Criterion for an Ideal in a Boolean algebra

Question

"Joy of Sets" on p.26 states [as one criterion] that a non-empty set $I$ of a Boolean algebra $B$ is called an ideal if and only if:

$b, c\in I\rightarrow b\vee c\in I$

If $b$ and $c$ are in $I$, how is it possible that $b\vee c$ not be an element of $I$?

Thanks

EDIT The text says:

It is possible to define a boolean algebra as a poset satisfying certain conditions. In this case $b\vee c$ turns out to be the unique least upper bound of $b$ and $c$, and $b\wedge c$ is the unique greatest lower bound.

Then the text goes into a set of problems on Ideals and Filters from which I asked my question.

Answer 1

In a Boolean algebra, or a lattice in general, $b\vee c$ need not be equal to $b$ or $c$. Only when the underlying partial order is total do we have this in general. For example, in the Boolean algebra of subsets of $\{0,1\}$, $\{0\}\vee\{1\}=\{0\}\cup\{1\}=\{0,1\}$ which is not equal to either $\{0\}$ or $\{1\}$. So the set $\{\emptyset,\{0\},\{1\}\}$ is a subset which is downward closed but is not an ideal of ${\mathcal P}(\{0,1\})$.

Answer 2

As you can see in Definitions of Boolean algebras, Boolean algebras can be defined in several ways. If the approach you're using is with two operation $\wedge$ and $\vee$, plus the complement, then you can show that the following conditions are equivalent, for two elements $a$ and $b$:

1. $a\wedge b=a$
2. $a\vee b=b$

If you define $a\le b$ when those equivalent conditions are satisfied, then you get an order relation (a partial order, in general), with respect to which $a\wedge b$ is the greatest lower bound of $\{a,b\}$ and $a\vee b$ is the lowest upper bound of $\{a,b\}$.

Thus, in general, $b\vee c$ need not be equal to either $b$ or $c$: it will be one of them if and only if either $b\le c$ or $c\le b$.

A nonempty subset $I$ of the Boolean algebra $B$ called an ideal when it satisfies

1. if $a\in I$ and $b\le a$, then $b\in I$;
2. if $b,c\in I$, then $b\vee c\in I$.

If you define two new operations

• $a+b=(a\vee b)\wedge(a\wedge b)'$
• $ab=a\wedge b$

then $B$ becomes a ring. More precisely a Boolean ring, where each element satisfies $a^2=a$. An ideal in the previous sense is exactly an ideal in ring theoretical sense.