Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.