Vietoris Topology

Question

‎let ‎‎$ X ‎‎‎‎$ ‎be a‎ ‎topological ‎space ‎and‎ ‎$ \operatorname{‎Exp}(X‎)‎‎‎ $ ‎is ‎the set of all ‎closed ‎non-empty subsets of $X$ .‎‎

If $ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎\ldots ‎V_{n}‎$ ‎are ‎the non-empty open subset ‎in ‎$ ‎X‎$‎‎‎, ‎define:‎ ‎

$$ ‎\langle U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, \ldots V_{‎n}‎ \rangle ‎ ‎ =‎ ‎\{ F‎ ‎\in \operatorname{‎Exp}(X‎)‎‎‎‎\mid F‎\subseteq‎ ‎U,‎ \forall 1‎‎ \leq i ‎‎‎\leq n ‎‎‎:‎ F‎\cap ‎V_{i} ‎\neq \emptyset\}‎$$‎

families ‎$ B‎ ‎‎ $‎i‎ncludes all sets of the form $\langle‎ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, \ldots, ‎V_{‎n} \rangle ‎$ ‎is ‎‎the basis for a topology for ‎$ ‎Exp (‎ X‎ ) ‎$‎‎ This topology is called the Vietoris topology.

My ‎question:‎

if ‎‎$ ‎X‎$ ‎is ‎‎a $T_{1}$ ‎space, ‎then ‎is the Vietoris ‎topology ‎ ‎‎$ ‎T_{1}‎$‎?‎ ‎ ‎

Answer 1

This topology is called the Vietoris topology on $\operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.

I normally use a standard subbase for this, for every non-empty open set $U \subset X$, define $[U] = \{F \in H(X): F \cap U \neq \emptyset\}$ and $\langle U \rangle = \{F \in H(X): F \subseteq U\}$. Then $\langle U, V_1, \ldots, V_n\rangle = \langle U \rangle \cap \cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = \langle X, U\rangle$, so is indeed open. The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.

As to the $T_1$ question: yes.

Suppose $A \neq B$ are two different points in $H(X)$. So we can assume that $\exists x \in A, x \notin B$ (by symmetry, or we rename our sets).

Then $X\setminus B$ is open, $A \in [X\setminus B] $ (as witnessed by $x$), $B \notin [X \setminus B]$ by definition, and $X \setminus \{x\}$ is open as $X$ is $T_1$ and $B \in \langle X\setminus \{x\}\rangle $ and $A \notin \langle X\setminus \{x\}\rangle $, as witnessed by $x$ again.

So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.

Answer 2

Brandsma's answer is superb, and I here only mean to add some detail in the way of clarification regarding compactness of the Vietoris Topology, and to provide a small amount of information regarding literature surrounding the question and the issue of compactness. According to Keesling (Theorem 3 in Normality and Compactness are Equivalent in Hyperspaces), published in 1970 in the Bulletin of the AMS, the under the assumption that the Continuum Hypothesis (CH) holds, $H(X)$ (with the Vietoris topology) is normal if and only if $X$ is compact. The team of Costantini, Levi, and Pelant discussed various compactness results in their article Compactness and Local Compactness in Hyperspaces, published in Topology and Its Applcations in 2002, and they mention the 1951 work of Michael, Topologies on Spaces of Subsets, published in the Transactions of the AMS, in which it is shown that $H(X)$ is compact if and only if $X$ is compact. It appears that CH is not required for this result.

Regarding the original question, the above mentioned article by Michael includes an excellent summary of several related questions (see Theorem 4.9), and in particular, Michael points out that the converse does not hold; that is, $X$ being $T_1$ does not imply that $H(X)$ is $T_1$.

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