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On page 87, Introduction to Boolean Algebras,Steven Givant,Paul Halmos(2000)

Give an example of a subalgebra $B$ of a Boolean algebra $A$ and of a subset $E$ of $B$ such that $E$ has a supremum in $B$ but not in $A$.

The example that occured to me is that $A = [0 ,1) \cup (1, 3]$, $B =[0,1) \cup[2,3]$ and $E =[0,1)$, which has a supremum in $B$, but dosn't have one in $A$.

However, I'm confused with the answer.

Consider the field $A$ of finite and cofinite sets of integers, and the class $B$ of those subsets of integers that are either finite sets of even integers, or else the complements of such sets.

I can't follow it. It seems to me if $E = \{ \text{finite subset of even integers}\}$, then it doesn't have a supremum in $B$.

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1 Answer 1

up vote 1 down vote accepted

Your $A$ and $B$ aren’t Boolean algebras: they’re just linear orders.

Let $\mathscr{A}$ be the field of finite and cofinite sets of integers. Let $\Bbb E$ be the set of even integers, and let

$$\mathscr{B}=\{A\subseteq\Bbb Z:\text{either }A\text{ or }\Bbb Z\setminus A\text{ is a finite subset of }\Bbb E\}\;.$$

Let $\mathscr{E}=\{B\in\mathscr{B}:B\text{ is finite}\}$. Then $\sup_{\mathscr{B}}\mathscr{E}=\Bbb Z$: $\Bbb Z$ is in fact the unique member of $\mathscr{B}$ that contains every element of $\mathscr{E}$. $\mathscr{E}$ has no supremum in $\mathscr{A}$, however: if $A\in\mathscr{A}$ is an upper bound for $\mathscr{E}$ in $\mathscr{A}$, then $A$ contains all but finitely many of the odd integers, so we may choose any odd integer $n\in A$ and observe that $A\setminus\{n\}$ is a smaller upper bound for $\mathscr{E}$.

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