---
tag: 17S8AM75AD
kind: example
name: Regular open algebra
msc: ["06E10", 54A05]
instance_of: [75BPV95AAS]
uses: [833XAA2Q65, 0CQHFPGJ45]
sources:
  - url: https://en.wikipedia.org/wiki/Regular_open_set
---

## Example

Let $X$ be a [[topological space|0CQHFPGJ45]], and let $\operatorname{RO}(X)$ be the set of its [[regular open sets|833XAA2Q65]], the open sets $U$ with $U = \operatorname{int} \overline{U}$.
Order $\operatorname{RO}(X)$ by inclusion and define:

1. $U \wedge V = U \cap V$;
2. $U \vee V = \operatorname{int} \overline{U \cup V}$;
3. $\neg U = \operatorname{int}(X \setminus U)$;
4. $0 = \emptyset$ and $1 = X$.

Then $\operatorname{RO}(X)$ is a [[complete Boolean algebra|75BPV95AAS]], called the regular open algebra of $X$.
The supremum of a family $(U_i)$ is $\operatorname{int} \overline{\bigcup_i U_i}$, and the infimum is $\operatorname{int} \bigcap_i U_i$.

Check of the key points: the intersection of two regular open sets is regular open.
Also $\operatorname{int} \overline{W}$ is regular open for every open $W$, so the operations stay in $\operatorname{RO}(X)$.
For $U$ regular open, $\neg U$ is regular open, $U \cap \neg U = \emptyset$, and $\operatorname{int} \overline{U \cup \neg U} = X$, because $U \cup \neg U$ is dense in $X$.
For any family, $\operatorname{int} \overline{\bigcup_i U_i}$ is regular open, contains every $U_i$, and lies in every regular open $V$ that contains all $U_i$, since then $\overline{\bigcup_i U_i} \subseteq \overline{V}$.

## Remarks

The join is not the union: in $\R$, $(0, 1) \vee (1, 2) = (0, 2)$.
For $X = \R$, the algebra has no atoms, since every nonempty regular open set contains two disjoint open intervals.
Every Boolean algebra $B$ embeds densely into the regular open algebra of its [[Stone space|ZV38HA5PAQ]], and this gives the completion of $B$.
In Mathlib, the regular elements of a Heyting algebra are `Heyting.Regular`; for the open sets of $X$ they are the regular open sets.

## Explanations

### L1 human

The open sets of a space are almost a Boolean algebra, but the complement of an open set is not open.
The fix is to keep only the regular open sets, which have no cracks, and to fill in the cracks after each operation.
For example, on the line, $(0, 1)$ "or" $(1, 2)$ is $(0, 2)$: the missing point $1$ is filled in.
The result is a Boolean algebra in which every family of elements has a least upper bound.