How could I proceed in proving that a Lindenbaum algebra is atomless?

Given a $$P$$ infinite set of propositional variables we consider the Lindenbaum algebra generated by $$P$$. Then is this algebra atomless?

• I think the third paragraph in this answer has everything you need. But it might not have all the required detail... – amrsa Apr 19 at 20:42

Let $$\varphi$$ be any propositional formula that you believe is an atom. In particular, $$\varphi$$ isn't a contradiction. Choose proposition variable $$p$$ such that $$p$$ does not occur in $$\varphi$$. Then $$p\land\varphi$$ isn't a contradiction but $$p\land\varphi$$ implies $$p$$ while $$\varphi$$ does not. Therefore $$\varphi$$ isn't an atom.
I'll leave it to you to show why $$p\land\varphi$$ can't be a contradiction and why $$\varphi$$ can't imply $$p$$; both under the assumptions above of course. It's pretty intuitively obvious though. The above also makes it clear why the finitely generated case is different.