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Let $(X, \tau)$ be a topological space and $\mathcal K(X)$ the collection of all non-empty closed subsets of $X$. Let $$ U^+ := \{K \in \mathcal K(X) \mid K \cap U \neq \emptyset\} \quad \text{and} \quad U^- := \{K \in \mathcal K(X) \mid K \subset U\} \quad \forall U \in \tau. $$

The Vietoris topology $\mathcal T$ of $\mathcal K(X)$ is defined as the one generated by the subbase $\mathcal B :=\{U^+, U^- \mid U \in \tau\}$. Clearly, $$ \begin{align} \mathcal B_1 &:= \left \{ \bigg ( \bigcap_{i=1}^m U_i^+ \bigg ) \cap \bigg ( \bigcap_{j=1}^n V_i^- \bigg ) \,\middle\vert\, m,n \in \mathbb N^*;U_1, \ldots, U_m,V_1, \ldots, V_n \in \tau \right\} \\ &= \left \{ \bigg ( \bigcap_{i=1}^m U_i^+ \bigg ) \cap \bigg ( \bigcap_{j=1}^n V_i \bigg )^- \,\middle\vert\, m,n \in \mathbb N^*;U_1, \ldots, U_m,V_1, \ldots, V_n \in \tau \right\} \\ &= \left \{ \bigg ( \bigcap_{i=1}^m U_i^+ \bigg ) \cap V^- \,\middle\vert\, m \in \mathbb N^*;U_1, \ldots, U_m,V \in \tau \right\} \end{align} $$ is a base of $\mathcal T$. It is mentioned in this answer that $$ \mathcal B_2 := \left \{ \bigg ( \bigcap_{i=1}^m U_i^+ \bigg ) \cap \bigg ( \bigcup_{i=1}^m U_i \bigg )^- \,\middle\vert\, m \in \mathbb N^*;U_1, \ldots, U_m \in \tau \right\} $$ is also a base of $\mathcal T$. Below is my attempt to verify this claim.

My attempt: It's clear that $\mathcal B_2 \subset \mathcal B_1$. Notice that $$ (U^+ \cap X^+) \cap (U \cup X)^- = U^+ \quad \text{and} \quad (U^+ \cap \emptyset^+) \cap (U \cup \emptyset)^- = U^-. $$

So $\mathcal B \subset \mathcal B_2$. It suffices to prove that $\mathcal B_2$ is stable under finite intersection. Let $m, n \in \mathbb N^*;U_1, \ldots, U_m, V_1, \ldots, V_n \in \tau$. We have $$ \begin{align} & \bigg ( \bigcap_{i=1}^m U_i^+ \bigg ) \cap \bigg ( \bigcup_{i=1}^m U_i \bigg )^- \cap \bigg ( \bigcap_{i=1}^n V_i^+ \bigg ) \cap \bigg ( \bigcup_{i=1}^n V_i \bigg )^- \\ =& \bigg ( \bigcap_{i=1}^m U_i^+ \cap \bigcap_{i=1}^n V_i^+ \bigg ) \cap \bigg [ \bigg (\bigcup_{i=1}^m U_i \bigg ) \cap \bigg ( \bigcup_{i=1}^n V_i \bigg ) \bigg ]^- \\ =& \bigg [ \bigcup_{\substack{i= 1, \ldots, m \\ j = 1, \ldots, n}} (U^+_i \cap V^+_j) \bigg ] \cap \bigg [ \bigcup_{\substack{i= 1, \ldots, m \\ j = 1, \ldots, n}} (U_i \cap V_j) \bigg ]^-. \end{align} $$

However, I'm stuck because $U^+_i \cap V^+_j \neq (U_i \cap V_j)^+$. Could you please elaborate on how to proceed?

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1 Answers1

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Let $(X, \tau)$ be a topological space. A collection $\mathcal B \subset \tau$ is called a base of $\tau$ if every open set can be represented as a union of some subfamily of $\mathcal B$. Equivalently, a family $\mathcal{B}$ of subsets of $X$ is a base for the topology $\tau$ if and only if $\mathcal{B} \subseteq \tau$ and for every $U \in \tau$ and $x \in U$ there is $B \in \mathcal{B}$ such that $x \in B \subseteq U$.

We want to prove that $\mathcal B_2$ is a base of our Vietoris topology $\mathcal T$. Clearly, $\mathcal B_2 \subset \mathcal B_1 \subset \mathcal T$. Now we fix $x \in U \in \mathcal T$. Because $\mathcal B$ is a subbase of $\mathcal T$, there is $V \in \mathcal B$ such that $x \in V$. The claim then follows from $\mathcal B \subset \mathcal B_2$.

Analyst
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