Topological limits / Kechris p26, Ex 4.23

See here for definitions.

Kechris, p. 26
(B) Exercise (4.23).  Let (X, d) be metric space with d \le 1. Show that for nonempty K, K_n \in K(X),
(i) \delta(K, K_n) \to 0 \Rightarrow K \subseteq \underline{\text{Tlim}}_n K_n.
Proof.
Suppose \delta(K, K_n) \to 0.

Thus, for all \epsilon > 0, there exists N, such that n \ge N implies \delta(K, K_n) < \epsilon.
That is, \max_{x \in K} d(x, K_n) < \epsilon.

Let x \in K.  Now, choose M_1 < M_2 < \ldots < M_m < \ldots such that n \ge M_i implies d(x, K_n) < 1/ i.
Now, choose a sequence (x_n) in X, such that x_n \in K_n, in the following way:  For n \in \mathbb{N},
M_i \le n < M_{i + 1}, for some i \in \mathbb{N}.  Pick any x_n \in K_n such that d(x, x_n) < 1 /i.

Then x_n \to x, so x \in \underline{\text{Tlim}}_n K_n. \blacksquare

(ii) \delta(K_n, K) \to 0 \Rightarrow K \supseteq \overline{\text{Tlim}}_n K_n.
Proof.
Assume \delta(K_n,K) \to 0.  Let x \in \overline{\text{Tlim}}_n K_n \cap K^c.  Since x \notin K
and since K is compact, hence closed, d(x,K) = \epsilon > 0.
x \in \overline{\text{Tlim}}_n K_n \Rightarrow \exists(x_n)[ x_n \in K_n \text{, for all n, and for some subsequence } (x_{n_i}), x_{n_i} \to x].
\exists some N such that n_i \ge N \Rightarrow d(x_{n_i}, x) < \epsilon / 2, and \delta(K_{n_i}, K) < \epsilon / 2.
Therefore, d(x_{n_i}, K) < \epsilon / 2.  Therefore \exists (y \in K)[d(x_{n_i}, y) < \epsilon / 2].
Therefore d(x,y) \le d(x, x_{n_i}) + d(x_{n_i}, y) < \epsilon, which is a contradiction.
Thus, \overline{\text{Tlim}}_n K_n \subseteq K. \blacksquare

(i) and (ii) show that d_H(K_n, K) \to 0 \Rightarrow K = \text{Tlim} K_n.
To show that the converse fails, consider the real numbers \mathbb{R} with the trivial metric.
Let K_n = \{ 1/n, 1 \}, which is clearly compact.  Then \text{Tlim} K_n = \{ x : \text{every open neighborhood of x meets } K_n \text{ for infinitely many n} \} = \{ 1 \}
And \delta(K_n, K) = \max_{x \in K_n} d(x, K) = 1 - 1/n = \frac{n - 1}{n} \to 1.

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