Let $X$ be a topological space. Denote by $K(X)$ the space of all non-empty compact subsets of $X$ equipped with the Vietoris topology, namely the one generated by the sets of the form:

$\{K\in K(X):K\subseteq U\}$ $\{K\in K(X):K\cap U\not=\emptyset\}$

for $U$ open in $X$.

It'easy to note that $K\subseteq U$ implies $K\cap U\not=\emptyset$. So why do we need the first set? Thank you-

  • $\begingroup$ Brian answered this question here math.stackexchange.com/questions/198777/… $\endgroup$ – Daniel Valenzuela Feb 16 '16 at 18:24
  • $\begingroup$ @DanielValenzuela That question seems different to me, the user asked about $K\subseteq X$ instead of $K\subseteq U$... $\endgroup$ – Richard Feb 16 '16 at 18:29
  • $\begingroup$ I know that's why I didn't refer to the question but to the answer. That is, I didn't say "somebody asked the same thing" but Brian's answer contains an answer to your question. I believe that reading the answer should clarify things for you. If not you are welcome to specify your question. $\endgroup$ – Daniel Valenzuela Feb 16 '16 at 18:42
  • $\begingroup$ You also need it to show $X$ compact then $K(X)$ compact as well, for instance, because you then have a nice subbase for the Alexander subbase lemma. $\endgroup$ – Henno Brandsma Feb 16 '16 at 22:05

If you take only the sets of the form $U^+=\{K\in K(X):K\cap U\ne\varnothing\}$ as generators, you’ll get a topology, but it won’t be the Vietoris topology, and in general it won’t be very nice. For instance, take $X=\Bbb N$ with the discrete topology. The Vietoris topology on $K(\Bbb N)$ is discrete; the topology generated by the sets $U^+$ for $U\subseteq\Bbb N$ is not even $T_1$, since every open nbhd of the point $\{0\}\in K(\Bbb N)$ also contains every point $\{0,n\}$ for $n\in\Bbb Z^+$.

  • $\begingroup$ thank you, but what if I take only the sets of the form $U^+$ as generators? I meant this. $\endgroup$ – Richard Feb 16 '16 at 18:48
  • $\begingroup$ @Richard: Sorry: I misread the question. The same problem arises: see the revised answer. $\endgroup$ – Brian M. Scott Feb 16 '16 at 18:49

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