Let $X$ be metrizable (not necessarily Polish), and consider the hyperspace of all compact subsets of $X$, $K(X)$, endowed with the Vietoris topology (subbasic opens: $\{K∈K(X):K\subseteq U \}$ and $\{K\in K(X):K\cap U\neq \emptyset\}$ for $U\subseteq X$ open), or equivalently, the Hausdorff metric. We want to show that $$K_{f}(X)=\{K\in K(X):K \text{is finite}\}$$ is $F_{\sigma}$ in $K(X)$.

I've been thinking about this exercise and I have not been able to solve.

Any ideas?

  • $\begingroup$ What are the results of your thoughts? Can you solve this for a special case? What is the simplest nontrivial example you can think of? Have you seen a similar statement? $\endgroup$ – Hans Engler Feb 20 '15 at 22:08

HINT: For $n\in\Bbb N$ let $\mathscr{K}_n(X)=\{K\in \mathscr{K}(X):|K|\le n\}$, and show that each $\mathscr{K}_n(X)$ is closed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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