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

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

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