Proving Lindenbaum's Theorem via the ultrafilter theorem Consider a first-order language $L$. Lindenbaum's Theorem states that if $S$ is a consistent set of $L$-sentences. Then there is a complete and consistent set $S^{*}$ of $L$-sentences such that $S\subseteq S^{*}$.
I am learning about filters/ultrafilters and I am looking for someone to point me in the right direction to deduce Lindenbaum's Theorem from the ultrafilter theorem.
 A: Here are some things to check:

*

*Let $\mathcal{S}$ be the set of all $L$-sentences. Write $\varphi\equiv \psi$ when sentences $\varphi$ and $\psi$ are logically equivalent. Then $\mathcal{B} = \mathcal{S}/{\equiv}$ has the structure of a Boolean algebra, called the Lindenbaum-Tarski algebra.

*A filter on the Boolean algebra $\mathcal{B}$ is the same thing as an $L$-theory (a set of $L$-sentences) which is closed under logical consequence.

*A filter is proper if and only if the corresponding $L$-theory is consistent.

*A filter is an ultrafilter if and only if the corresponding $L$-theory is complete and consistent.

Once you've internalized the above correspondences, the equivalence between the ultrafilter theorem and Lindenbaum's theorem is immediate. Extending a filter to an ultrafilter is the same thing as extending a consistent theory to a consistent complete theory.
In more detail: Given a consistent $L$-theory $T$, form the filter $F\subseteq \mathcal{B}$ corresponding to the set of all logical consequences of $T$. This is a proper filter since $T$ is consistent. Then use the ultrafilter theorem to extend $F$ to an ultrafilter $U$. Finally, let $T^*$ be the theory corresponding to $U$. We have $T\subseteq T^*$, and $T^*$ is a complete consistent $L$-theory.
