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I'm doing Exercise 11 from chapter 2 of "Introduction to Mathematical Logic" from Cori and Lascar. In this context, 'dense' means for all $a < b$, there's a $c$ with $a < c < b$.

The solutions on the back of the book proceed as follows: (paraphrased)

If the boolean algebra is dense, for any $a$, there's a $b$ with $0 < b < a$, so $a$ can't be an atom, and there are no atoms.

Assume there are no atoms, let $a < b$. $a + b$ is not zero, and not an atom, so there's a $d$ with $$0 < d < a + b$$ Set $c = a \cup d$. "It's immediate to verify that we then have $a < c < b$, so the boolean algebra is dense."

Now, I get that if $x < y$, $x < y \leq y \cup z$, $z \leq y \cup z$, so $x \cup z \leq y \cup z$ since it's the least upper bound. However, this isn't a strict inequality -- just set $z = 1$ or $x \cup y$. Then, I get how we got

$$a = a \cup 0 \leq a \cup d \leq a \cup (a + b) = a + (a + b) + a(a+b) = b + a + ab = a \cup b = b$$ Where the last equality holds since $a < b$, but I can't figure out why the equalities would be strict here (In other words, why $a \neq a \cup d$ and $a \cup d \neq b$).

atreju
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  • Because $+$ is an abelian group operation, for which every element is its own inverse, the cancellation law holds: $(\forall a, x, y)$ $(a + x = a + y \iff x = y).$ Equivalently: $(\forall a, x, y)$ $(a + x \ne a + y \iff x \ne y).$ We are given $d \ne 0$ and $d \ne a + b,$ therefore $a + d \ne a + 0 = a$ and $a + d \ne a + (a + b) = (a + a) + b = b.$ As observed in Eric Wofsey's answer, $c = a + d.$ Therefore $c \ne a$ and $c \ne b.$ – Calum Gilhooley Sep 25 '22 at 20:48

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Note first that $a(a+b)=a+ab=a+a=0$ since $a\leq b$. It follows that $ad=ad(a+b)=0$ since $d\leq a+b$. Thus $a\cup d=a+d$. If we had $a\cup d=a$, this would thus mean $d=0$, and if we had $a\cup d=b$, this would mean $d=a+b$.

(In general, a good way to think about these things is to imagine that the elements of your Boolean algebra are sets, with $+$ as symmetric difference and $\cdot$ as intersection. When $a\subseteq b$, the symmetric difference $a+b$ will be just $b\setminus a$, and so the idea is that you're forming $c$ by taking part but not all of the difference $b\setminus a$ and adding it onto $a$, to get a set strictly between $a$ and $b$ (draw a Venn diagram if this is not immediately obvious to you). Of course, this on its own does not prove the statement in a general Boolean algebra, but it can give you some clues about how you might try to prove it. And in fact, by Stone's representation theorem, every Boolean algebra is isomorphic to an algebra of sets like this, so if that theorem is available to you then arguments like this are actually complete proofs.)

Eric Wofsey
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