It is a student exercise that no group can be represented as a set-theoretic union of its two proper subgroups. The same also can be shown for Boolean algebras. On the other hand, it's not hard to show that any infinite Boolean algebra $\mathcal{A}$ can be covered by its $k$ proper subalgebras $\mathcal{A}_0,\ldots,\mathcal{A}_{k-1}$, where $k\in\mathbb{N}\setminus\{0,1,2,4\}$, such that $\mathcal{A}_i\not\subseteq\bigcup_{j\neq i}\mathcal{A}_j$ for every $i<k$. However, I am not sure if $k=4$ should really be excluded here. Thus, my question is the following:

**Q**: Let $\mathcal{A}$ be an infinite Boolean algebra. Do there always exist its proper subalgebras $\mathcal{A}_0,\mathcal{A}_1,\mathcal{A}_2,\mathcal{A}_3$ such that $\mathcal{A}=\bigcup_{i<4}\mathcal{A}_i$ and $\mathcal{A}_i\not\subseteq\bigcup_{j\neq i}\mathcal{A}_j$ for every $i<4$?

I suspect that a careful (and tedious) analysis of cases could show that the answer is negative (likewise for $k=2$), but perhaps there is some clever proof (or refutal) of it. I am also asking you about possible references to papers or books in which such problems are studied.

(I asked the same question at the Math Stack Exchange, but the interest was literally null.)

The same also can be shown for Boolean algebras": I wouldn't say it can "also be shown", rather it's a particular case of the group case, just applied to the underlying additive group. $\endgroup$ – YCor Mar 24 at 6:47