0
$\begingroup$

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-

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $n\in\Bbb Z^+$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.