Prove that if $\alpha \neq 0,$ then $\exists \beta$ such that $\beta \neq 0$ and $\beta < \alpha,$

Question

Prove that if $\alpha \neq 0,$ then $\exists \beta$ such that $\beta \neq 0$ and $\beta < \alpha,$which means $\beta \rightarrow \alpha$ but not $\alpha \rightarrow \beta$

My trial: I got a hint that it is a proof that the given boolean Algebra is atomless, but still I do not know how to proof this. could anyone help me please?

Answer 1

$\alpha$ is an atom by definition iff

$$\lnot(\exists \beta: 0 \neq \beta \text{ and } \beta < \alpha)\tag{1}$$

So if the algebra in question is atomless, $(1)$ never holds for any $\alpha\neq 0$, and we always have a $\beta$ as required.

So the hint you were given is the answer (or you have to prove the hint too, looks like you're in the Lindenbaum algebra (IIRC)?)