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Let $X$ be a topological space, by $2^X$ we denote the family of all nonempty closed subsets of $X$, the Vietoris topology endowed to $2^X$ is the one whose base consists of sets of the form $〈U_0,...,U_n〉=\{F∈2^X|F⊆ \bigcup_{i≤n}U_i\text{ and }F∩U_i\neq∅\text{ for }i≤n\}$,where $U_0,...,U_n$ are open sets in $X$.

Is there anyway of characterizing the sets $A\in 2^X$ such that $A\in \text{Cl}_{2^X}(<U_0,...,U_n>)$. I have the feeling that it should go somewhat like this:

$A\in \text{Cl}_{2^X}(<U_0,...,U_n>)$ if and only if $A\subseteq \bigcup_{i\leq n} \bar{U_i}$ and $A \cap \bar{U_i}$ for every $i\leq n$.

But I have got anywhere close to proving this, nor a can come up with a counter example. Do you know any characterization of this kind, or am I just looking in the wrong direction?

Thanks in advance.

Henno Brandsma
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Victor Gustavo May
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1 Answers1

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If $X$ is a $T_1$ space, then $\text{cl}\langle U_0,\dots,U_n\rangle=\langle \text{cl }U_0,\dots, \text{cl }U_n\rangle$.

To see that $\subseteq$ holds, first notice that clearly, $\langle U_0,\dots,U_n\rangle\subseteq \langle \text{cl }U_0,\dots,\text{cl }U_n\rangle$. Now we need to see that $ \langle \text{cl }U_0,\dots,\text{cl }U_n\rangle$ is closed. If $F \in 2^X$ is not an element of the former set, then either $F\cap (X\setminus \bigcup_{i\leq n}\text{cl }U_i)$ is not empty, or there exists $i\leq n$ such that $F\cap \text{cl } U_i=\emptyset$. In the first case, $\langle X, (X\setminus \bigcup_{i\leq n}\text{cl }U_i)\rangle$ is an open nhood of $F$ disjoint from $ \langle \text{cl }U_0,\dots,\text{cl }U_n\rangle$, and in the second case $\langle X\setminus \text{cl } U_i\rangle$ is an open nhood of $F$ disjoint from $ \langle \text{cl }U_0,\dots,\text{cl }U_n\rangle$.

This proved $\subseteq$. Notice that, up to now, we didn't need to use that $X$ is $T_1$.

For the other side, suppose $F \in \langle \text{cl }U_0,\dots, \text{cl }U_n\rangle$. Let $\langle V_0, \dots, V_m\rangle$ be a basic open nhood of $F$. We must see that it intersects $\langle U_0, \dots, U_n\rangle$. For each $i\leq n$, there exists $x_i \in F\cap \text{cl }U_i$. Notice that $\bigcup_{j\leq m} V_j$ is a open nhood of $x_i$, so there exists $y_i \in U_i\cap \bigcup_{j\leq m} V_j$. Also, there exists $z_j\in F\cap V_j\subseteq \bigcup_{i\leq n}\text{cl }U_i\cap V_j$ for each $j\leq m$. Given $j$, fix $i'$ such that $z_j\in \text{cl }U_{i'}\cap V_j$. Since $V_j$ is an open nhood of $z_j$, there exists $w_j \in V_j\cap U_{i'}\subseteq V_j \cap \bigcup_{i\leq n}U_i$.

Then $\{y_0, \dots, y_n, w_0, \dots, w_m\}$ is in the intersection.

This theorem is stated here. I believe this paper is a very nice source of information about the Vietoris topology.

  • By the way, the $T_1$ hypothesis is needed. I remember to have constructed a (simple) counterexample once, I will edit the answer if I remember. –  Jan 31 '20 at 06:21
  • It is the paper where the Vietoris topology was first defined. A classic. – Henno Brandsma Jan 31 '20 at 07:14
  • Hello Vinicius, this was really helpful. Thanks a lot! – Victor Gustavo May Jan 31 '20 at 07:33
  • A trivial example for a $T_0$ space where this fails: $X={0,1}$ in the Sierpinski topology with open $0$. Then $2^X={{1},X}$, taking $U={0}$ we have $\langle U \rangle = \emptyset$ while $\langle \overline{U} \rangle = \langle X \rangle = 2^X$ e.g. – Henno Brandsma Jan 31 '20 at 12:59