Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $B$.

share|cite|improve this question
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 $\mathscr{E}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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