# 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 :) Commented 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. Commented 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 $X$. Commented 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. Commented 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). Commented 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 $$n\in\omega$$. For each finite open cover $$\mathscr{U}$$ of $$K$$ by sets of diameter less than $$2^{-n}$$ pick distinct points $$p(U),q(U)\in K\cap U$$ for each $$U\in\mathscr{U}$$ that intersects $$K$$. 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? Commented Jun 19, 2012 at 17:28
• @ismythe: Yes, yes, and yes. I think that I’ve fixed everything now. Commented Jun 19, 2012 at 17:49
• @BrianM.Scott "Pick distinct points $p(U), q(U)\in K\cap U$ for each $U\in\mathscr{U}$." This can't be done if $|K\cap U|\leq1$. Commented May 31, 2023 at 5:22
• @DickGrayson: $|K\cap U|$ cannot be $1$: $K$ is a non-empty compact set with no isolated points in a metric space, so if $U$ is open and $K\cap U\ne\varnothing$, then $K\cap U$ is infinite. It could be $0$ if $\mathscr{U}$ contains an open set disjoint from $K$, but such a might as well not be in $\mathscr{U}$ and can be ignored. I didn’t expect that to be a problem, but since it’s come up, I’ll reword it to make this explicit. Commented May 31, 2023 at 7:27
• @DickGrayson: $0$ most certainly is an isolated point of the set $K=\{0\}$. The unique member of a singleton set is always an isolated point of that set. The perfect sets in the space $X$ of the question are precisely the non-empty compact sets $K$ such that every point of $K$ is a limit point of $K$, i.e., such that no point of $K$ is an isolated point of $K$. Equivalently, if $x\in K$ and $U$ is an open nbhd of $x$ (in $X$ or in $K$, it doesn’t matter), then $U\cap(K\setminus\{x\})\ne\varnothing$. Commented Jun 1, 2023 at 2:02