Boolean subalgebra
Boolean Subalgebras
Let $A$ be a Boolean algebra^{} and $B$ a nonempty subset of $A$. Consider the following conditions:

1.
if $a\in B$, then ${a}^{\prime}\in B$,

2.
if $a,b\in B$, then $a\vee b\in B$,

3.
if $a,b\in B$, then $a\wedge b\in B$,
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 $1=a\vee {a}^{\prime}\in B$ by picking an element $a\in B$. As a result, $0={1}^{\prime}\in B$ as well.
A nonempty subset $B$ of a Boolean algebra $A$ satisfying conditions 1 and 2 (or equivalently 1 and 3) is called a Boolean subalgebra of $A$.
Examples.

•
Every Boolean algebra contains a unique twoelement Boolean algebra consisting of just $0$ and $1$.

•
If $A$ contains a nontrivial element $a$ ($\ne 0$), then $0,1,a,{a}^{\prime}$ form a fourelement Boolean subalgebra of $A$.

•
Here is a concrete example. Let $A$ be the field of all sets (http://planetmath.org/RingOfSets) of a set $X$ (the power set^{} of $X$). Let $B$ be the set of all finite and all cofinite subsets of $A$. Then $B$ is a subalgebra^{} of $A$.

•
Continuing from the example above, if $X$ is equipped with a topology $\mathcal{T}$, then the set of all clopen sets in $X$ is a Boolean subalgebra of the algebra^{} of all regular open sets of $X$.
Remarks. Let $A$ be a Boolean algebra.

•
Arbitrary intersections^{} 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 $a\in A$, there is a $b\in B$ such that $b\le a$. It can be easily shown that $B$ is dense in $A$ is equivalent^{} to any one of the following:

(a)
for any $a\in A$, there is $b\in B$ such that $a\le b$;

(b)
for any $x,y\in A$ with $x\le y$, there is $z\in B$ such that $x\le z\le y$.
To see the last equivalence, notice first that by picking $x=0$ we see that $B$ is dense, and conversely, if $x\le y$, then there is $r\in B$ such that $r\le y$, so that $x\le x\vee r\le y$.

(a)
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 $\u27e8X\u27e9$. As indicated by the remark above, $\u27e8X\u27e9$ is a Boolean subalgebra of $A$, the smallest subalgebra containing $X$. If $\u27e8X\u27e9=A$, then we say that the Boolean algebra $A$ itself is generated by $X$.
Proposition 1.
Every element of $\mathrm{\u27e8}X\mathrm{\u27e9}\mathrm{\subseteq}A$ has a disjunctive normal form^{} (DNF). In other words, if $a\mathrm{\in}\mathrm{\u27e8}X\mathrm{\u27e9}$, then
$$a=\underset{i=1}{\overset{n}{\bigvee}}\underset{j=1}{\overset{\varphi (i)}{\bigwedge}}{a}_{ij}=({a}_{11}\wedge \mathrm{\cdots}\wedge {a}_{1\varphi (1)})\vee ({a}_{21}\wedge \mathrm{\cdots}\wedge {a}_{2\varphi (2)})\vee \mathrm{\cdots}\vee ({a}_{n1}\wedge \mathrm{\cdots}\wedge {a}_{n\varphi (n)}),$$ 
where either ${a}_{i\mathit{}j}$ or ${a}_{i\mathit{}j}^{\mathrm{\prime}}$ belongs to $X$.
Proof.
Let $B$ be the set of all elements written in DNF using elements $X$. Clearly $B\subseteq \u27e8X\u27e9$. What we want to show is that $\u27e8X\u27e9\subseteq B$. First, notice that the join of two elements in $B$ is in $B$. Second, the complement^{} of an element in $B$ is also in $B$. We prove this in three steps:

1.
If $a\in B$ and either $b$ or ${b}^{\prime}$ is in $X$, then $b\wedge a\in B$.
If $a=({a}_{11}\wedge \mathrm{\cdots}\wedge {a}_{1\varphi (1)})\vee ({a}_{21}\wedge \mathrm{\cdots}\wedge {a}_{2\varphi (2)})\vee \mathrm{\cdots}\vee ({a}_{n1}\wedge \mathrm{\cdots}\wedge {a}_{n\varphi (n)})$, then
$b\wedge a$ $=$ $b\wedge (({a}_{11}\wedge \mathrm{\cdots}\wedge {a}_{1\varphi (1)})\vee ({a}_{21}\wedge \mathrm{\cdots}\wedge {a}_{2\varphi (2)})\vee \mathrm{\cdots}\vee ({a}_{n1}\wedge \mathrm{\cdots}\wedge {a}_{n\varphi (n)}))$ $=$ $(b\wedge {a}_{11}\wedge \mathrm{\cdots}\wedge {a}_{1\varphi (1)})\vee (b\wedge {a}_{21}\wedge \mathrm{\cdots}\wedge {a}_{2\varphi (2)})\vee \mathrm{\cdots}\vee (b\wedge {a}_{n1}\wedge \mathrm{\cdots}\wedge {a}_{n\varphi (n)})$ which is in DNF using elements of $X$.

2.
If $a,b\in B$, then $a\wedge b\in B$.
In the last expression of the previous step, notice that each term $b\wedge {a}_{i1}\wedge \mathrm{\cdots}\wedge {a}_{i\varphi (i)}$ 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.
If $a\in B$, then ${a}^{\prime}\in B$.
If $a=({a}_{11}\wedge \mathrm{\cdots}\wedge {a}_{1\varphi (1)})\vee ({a}_{21}\wedge \mathrm{\cdots}\wedge {a}_{2\varphi (2)})\vee \mathrm{\cdots}\vee ({a}_{n1}\wedge \mathrm{\cdots}\wedge {a}_{n\varphi (n)})$, then
$${a}^{\prime}=({a}_{11}^{\prime}\vee \mathrm{\cdots}\vee {a}_{1\varphi (1)}^{\prime})\wedge ({a}_{21}^{\prime}\vee \mathrm{\cdots}\vee {a}_{2\varphi (2)}^{\prime})\wedge \mathrm{\cdots}\wedge ({a}_{n1}^{\prime}\vee \mathrm{\cdots}\vee {a}_{n\varphi (n)}^{\prime}),$$ which is the meet of $n$ elements in $B$. Consequently, by step 2, ${a}^{\prime}\in B$.
Since $B$ is closed under^{} join and complementation, $B$ is a Boolean subalgebra of $A$. Since $B$ contains $X$, $\u27e8X\u27e9\subseteq B$. ∎
Similarly, one can show that $\u27e8X\u27e9$ is the set of all elements in $A$ that can be written in conjunctive normal form^{} (CNF) using elements of $X$.
Corollary 1.
Let $\mathrm{\u27e8}B\mathrm{,}x\mathrm{\u27e9}$ be the Boolean subalgebra (of $A$) generated by a Boolean subalgebra $B$ and an element $x\mathrm{\in}A$. Then every element of $\mathrm{\u27e8}B\mathrm{,}x\mathrm{\u27e9}$ has the form
$$({b}_{1}\wedge x)\vee ({b}_{2}\wedge {x}^{\prime})$$ 
for some ${b}_{\mathrm{1}}\mathrm{,}{b}_{\mathrm{2}}\mathrm{\in}B$.
Proof.
Let $X$ be a generating set for $B$ (pick $X=B$ if necessary). By the proposition^{} above, every element in $\u27e8B,x\u27e9$ is the join of elements of the form ${a}_{1}\wedge {a}_{2}\wedge \mathrm{\cdots}\wedge {a}_{n}$, where each ${a}_{i}$ or ${a}_{i}^{\prime}$ is in $X$ or is $x$. By the absorption laws, the form is reduced to one of the three forms $a$, $a\wedge x$, or $a\wedge {x}^{\prime}$, where $a\in B$. 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 $({a}_{1}\wedge x)\vee ({a}_{2}\wedge x)=({a}_{1}\vee {a}_{2})\wedge x$), and similarly for the last case. Therefore, any element of $\u27e8B,x\u27e9$ has the form $a\vee ({a}_{1}\wedge x)\vee ({a}_{2}\wedge {x}^{\prime})$. Since $a=(a\wedge x)\vee (a\wedge {x}^{\prime})$, by setting ${b}_{1}={a}_{1}\vee a$ and ${b}_{2}={a}_{2}\vee a$, we have the desired form. ∎
Remarks.

•
If $a=({b}_{1}\wedge x)\vee ({b}_{2}\wedge {x}^{\prime})$, then ${a}^{\prime}=({b}_{2}^{\prime}\wedge x)\vee ({b}_{1}^{\prime}\wedge {x}^{\prime})$.

•
Representing $a$ in terms of $({b}_{1}\wedge x)\vee ({b}_{2}\wedge {x}^{\prime})$ is not unique. Actually, we have the following fact: $({b}_{1}\wedge x)\vee ({b}_{2}\wedge {x}^{\prime})=({c}_{1}\wedge x)\vee ({c}_{2}\wedge {x}^{\prime})$ iff ${b}_{2}\mathrm{\Delta}{c}_{2}\le x\le {b}_{1}\leftrightarrow {c}_{1}$, where $\mathrm{\Delta}$ and $\leftrightarrow $ are the symmetric difference^{} and biconditional^{} operators.
Proof.
$(\Rightarrow )$. The LHS can be rewritten as $({b}_{1}\vee {b}_{2})\wedge ({b}_{2}\vee x)\wedge ({b}_{1}\vee {x}^{\prime})$. As a result, ${c}_{2}\wedge {x}^{\prime}\le {b}_{2}\vee x$ so that ${c}_{2}\vee x=({c}_{2}\wedge {x}^{\prime})\vee x\le ({b}_{2}\vee x)\vee x={b}_{2}\vee x$. Therefore, ${c}_{2}{b}_{2}={c}_{2}\wedge {b}_{2}^{\prime}\le ({c}_{2}\vee x)\wedge {b}_{2}^{\prime}\le ({b}_{2}\vee x)\wedge {b}_{2}^{\prime}={b}_{2}^{\prime}\wedge x\le x$. Similarly, ${b}_{2}{c}_{2}\le x$. Taking the join and we get ${b}_{2}\mathrm{\Delta}{c}_{2}\le x$. Dually, ${b}_{1}\mathrm{\Delta}{c}_{1}\le {x}^{\prime}$, and complementing this to get $x\le {({b}_{1}\mathrm{\Delta}{c}_{1})}^{\prime}={b}_{1}\leftrightarrow {c}_{1}$.
$(\Leftarrow )$. If ${b}_{2}{c}_{2}\le x$, then ${b}_{2}\vee {c}_{2}=({b}_{2}{c}_{2})\vee {c}_{2}\le x\vee {c}_{2}$, so that $({b}_{2}\vee {c}_{2})x\le {c}_{2}x$, or $({b}_{2}x)\vee ({c}_{2}x)\le {c}_{2}x$, which implies that ${b}_{2}x\le {c}_{2}x$. Similarly, ${c}_{2}x\le {b}_{2}x$. Putting the two inequalities^{} together and we have ${b}_{2}x={c}_{2}x$. Since $x\le {b}_{1}\leftrightarrow {c}_{1}$ is equivalent to ${b}_{1}\mathrm{\Delta}{c}_{1}\le {x}^{\prime}$ (dual statements), we also have ${b}_{1}{x}^{\prime}={c}_{1}{x}^{\prime}$ or ${b}_{1}\wedge x={c}_{1}\wedge x$. Combining (via meet) this equality with the last one, we get $({b}_{1}\wedge x)\wedge ({b}_{2}\wedge {x}^{\prime})=({c}_{1}\wedge x)\wedge ({c}_{2}\wedge {x}^{\prime})$. ∎
Title  Boolean subalgebra 

Canonical name  BooleanSubalgebra 
Date of creation  20130322 17:57:55 
Last modified on  20130322 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 