Atomlessness of the Lindenbaum-Tarski $\sigma$-algebra of infinitary language $L_{\omega_1}$

Question

Consider the Lindenbaum-Tarski $\sigma$-algebra $A$ of the infinitary language $L_{\omega_1}$ of propositional calculi. Since $L_{\omega_1}$ has $\omega_1$ propositional variables and allows disjunctions and conjunctions of all lengths strictly less than $\omega_1$, does that make $A$ atomless?

Answer 1

(You do specify the number of propositional atoms, but let me explain why that matters:)

For simplicity, I'll assume that $0$ is $\top$ and $1$ is $\perp$ in the Lindenbaum algebra, so that an atom is a non-tautologous formula which doesn't strictly imply any formulas other than $\top$. Of course, swapping $0$ and $1$ here won't change things, so this doesn't really matter.

The answer to your question depends on how many propositional atoms you have. If you only have countably many - say, $(p_i)_{i\in \omega}$ - then $$(*)\quad\bigvee_{i\in\omega}p_i$$ is an infinitary propositional formula, and (up to semantic equivalence) the only thing it strictly implies is $\top$: one formula strictly implies another iff the latter is true in properly more valuations than the former, and $(*)$ is false in exactly one valuation.

If, however, you have uncountably many propositional atoms - as your question states - then the answer is yes: given a non-tautologous $\varphi$, let $p$ be a propositional atom not occurring in $\varphi$, and consider $\varphi\vee p$. Clearly $\varphi$ implies $\varphi\vee p$, but since $\varphi$ isn't a tautology we can $(i)$ find a valuation making $\varphi$ false and $p$ (hence $\varphi \vee p$) true since $p$ doesn't occur in $\varphi$, and $(ii)$ find a valuation making $\varphi$ and $p$ (hence, $\varphi\vee p$) both false - so $\varphi\vee p$ is neither a tautology nor a formula implying $\varphi$.