# suprema in Boolean algebras

Let $(\mathbb{A},\lor, \land, ', 0,1)$ be a boolean algebra. Recall that a subalgebra $\mathbb{B}$ of a complete Boolean algebra $\mathbb{A}$ is a complete subalgebra, if for each subset $E\subseteq \Bbb{B}$, finite or infinite, $\lor^{\Bbb{A}}E\in \Bbb{B}.$

I have found that the set of all clopen subsets of $\beta\Bbb{N}$, which is denoted by $CLO(\beta\Bbb{N})$, is a subalgebra of $\mathcal{P}(\beta\Bbb{N})$ which is not a complete subalgebra but is a complete boolean algebra. But I think there is rather easy example. So my first question is:

1. Find an easy example of a complete Boolean algebra $\Bbb{A}$ with a subalgebra $\Bbb{B}$ which is not a complete subalgebra but is complete as a Boolean algebra.

For another question let me recall that a subalgebra $\Bbb{B}$ of a Boolean algebra $\Bbb{A}$ is regular if for any subset $E$ of $\Bbb{B}$ which has $\lor^{\Bbb{B}}E$ as its supremum in $\Bbb{B}$, then $\lor^{\Bbb{A}}E$ exists in $\Bbb{A}$ and $\lor^{\Bbb{A}}E=\lor^{\Bbb{B}}E$. So here is the second question:

1. Find an example of a Boolean algebra $\Bbb{A}$ with a subalgebra $\Bbb{B}$ which has a nonempty set $E\subseteq \Bbb{B}$ such that $E$ has a supremum in $\Bbb{B}$ but not in $\Bbb{A}$.

Thank you

• $\operatorname{clop}({\beta \mathbb{N}})$ is complete as $\beta \mathbb{N}$ is extremally disconnected. – Henno Brandsma Nov 12 '17 at 11:10