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"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.

  • That's a pretty weird question...why would you think that $b\vee c$ has to be an element of $I$? If a set contains $2$ and $3$, does it have to contain $2+3=5$? – Eric Wofsey Jun 06 '18 at 22:28
  • But certainly that's not the only condition in that criterion, is it? See Ideals and Filters in a Boolean Algebra. If it was just that, any singleton would be an ideal... – amrsa Jun 07 '18 at 18:48
  • @amrsa What I posted, as I had mentioned, was only one of two criteria. The other was:$[b\in I \text{ and }c\in B]\rightarrow b\wedge c\in I$ –  Jun 07 '18 at 19:00
  • Right, but the criterion can't be just one of the conditions, but the conjunction of them. Each of those criteria gives different properties, and an ideal has both. – amrsa Jun 08 '18 at 08:13

2 Answers2

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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 ${\mathcal P}(\{0,1\})$.

Mario Carneiro
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  • Maybe I could ask a follow-up. Ideals and filters are introduced in a problem section in "Joy of Sets," which I am studying. He says, "We may define a binary relation on a boolean algebra $\mathcal{B}$ by $b\leq c$ iff $b=b\wedge c$. Then come a series of questions and the quote I added above. Does this binary relation hold in any boolean algebra in general? He does not mention union or intersection (I can see how they relate, as in your answer. Or $+$ as in a comment above.) If it doesn't hold, I don't see how I might otherwise approach the problems. Thanks. With regards, –  Jun 08 '18 at 16:41
  • For you convenience (if the link works) this starts on p. 26. I entered Ideals in the search window. Thanks yet again. https://books.google.com/books?id=hCv-vFu4jskC&q=ideal#v=onepage&q=ideal&f=false –  Jun 08 '18 at 16:47
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    @Andrew There is a canonical ordering relation that goes with any boolean algebra or lattice. Given a partial order $(B,\le)$, you can define the meet $a\wedge b$ to be the unique value $c$ such that $d\le c$ if and only if $d\le a$ and $d\le b$, when it exists, and similarly for join; when these exist for all pairs we get a lattice structure. Conversely, given a lattice structure $(B,\wedge,\vee)$, we can define $b\le c$ iff $b=b\wedge c$ and recover a partial order satisfying the meet and join conditions in this sense. So the two approaches are really the same. – Mario Carneiro Jun 09 '18 at 00:16
  • Thank's for taking the time to give such a helpful answer. –  Jun 09 '18 at 00:30
  • As for union and intersection, these are just the values of the meet and join in a powerset lattice. For sets $A,B\subseteq X$, it has $A\le B$ iff $A\subseteq B$, $A\wedge B=A\cap B$ and $A\vee B=A\cup B$. – Mario Carneiro Jun 09 '18 at 00:33
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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.

egreg
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  • Thank you for your most kind reply. I was under the impression that a Boolean algebra is a complemented, distributive lattice in which case $\vee$ and $\wedge$ are the lub and glb respectively. But maybe this premise and conclusion are wrong. Sorry to take so much of you time and impose on your generosity. With regards, –  Jun 09 '18 at 18:41