Apr 25 '15 at 3:03
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    $\begingroup$ We only have a canonical set-theoretical inclusion $\ (2^X)^Y\rightarrow 2^{X\times Y}.\ $ In general, it is not an equality. Thus I don't see any natural homeomorphism. $\endgroup$ – Włodzimierz Holsztyński Apr 25 '15 at 5:15
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    $\begingroup$ As to the first question, you may also observe that if $(X,d)$ is a compact metric space, its hyperspace $H(X)$ (the set of closed sets of $X$ endowed with the Hausdorff distance) is a compact metric space too; moreover the construction extends to a functor from the category of compact metric spaces & continuous functions into itself, precisely : if $f:X\to Y$ is continuous the map $K\to f(K)$ is continuous from $H(X)$ to $H(Y)$ (just because $f$ is uniformly continuous). In particular, if $(X,d)$ and $(X',d')$ are homeomorphic compact metric spaces, so are their hyperspaces. $\endgroup$ – Pietro Majer Apr 25 '15 at 8:01

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