# on the Vietoris topology

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-

• Brian answered this question here math.stackexchange.com/questions/198777/… – Daniel Valenzuela Feb 16 '16 at 18:24
• @DanielValenzuela That question seems different to me, the user asked about $K\subseteq X$ instead of $K\subseteq U$... – Richard Feb 16 '16 at 18:29
• 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. – Daniel Valenzuela Feb 16 '16 at 18:42
• 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. – 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 $n\in\Bbb Z^+$.
• thank you, but what if I take only the sets of the form $U^+$ as generators? I meant this. – Richard Feb 16 '16 at 18:48