The Hyperspace of Compact Sets is Hausdorff?

Question

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).

Answer 1

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 $K_2$.

Answer 2

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!

Answer 3

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.