Equivalent definitions of atom in a Boolean Algebra

Question

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

1) for all $x\in\mathcal{B},a\leq x$ or $a\leq x'$

2) for all $x,y\in\mathcal{B},a\leq x\sqcup y\Rightarrow a\leq x$ or $a\leq y$

3) $a$ is minimal among nonzero elements of $\mathcal{B}$

I can't show any of the implications. Could you give me some hint?

Answer 1

For (1)$\implies$(2), suppose that $a\not\le x$ and $a\not\le y,$ and use (1) to show that we must have $a\not\le x\sqcup y,$ thus proving the contrapositive of (2), so proving (2).

For (2)$\implies$(1), take any $x\in\mathcal B$ and observe that $x\sqcup x'=1,$ so....

For (1)$\implies$(3), suppose $x\in\mathcal B$ such that $x<a.$ It follows that $a\le x'$ by hypothesis, so since $a'<x'$ (why?), then $1=a\sqcup a'\le x'$ (why?), and so....

For (3)$\implies$(1), take any $x\in\mathcal B$ and note that $a\sqcap x\le a$ and $a\sqcap x'\le a.$ Note further that $a\sqcap x$ and $a\sqcap x'$ cannot both be $0$ (why?), so without loss of generality, $0<a\sqcap x\le a,$ so by (3) we have $a\sqcap x=a,$ and so....

Answer 2

(1)$\implies$(2) If $a\le x\sqcup y$ and $a\not\le x$, then $a\le x'$. Thus $a\le(x\sqcup y)\sqcap x'=\ldots\le\ldots$

(2)$\implies$(3) Let $b\le a$; then $a\le b\sqcup b'$, so $a\le b$ or $a\le b'$. In the first case $a=b$; in the second case $b\le b'$, so …

(3)$\implies$(1) …