I was reading this page and, in the third part of the first remark I found the definition of dense sub-algebra of a Boolean algebra. It is stated that there are various equivalent definitions of this property but I am not able to find an argument to prove equivalence (and the argument given in the link is not convincing me).
I copy here part of the text in the link (with minor changes) to make the question self-contained:
A sub-algebra $B$ of a Boolean algebra $A$ is said to be dense in $A$ if for any $0\neq a\in A$, there is a $0\neq b\in B$ such that $b\le a$. It can be easily shown that $B$ is dense in $A$ is equivalent to say that for any $x,y\in A$ with $x\lneq y$, there is $z\in B$ such that $x\le z\le y$.
Could you prove the equivalence in the above sentence?
Of course it is trivial to see that the second condition implies the first (just let $x=0$ in the second condition and you get the first one).
The claim is not true.
I posted a wrong proof, realized that the proof in wrong and then immediately found a counterexample.
Let $A=2^{\mathbb N}$ and let $B$ be the system of all finite and cofinite (cofinite: with finite complement) subsets of $\mathbb N$. Then $B$ is a subalgebra of $A$. Clearly, it satisfies the first condition and the second condition fails. To see this, let $x$ be the set of all even numbers and let $y=x\cup\{1\}$.