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Here is what Kechris says about the Vietoris topology,

Let X be a top. space. We denote by K(X) the space of all compact subsets of X equipped with the Vietoris topology, i.e., the one generated by the sets of the form $\{K \in K(X):K \subseteq U\}$ and $\{K \in K(X): K \cap U \neq \emptyset\}$ for U open in X.

1) Isn't $K \subseteq U$ redundant? Can we not just say $K \subseteq X$?

EDIT: Is the definition above "for all U", or "for a fixed U"?

EDIT2: The problem that is seeming to stump me is that if we take U=X then it seems we will get all the compact sets.

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1 Answer 1

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No, it’s not at all redundant. Suppose that $X=\Bbb R$. $\{K\in K(X):K\subseteq(0,1)\}$ is clearly very different from $\{K\in K(X):K\subseteq\Bbb R\}$.

For each open $U\subseteq X$ let $U^+=\{K\in K(X):K\cap U\ne\varnothing\}$ and $U^-=\{K\in K(X):K\subseteq U\}$; the set of all such $U^+$ and $U^-$ as $U$ ranges over open sets in $X$ is a subbase for the Vietoris topology on $K(X)$. It may be helpful to see a base for the topology. For any open sets $U_1,\dots,U_n$ in $X$ let

$$\begin{align*} \langle U_1,\dots,U_n\rangle&=\left\{K\in K(X):K\subseteq\bigcup_{k=1}^nU_k\text{ and }K\cap U_k\ne\varnothing\text{ for }k=1,\dots,n\right\}\\ &=\left(\bigcup_{k=1}^nU_k\right)^-\cap\bigcap_{k=1}^nU_k^+\;; \end{align*}$$

the collection of all such $\langle U_1,\dots,U_n\rangle$ for $n\in\Bbb Z^+$ and $U_1,\dots,U_n$ open sets in $X$ is a base for the Vietoris topology.

Added: It’s perfectly true that $X^+=X^-=K(X)$; after all, we expect $K(X)$ to be an open set in itself! Similarly, $\varnothing^-=\varnothing$, which is indeed an open set in any topology.

You might find it helpful to investigate a very simple Vietoris topology in some detail. Let $X=\Bbb N$ with the discrete topology, so that $K(X)$ is just the family of finite subsets of $X$, and every subset of $X$ is open in $X$. For any $U\subseteq\Bbb N$, $U^-$ is the set of finite subsets of $U$, and $U^+$ is the set of finite sets of natural numbers that include at least one member of $U$. For instance, if $U$ is the set of even natural numbers, $U^-$ is the set of finite sets of even natural numbers, and $U^+$ is the set of finite sets of natural numbers that include at least one even number. Try to prove that the Vietoris topology on $K(\Bbb N)$ is discrete; it’s very easy if you use the base that I described, but it’s not at all hard to do directly from the given subbase.

The next step up would be to let $X=\{0\}\cup\{1/n:n\in\Bbb Z^+\}$ with the topology that it inherits from $\Bbb R$, and to investigate the Vietoris topology on $K(X)$. Now $K(X)$ includes not only all finite subsets of $X$, but also all infinite sets that include $0$.

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I still do not quite understand the topology. Does the generating set span over all U open in X, or a fixed U open in X? –  user41728 Sep 18 '12 at 23:14
    
@user41728: It ranges over all open subsets of $X$. I’ve considerably expanded my answer; see if it helps. –  Brian M. Scott Sep 18 '12 at 23:46
    
Thanks, it helps a lot. I'm working on the exercises you gave me. –  user41728 Sep 18 '12 at 23:53
    
(1) Let $H$ be in $K(N)$. If $H$ is empty then $H \in U^-$ for any $U$. Otherwise $H$ is nonempty and intersects itself and since $H$ is open in $N$, $H \in H^+$. Either way $H$ is open in $(K(X),Vietoris)$. –  user41728 Sep 19 '12 at 0:11
    
@user41728: Not quite. You need to show that $\{H\}$ is open in the Vietoris topology. $\{\varnothing\}=\varnothing^-$, so the empty set is taken care of. Now let $H=\{h_1,\dots,h_n\}$. Then $\{H\}=H^-\cap\bigcap_{k=1}^n\{h_k\}^+$. –  Brian M. Scott Sep 19 '12 at 0:19

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