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Source: Kechris, pg 26, Theorem 4.25

If $X$ is completely metrizable, so is $K(X)$.

Fix a complete compatible metric $d \leq 1$ on $X$. Let $(K_n)$ be Cauchy in $(K(X),d_H)$, where WLOG we can assume $K_n \neq \emptyset$. Let $K=\overline{T \lim_n} K_n$. We will show that $K \in K(X)$ and $d_H(K_n,K) \rightarrow 0$. Note first that $K= \bigcap_n (\overline{\bigcup_{i=n}^{\infty}K_i})$ and that $K$ is closed and nonempty.

My question: Why is $K$ nonempty?

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up vote 2 down vote accepted

In Claim 1 afterwards, Kechris proves that $L_n = \bigcup_{i=n}^\infty K_i$ is totally bounded, hence $\bar L_n = \overline{\bigcup_{i=n}^\infty K_i}$ is compact. But then $K = \bigcap_{n=1}^\infty \bar L_n$ is an intersection of a nested sequence of compact sets, hence it is nonempty.

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