It is a student exercise that no group can be represented as a union of the set theory of its two appropriate subgroups. The same can be shown for Boolean algebras. On the other hand, it is not difficult to prove that any infinite Boolean algebra $ mathcal {A} $ can be covered by your $ k $ appropriate 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 each $ i <k $. However, I'm not sure if $ k = 4 $ I really should be excluded here. So, my question is this:

**Q**: Leave $ mathcal {A} $ Be an infinite Boolean algebra. Do your own subalgebras always exist? $ mathcal {A} _0, mathcal {A} _1, mathcal {A} _2, mathcal {A} _3 $ such that $ mathcal {A} = bigcup_ {i <4} mathcal {A} _i $ Y $ mathcal {A} _i not subseteq bigcup_ {j neq i} mathcal {A} _j $ for each $ i <4 $?

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

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