6
$\begingroup$

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.)

$\endgroup$
  • 1
    $\begingroup$ MathSE link: math.stackexchange.com/questions/3158159/… $\endgroup$ – YCor Mar 23 at 18:06
  • 2
    $\begingroup$ "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
4
$\begingroup$

It's a good idea for such questions to think about the finite case, since it indeed allows to solve the general case.

Proposition: every unital Boolean algebra $A$ of (possibly infinite) cardinal $\ge 16$ (i.e., whose spectrum has cardinal $\ge 4$) is a non-redundant union of 4 unital subalgebras.

Proof: by pull-back, it's enough to suppose that $A=\mathbf{Z}/2\mathbf{Z}^4$. Its unital subalgebras are the $A_R$, where $R$ ranges over equivalence relations on $\{1,2,3,4\}$, and $A_R$ is the set of quadruples that are constant on $R$-equivalence classes.

Define $A_{ij}=\{a\in A:a_i=a_j\}$ and $A_{ijk}=\{a\in A:a_i=a_j=a_k\}$.

We consider the four unital subalgebras $A_{12}$, $A_{13}$, $A_{14}$, $A_{234}$. If we forget the first, the three others do not cover $A$, since $(0,0,1,1)\notin A_{13}\cup A_{14}\cup A_{234}$; similarly for the second and third; also forgetting the fourth one does not give a cover since $(0,1,1,1)\notin A_{12}\cup A_{13}\cup A_{14}$. Still the four form a cover: indeed the only elements not in $A_{12}\cup A_{13}\cup A_{14}$ are $(0,1,1,1)$ and $(1,0,0,0)$ and they belong to $A_{234}$. This concludes that $A=\mathbf{Z}/2\mathbf{Z}^4$ is the non-redundant union of $A_{12}$, $A_{13}$, $A_{14}$, $A_{234}$.

If take for granted your claim for other values [edit: justified by Emil Jeřábek as a comment below, by generalizing the above argument], this yields:

Corollary: for every $k\in\mathbf{N}\smallsetminus\{0,2\}$, every infinite unital Boolean algebra can be written as non-redundant union of $k$ unital subalgebras.

(No reason to exclude 1 by restricting to proper subalgebras: assuming non-redundancy assumption excludes the whole subalgebra for $k>1$.)


Addendum: a classical lemma of B-H. Neumann (for groups) implies that whenever a Boolean algebra $A$ is written as non-redundant union of $k$ subalgebras ($k<\infty$), then each of those subalgebras has finite index in $(A,+)$.

Also, in a unital Boolean algebra $A$, every subalgebra of finite index contains an ideal of finite index (with some explicit bound). And hence is obtained by pullback from a subalgebra in a finite quotient Boolean algebra.

$\endgroup$
  • $\begingroup$ What do you mean by unital? $\endgroup$ – Emil Jeřábek Mar 24 at 11:40
  • $\begingroup$ @EmilJeřábek I mean, with a unit (I'm aware that most people assume that Boolean algebra are unital, that homomorphisms are required to be unital, etc., but I wrote it to avoid ambiguity, and also because for some statements, unital doesn't matter.) $\endgroup$ – YCor Mar 24 at 11:42
  • 1
    $\begingroup$ I don’t understand. The unit of a Boolean algebra is definable from $\lor$ (or $\land$) and complement, hence all Boolean algebras are automatically unital, and Boolean algebra homomorphisms preserve unity even you don’t assume it. What would be your example of a nonunital Boolean algebra? $\endgroup$ – Emil Jeřábek Mar 24 at 17:10
  • 1
    $\begingroup$ I see. Thank you for the explanation. $\endgroup$ – Emil Jeřábek Mar 24 at 19:15
  • 2
    $\begingroup$ I think you do not need to take anything for granted: your construction readily generalizes to all finite $k\ge3$ (for $|A|\ge2^k$), using $A_{1,2},\dots,A_{1,k},A_{2,\dots,k}$. $\endgroup$ – Emil Jeřábek Mar 25 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.