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. Is it true to say that $$(2^{X})^{Y}\simeq 2^{X \times Y}$$

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 $Y$.

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 $Y$.