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I want to show that the following definitions for an atom $a$ are equivalent for a nonzero element $a$ in a Boolean algebra $\mathcal{B}$:

  1. for all $x\in\mathcal{B},a\leq x$ or $x\land a=0$

  2. for all $x\in\mathcal{B},x\leq a\Rightarrow x=0$ or $x=a$

I can proof that the first statement indicates the second, but I could not figure out how to proof the first statement starting from the second.

What I have so far is (proof that first implies second):

$a\leq x\Rightarrow x\land a=a$, so we know that either $x\land a=a$ or $x\land a=0$

First possibility: $x\land a=a$

Since $x\leq a$ implies $x\land a=x$, we find $x=a$

Second possibility: $x\land a=0$

Since $x\leq a$ implies $x\land a=x$, we find $x=0$

Can someone find the second part of the proof?

Sven BandSomething
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1 Answers1

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Suppose that $a \in B$ is such that for all $x \in B$, if $x \leq a$ then $x=0$ or $x=a$.
Take $x \in B$, whatever.
Given that $x \wedge a \leq a$, we conclude that $x \wedge a = 0$ or $x \wedge a = a$; in the second case, $a \leq x$.

Update. I found your proof that (1) implies (2) rather confusing (starting in the first sentence), so I'm adding my own version too.

Suppose that $a \in B$ is such that for all $x \in B$, we have that $a \leq x$ or $a \wedge x = 0$.
In order to prove that for all $x \in B$, if $x \leq a$ then $x=0$ or $x=a$, take $x \leq a$, in $B$.
By hypothesis, either $a \leq x$ or $a \wedge x = 0$.
In the first case, $x=a$; in the second, since $x = a \wedge x$ (because $x \leq a$), we obtain $x=0 $.

amrsa
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