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If $X$ is Hausdorff then $K(X)$ is Hausdorff where $K(X)$ is the Hyperspace of Compact Sets equipped with the topology from the Hausdorff metric.(subbasic opens: $\{K\in K(X):K\subseteq U\}$ and $\{K\in K(X):K∩U\neq \emptyset \}$ for $U\subseteq X$ open).

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  • $\begingroup$ Yes, it's Hausdorff. If $A\ne B$, then one contains a point that's not in the other set, and anything sufficiently close will share this property. $\endgroup$ – user147263 Feb 25 '15 at 20:24
  • $\begingroup$ You need to say that $X$ is metric, as well, for the "Hausdorff metric" to make sense. $\endgroup$ – GEdgar Feb 25 '15 at 23:28
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You have to check whether distinct compact sets $K_1$ and $K_2$ have disjoint neighbourhoods. If, say, $x \in K_1 \backslash K_2$, then there are disjoint open sets $U_1$, $U_2$ so $x \in U_1$ and $K_2 \subseteq U_2$. These correspond to disjoint subbasic open $\{K: K \cap U_1 \ne \emptyset\}$ containing $K_1$ and $\{K: K \subseteq U_2\}$ containing $K_2$.

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By slightly cheating, if you assume that the Hausdorff metric is indeed a metric, then the space is immediately $T_2$ since metric spaces are!

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In fact it's and iff:

If $X$ is a $T_1$ space (not even metric), then:

a) $\mathcal K(X)$ is $T_2$ iff $X$ is $T_2$.

b) $\mathcal K(X)$ is $T_3$ iff $X$ is $T_3$.

c) $\mathcal K(X)$ is $T_{3\frac{1}{2}}$ iff $X$ is $T_{3\frac{1}{2}}$.

where $T_3$ is regular hausdorff, and $T_{3\frac{1}{2}}$ is tychonoff.

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