# example of Boolean algebras

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In $P(A)$, let $F(A)$ be the collection of all finite subsets of $A$, and $cF(A)$ the collection of all cofinite subsets of $A$. Then $F(A)\cup cF(A)$ is a Boolean algebra.

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More generally, any field of sets is a Boolean algebra. In particular, any sigma algebra $\sigma$ in a set is a Boolean algebra.

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(product     of algebras  ) Let $A$ and $B$ be Boolean algebras. Then $A\times B$ is a Boolean algebra, where

 $\displaystyle(a,b)\vee(c,d)$ $\displaystyle:=$ $\displaystyle(a\vee c,b\vee d),$ (1) $\displaystyle(a,b)\wedge(c,d)$ $\displaystyle:=$ $\displaystyle(a\wedge c,b\wedge d),$ (2) $\displaystyle(a,b)^{\prime}$ $\displaystyle:=$ $\displaystyle(a^{\prime},b^{\prime}).$ (3)
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More generally, if we have a collection of Boolean algebras $A_{i}$, indexed by a set $I$, then $\prod_{i\in I}A_{i}$ is a Boolean algebra, where the Boolean operations are defined componentwise.

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In particular, if $A$ is a Boolean algebra, then set of functions from some non-empty set $I$ to $A$ is also a Boolean algebra, since $A^{I}=\prod_{i\in I}A$.

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(subalgebras  ) Let $A$ be a Boolean algebra, any subset $B\subseteq A$ such that $0\in B$, $a^{\prime}\in B$ whenever $a\in B$, and $a\vee b\in B$ whenever $a,b\in B$ is a Boolean algebra. It is called a Boolean subalgebra of $A$. In particular, the homomorphic image   of a Boolean algebra homomorphism is a Boolean algebra.

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(quotient algebras  ) Let $A$ be a Boolean algebra and $I$ a Boolean ideal in $A$. View $A$ as a Boolean ring  and $I$ an ideal in $A$. Then the quotient ring $A/I$ is Boolean, and hence a Boolean algebra.

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Let $A$ be a set, and $R_{n}(A)$ be the set of all $n$-ary relations on $A$. Then $R_{n}(A)$ is a Boolean algebra under the usual set-theoretic operations. The easiest way to see this is to realize that $R_{n}(A)=P(A^{n})$, the powerset of the $n$-fold power of $A$.

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Let $X$ be a topological space and $A$ be the collection of all regularly open  sets in $X$. Then $A$ has a Boolean algebraic structure  . The meet and the constant operations follow the usual set-theoretic ones: $U\wedge V=U\cap V$, $0=\varnothing$ and $1=X$. However, the join $\wedge$ and the complementation ${}^{\prime}$ on $A$ are different. Instead, they are given by

 $\displaystyle U^{\prime}$ $\displaystyle:=$ $\displaystyle X-\overline{U},$ (4) $\displaystyle U\vee V$ $\displaystyle:=$ $\displaystyle(U\cup V)^{\prime\prime}.$ (5)
Title example of Boolean algebras ExampleOfBooleanAlgebras 2013-03-22 17:52:33 2013-03-22 17:52:33 CWoo (3771) CWoo (3771) 15 CWoo (3771) Example msc 06B20 msc 03G05 msc 06E05 msc 03G10