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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?

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    I'll enjoy reading the answers, this looks like a very nice theorem :)2012-06-18
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    @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.2012-06-18
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    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 $X$.2012-06-19
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    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.2012-06-19
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    @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).2012-06-19
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    @ismythe: In fact, let $X$ be a complete separable metric space, let $h$ be a Hausdorff measure function (for defining generalized Hausdorff measures), and let $Z(X)$ be the collection of compact subsets of $X$ having Hausdorff $h$-measure zero. Then $Z(X)$ has a first category (i.e. meager) complement in $K(X).$ Similarly, in appropriate spaces (${\mathbb R}^n$ certainly qualifies), the same is true for Borel measures in which singletons have measure zero, generalized capacity notions in which singletons have zero capacity, most any kind of upper porosity notion (but not lower porosity), etc.2012-06-19
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    In case anyone is interested in some references for the statements I made in my previous comment . . . MR 94d:54056 (p. 14); MR 94g:26004 (p. 534); MR 97c:26001 (theorem 2); MR 89e:28018 (p. 769); MR 95i:26006 (p. 800); MR 30 #4887 (p. 1216); MR 99g:28009; MR 86h:41036; MR 90k:54052 (p. 156, p. 158); MR 94e:28004 (section 5); MR 96e:03057 (30.15 on p. 237); MR 98b:54042; MR 53 #1544 (p. 122); MR 96b:30074 (theorem 1); MR 94e:26041 (p. 298); MR 88c:540292012-06-19

1 Answers 1

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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 $G=\{K\in\mathscr{K}:K\text{ is perfect}\}$.

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    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?2012-06-19
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    @ismythe: Yes, yes, and yes. I think that I’ve fixed everything now.2012-06-19