Boolean subalgebra

Boolean Subalgebras

Let A be a Boolean algebraMathworldPlanetmath and B a non-empty subset of A. Consider the following conditions:

  1. 1.

    if aB, then aB,

  2. 2.

    if a,bB, then abB,

  3. 3.

    if a,bB, then abB,

It is easy to see that, by de Morgan’s laws, conditions 1 and 2 imply conditions 1 and 3, and vice versa. Also, If B satisfies 1 and 2, then 1=aaB by picking an element aB. As a result, 0=1B as well.

A non-empty subset B of a Boolean algebra A satisfying conditions 1 and 2 (or equivalently 1 and 3) is called a Boolean subalgebra of A.


  • Every Boolean algebra contains a unique two-element Boolean algebra consisting of just 0 and 1.

  • If A contains a non-trivial element a (0), then 0,1,a,a form a four-element Boolean subalgebra of A.

  • Here is a concrete example. Let A be the field of all sets ( of a set X (the power setMathworldPlanetmath of X). Let B be the set of all finite and all cofinite subsets of A. Then B is a subalgebraPlanetmathPlanetmath of A.

  • Continuing from the example above, if X is equipped with a topology 𝒯, then the set of all clopen sets in X is a Boolean subalgebra of the algebraMathworldPlanetmath of all regular open sets of X.

Remarks. Let A be a Boolean algebra.

  • Arbitrary intersectionsMathworldPlanetmath of Boolean subalgebras of A is a Boolean subalgebra.

  • An arbitrary union of an increasing chain of Boolean subalgebras of A is a Boolean subalgebra.

  • A subalgebra B of A is said to be dense in A if it is dense as a subset of the underlying poset A. In other words, B is dense in A iff for any aA, there is a bB such that ba. It can be easily shown that B is dense in A is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to any one of the following:

    1. (a)

      for any aA, there is bB such that ab;

    2. (b)

      for any x,yA with xy, there is zB such that xzy.

    To see the last equivalence, notice first that by picking x=0 we see that B is dense, and conversely, if xy, then there is rB such that ry, so that xxry.

Subalgebras Generated by a Set

Let A be a Boolean algebra and X a subset of A. The intersection of all Boolean subalgebras (of A) containing X is called the Boolean subalgebra generated by X, and is denoted by X. As indicated by the remark above, X is a Boolean subalgebra of A, the smallest subalgebra containing X. If X=A, then we say that the Boolean algebra A itself is generated by X.

Proposition 1.

Every element of XA has a disjunctive normal formMathworldPlanetmath (DNF). In other words, if aX, then


where either aij or aij belongs to X.


Let B be the set of all elements written in DNF using elements X. Clearly BX. What we want to show is that XB. First, notice that the join of two elements in B is in B. Second, the complementPlanetmathPlanetmath of an element in B is also in B. We prove this in three steps:

  1. 1.

    If aB and either b or b is in X, then baB.

    If a=(a11a1ϕ(1))(a21a2ϕ(2))(an1anϕ(n)), then

    ba = b((a11a1ϕ(1))(a21a2ϕ(2))(an1anϕ(n)))
    = (ba11a1ϕ(1))(ba21a2ϕ(2))(ban1anϕ(n))

    which is in DNF using elements of X.

  2. 2.

    If a,bB, then abB.

    In the last expression of the previous step, notice that each term bai1aiϕ(i) can be written in DNF by iteratively using the result of the previous step. Hence, the join of all these terms is again in DNF (using elements of X).

  3. 3.

    If aB, then aB.

    If a=(a11a1ϕ(1))(a21a2ϕ(2))(an1anϕ(n)), then


    which is the meet of n elements in B. Consequently, by step 2, aB.

Since B is closed underPlanetmathPlanetmath join and complementation, B is a Boolean subalgebra of A. Since B contains X, XB. ∎

Similarly, one can show that X is the set of all elements in A that can be written in conjunctive normal formMathworldPlanetmath (CNF) using elements of X.

Corollary 1.

Let B,x be the Boolean subalgebra (of A) generated by a Boolean subalgebra B and an element xA. Then every element of B,x has the form


for some b1,b2B.


Let X be a generating set for B (pick X=B if necessary). By the propositionPlanetmathPlanetmath above, every element in B,x is the join of elements of the form a1a2an, where each ai or ai is in X or is x. By the absorption laws, the form is reduced to one of the three forms a, ax, or ax, where aB. Joins of elements in the form of the first kind is an element of B, since B is a subalgebra. Joins of elements in the form of the second kind is again an element in the form of the second kind (for (a1x)(a2x)=(a1a2)x), and similarly for the last case. Therefore, any element of B,x has the form a(a1x)(a2x). Since a=(ax)(ax), by setting b1=a1a and b2=a2a, we have the desired form. ∎


  • If a=(b1x)(b2x), then a=(b2x)(b1x).

  • Representing a in terms of (b1x)(b2x) is not unique. Actually, we have the following fact: (b1x)(b2x)=(c1x)(c2x) iff b2Δc2xb1c1, where Δ and are the symmetric differenceMathworldPlanetmathPlanetmath and biconditionalMathworldPlanetmath operators.


    (). The LHS can be rewritten as (b1b2)(b2x)(b1x). As a result, c2xb2x so that c2x=(c2x)x(b2x)x=b2x. Therefore, c2-b2=c2b2(c2x)b2(b2x)b2=b2xx. Similarly, b2-c2x. Taking the join and we get b2Δc2x. Dually, b1Δc1x, and complementing this to get x(b1Δc1)=b1c1.

    (). If b2-c2x, then b2c2=(b2-c2)c2xc2, so that (b2c2)-xc2-x, or (b2-x)(c2-x)c2-x, which implies that b2-xc2-x. Similarly, c2-xb2-x. Putting the two inequalitiesMathworldPlanetmath together and we have b2-x=c2-x. Since xb1c1 is equivalent to b1Δc1x (dual statements), we also have b1-x=c1-x or b1x=c1x. Combining (via meet) this equality with the last one, we get (b1x)(b2x)=(c1x)(c2x). ∎

Title Boolean subalgebra
Canonical name BooleanSubalgebra
Date of creation 2013-03-22 17:57:55
Last modified on 2013-03-22 17:57:55
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 9
Author CWoo (3771)
Entry type Definition
Classification msc 06E05
Classification msc 03G05
Classification msc 06B20
Classification msc 03G10
Synonym dense subalgebra
Defines dense Boolean subalgebra