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

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