Definition (Dense Boolean SubAlgebra): Let $A$ be a Boolean algebra and $B\subseteq A$ a Boolean subalgebra. We say that $B$ is dense in $A$ if for every $a\in A$ there exist $b\in B$ such that $b \leq a$.
I have found here (https://planetmath.org/booleansubalgebra) stated the following claim (with proof?):
Claim: $B$ is dense in $A$ iff for every $a_1\leq a_2\in A$, there exists $b\in B$ such that $a_1 \leq b \leq a_2$.
The direction $(\Leftarrow)$ is clear, taking $a_1=0$ and $a_2=a$. For $(\Rightarrow)$ the proposed proof is the following:
Proof (?) of ($\Rightarrow$) : Assume $a_1\leq a_2$. Since $B$ is dense in $A$ by hp., we can find $r\in B$ such that $r\leq a_2$. We conclude by picking $b= r\vee a_1$ so that $a_1 \leq b \leq a_2$ as desired.
The problem I see with this proof is that the candidate $b= r\vee a_1$ is not guaranteed to be in $B$ since $a_1$ is not in $B$.
QUESTION: Is the property stated in the link actually true?
- ) If yes, is the proof above correct and am I missing something obvious?
- ) If no, what is the correct(ed) property that they might have had in their mind?
The statement is simply incorrect (also, the definition of "dense subalgebra" is wrong: you need to require both $a$ and $b$ to be nonzero, or else trivially every subalgebra is dense, by taking $b=0$). First of all, it would trivially imply that $B$ has to be all of $A$ since you could have $a_1=a_2$. Even if you modify the statement to require $a_1<a_2$ it is still incorrect. For instance, consider the example where $A$ is the power set of some infinite set and $B$ is the subalgebra of finite or cofinite sets. Then $B$ is dense in $A$ but if $S\subseteq T$ are infinite coinfinite sets then there is no element of $B$ in between them.
A correct statement along these lines would be that $B$ is dense in $A$ iff whenever $a_1<a_2$ for $a_1\in B$ and $a_2\in A$, then there exists $b\in B$ such that $a_1<b\leq a_2$. To prove the forward direction for this statement, by density of $B$ there exists $c\in B$ such that $0<c\leq a_2\wedge \neg a_1$, and then you can take $b=a_1\vee c$.