## Hausdorff metric compatible with Vietoris topology / Kechris p25 B Ex 4.21

Kechris, p. 25
(B) Exercise (4.21).
Show that the Hausdorff metric is compatible with the Vietoris topology.

Proof.
Let $(X, d)$ be a metric space with $d \le 1$. Let $\tau_H$ denote the topology from the Hausdorff metric on $K(X)$, with basis $\mathcal{B}_H = \{ B^H(K, \epsilon) : K \in K(X), \epsilon > 0 \}$, $\tau_V$ denote the Vietoris topology on $K(X)$, with basis
$\mathcal{B}_V = \{ V_{U_i}^{0,n} = \{ K \in K(X) : K \subseteq U_0 \ \& \ K \cap U_1 \neq \emptyset \ \& \ \ldots \ \& \ K \cap U_n \neq \emptyset \} : U_i \text{ open in } X, 0 \le i \le n \}$.

If $K \subset X$, $B^X(K, \epsilon) = \{ x \in X : d(x, K) < \epsilon \}$.

(i) $\tau_H \subseteq \tau_V$.
Proof.
Let $B = B(K, \epsilon) \in \mathcal{B}_H$.  Suppose $K = \emptyset$.  Then, since $\{ \emptyset \} \in \mathcal{B}_V$ and $\{ \emptyset \} \subset B$, we’re done.

So, suppose now that $K \neq \emptyset$. Cover $K$ with finitely many balls $U_i$, $1 \le i \le n$, of radius $\epsilon$, such that, for all $i$, $U_i \cap K \neq \emptyset$.  Let $U_0 = \bigcup_{i = 1}^n U_i$.

I will show that $V = V_{U_i}^{0,n} \in \mathcal{B}_V$ satisfies $V \subseteq B$.

Let $L \in V$.  That is $L \subseteq U_0$, and for all $i$, $L \cap U_i \neq \emptyset$.  To see that $d_H(L,K) < \epsilon$, I check that $L \subseteq B^X(K, \epsilon)$ and $K \subseteq B^X(L, \epsilon)$.  Let $l \in L$.  Then $l \in U_0$, so $l \in U_i$ for some $i \in \{ 1, \ldots, n \}$, so $d(l, K) < \epsilon$.  Similarly, if $k \in K$, $d(k, L) < \epsilon$.

So $V \subseteq B$. $\blacksquare$

(ii) $\tau_V \subseteq \tau_H$.
Proof.
Let $V = V_{U_i}^{0,n} \in \mathcal{B}_V$, $K \in V$.  If $K = \emptyset$, then $B^H( \emptyset, 1/2) \subset V$. So suppose that $K \neq \emptyset$.  I have that $K \subseteq U_0$, and $K \cap U_i \neq \emptyset$, $i \in \{1, \ldots, n\}$.

Claim 1: $\inf \{ d(x, K) : x \in X \backslash U_0 \} = \epsilon_0 > 0$.
Proof.
Suppose $\epsilon_0 = 0$.  Then, I can pick $(x_n) \subset X \backslash U_0$ such that $\lim_n d(x_n, K) = 0$.  Let $(k_n) \subseteq K$ such that $d(x_n, k_n) \to 0$.  Then, $(k_n)$ has a convergent subsequence, $k_{n_i} \to k$.  But then $x_{n_i} \to k$, which is a contradiction since $X \backslash U_0$ is closed. $\blacksquare$

Let $i \in \{1, \ldots, n \}$.  Let $x_i \in K \cap U_i$.  Then $\exists \, \epsilon_i$ such that $B^X(x_i, \epsilon_i) \subseteq U_i$. Then let $\epsilon = \min \{ \epsilon_0, \ldots, \epsilon_n \}$.

Claim 2: $B = B^H(K, \epsilon) \subseteq V$.
Proof.
Let $L \in B$.  That is, $d_H(K,L) < \epsilon$.  Then, $L \subseteq B^X(K, \epsilon)$, $K \subseteq B^X(L, \epsilon)$. Then, for each $i \in \{1, \ldots, n \}$, $d(x_i, L) < \epsilon$.  So there exists $l_i \in L$ such that $d(x_i, l_i) < \epsilon$.  Hence $l_i \in U_i$, and $L \cap U_i \neq \emptyset$. $\blacksquare$

Finally, let $l \in L$.  Then $d( l, K) < \epsilon$, so $l \notin X \backslash U_0$ (see Claim 1).  So $L \subseteq U_0$, and $L \in V$. $\blacksquare$

So $\tau_V \subseteq \tau_H$. $\blacksquare$

And hence $\tau_V = \tau_H$. $\blacksquare$