## 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 $\delta(K_n, K) = \max_{x \in K_n} d(x, K) = 1 - 1/n = \frac{n - 1}{n} \to 1$.