# The collection of all compact perfect subsets is $G_\delta$ in the hyperspace of all compact subsets

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: $\{K\in K(X):K\subset U\}$ and $\{K\in K(X):K\cap U\neq\emptyset\}$ for $U\subset X$ open), or equivalently, the Hausdorff metric. We want to show that $K_p(X)=\{K\in K(X):K \text{ is perfect}\}$ is $G_\delta$ in $K(X)$. (This is another question from Kechris, Classical Descriptive Set Theory, Exercise 4.31.)

A possible approach: $K_p(X) = \bigcap_{n=1}^\infty \{K\in K(X): \forall x\in K, (B(x,1/n)\setminus\{x\})\cap K\neq\emptyset\}$. What can we say about the complexity of $\{K\in K(X): \forall x\in K, (B(x,1/n)\setminus\{x\})\cap K\neq\emptyset\}$? Note that for fixed $x$, the set $\{K\in K(X): (B(x,1/n)\setminus\{x\})\cap K\neq\emptyset\}$ is open in $K(X)$. Also, the set $\{(x,K)\in X\times K(X):x\in K\}$ is closed in $X\times K(X)$, but I don't think this helps since the projection of a $G_\delta$ set need not be $G_\delta$.

Any ideas?

• I'll enjoy reading the answers, this looks like a very nice theorem :) Jun 18, 2012 at 23:05
• @Olivier: It gets better. If $X$ is perfect Polish, then $K_p(X)$ is a dense $G_\delta$ (hence generic) in $K(X)$. Working through this part is something I'll do in the near future. Jun 18, 2012 at 23:26
• Why is $\{K \in K(X) : (B(x, \frac{1}{n}) - \{x\}) \cap K \neq \emptyset\}$ open in the Vietoris Topology. I don't see why $B(x, \frac{1}{n}) - \{x\}$ needs to be open in $X$. Jun 19, 2012 at 5:37
• My idea is that separable metric spaces are second countable. Let $\{U_n\}$ denote the countably many open set. You want to try to make $\mathcal{V}_n$ an open subset of $K(X)$ contains all compact subset of $X$ except those $K \subset X$ which contains an isolated point $x$ such that $U_n \cap K = \{x\}$, i.e. $U_n$ witnesses that $K$ has an isolated point. Then intersect all that $\mathcal{V}_n$. However, I was not able to make this work. Hopefully, this is some inspiration for you. Jun 19, 2012 at 6:15
• @William: $B(x,1/n)\setminus\{x\} = B(x,1/n)\cap\{x\}^c$, and since a metric space is in particular T1, $\{x\}^c$ is open. As for your suggestion, I will think about that, though the question does not specify separable (well, actually, it does, but Kechris says in the errata to the book that this condition should be dropped). Jun 19, 2012 at 12:30

For any $n\in\Bbb Z^+$ and open $U_1,\dots,U_n$ in $X$ define

$$B(U_1,\dots,U_n)=\left\{K\in\mathscr{K}(X):K\subseteq\bigcup_{k=1}^nU_k\text{ and }K\cap U_k\ne\varnothing\text{ for }k=1,\dots n\right\}\;;$$

the collection $\mathscr{B}$ of these sets is a base for the topology of $\mathscr{K}(X)$.

For $n\in\omega$ let $\mathfrak{U}_n$ be the collection of all finite families of open sets of diameter less than $2^{-n}$. For each $\mathscr{U}\in\mathfrak{U}$ and $p,q:\mathscr{U}\to\bigcup\mathscr{U}$ such that for each $U\in\mathscr{U}$, $p(U)$ and $q(U)$ are distinct points of $U$, fix disjoint open sets $V_{\mathscr{U},p,q}(U)$ and $W_{\mathscr{U},p,q}(U)$ for $U\in\mathscr{U}$ such that $p\in V_{\mathscr{U},p,q}(U)\subseteq U$ and $q\in W_{\mathscr{U},p,q}(U)\subseteq U$. Then let

$$G(\mathscr{U},p,q)=B(\mathscr{U})\cap\bigcap_{U\in\mathscr{U}}B\big(V_{\mathscr{U},p,q}(U),W_{\mathscr{U},p,q}(U),X\big)\;,$$

let $\mathscr{G}_n$ be the set of all such $G(\mathscr{U},p,q)$ for $\mathscr{U}\in\mathfrak{U}_n$, and let $G_n=\bigcup\mathscr{G}_n$; clearly each $G_n$ is open in $\mathscr{K}(X)$.

Let $K\subseteq X$ be a non-empty compact set without isolated points. Fix $n\in\omega$. Let $\mathscr{U}$ be a finite open cover of $K$ by sets of diameter less than $2^{-n}$. Pick distinct points $p(U),q(U)\in K\cap U$ for each $U\in\mathscr{U}$. Then $G(\mathscr{U},p,q)\in\mathscr{G}_n$ is an open nbhd of $K$ in $\mathscr{K}(X)$, so $K\in G_n$.

Now suppose that $K\subseteq X$ is compact but has an isolated point $x$. Fix $m\in\omega$ such that $$B(x,2^{-m})\cap K=\{x\}\;,$$ where $B(x,\epsilon)$ is the open ball of radius $\epsilon$ centred at $x$.

Suppose that $n\ge m$ and $K\in G(\mathscr{U},p,q)\in\mathscr{G}_n$. Some $U\in\mathscr{U}$ contains $x$, and $$K\in B\big(V_{\mathscr{U},p,q}(U),W_{\mathscr{U},p,q}(U),X\big)\;,$$ so there are distinct points $y\in K\cap V_{\mathscr{U},p,q}(U)$ and $z\in K\cap W_{\mathscr{U},p,q}(U)$. But $$y,z\in U\subseteq B(x,2^{-n})\subseteq B(x,2^{-m})\;,$$ so $y,z\in B(x,2^{-m})\cap K=\{x\}$, which is impossible. Thus, $K\notin G_n$ for $n\ge m$.

Finally, let $G=\bigcap_{n\in\omega}G_n$. Clearly $G$ is a $G_\delta$-set in $\mathscr{K}(X)$, and we’ve just shown that $G=\{K\in\mathscr{K}:K\text{ is perfect}\}$.

• This looks like it works. A couple questions (minor typos): Should $G_n=\bigcup\mathscr{G}_n$? Also, we need the $p_k, q_k\in U_k$ chosen in the third paragraph to also be in $K$, right? (Which we can do, because $K$ is perfect and $U_k\cap K\neq\emptyset$.) And must the corresponding $V_k$ and $W_k$ be those which are specified in the second paragraph? Jun 19, 2012 at 17:28
• @ismythe: Yes, yes, and yes. I think that I’ve fixed everything now. Jun 19, 2012 at 17:49