I was reading this article by Alexander Paseau on "Proofs of the Compactness Theorem" and, when explaining Henkin's construction for propositional logic, he makes a comment on how it's not necessary to use Zorn's Lemma to obtain the desired maximal set, but only the Ultrafilter Lemma. Now, the proofs of compactness via ultrafilters that I'm aware of make a detour by Lindenbaum's algebra in order to obtain a partial ordering from the preorder defined by $\phi \leq \psi$ iff $\phi \models \psi$. Surprisingly, Paseau bypasses this detour and uses directly the preorder so defined to obtain a maximal set. More formally, let $\Gamma$ be a finitely satisfiable set. He defines
$\Gamma^* = \{ \phi \; | \; \exists \gamma_1, \dots, \gamma_n \in \Gamma ((\gamma_1 \wedge \dots \wedge \gamma_n) \models \phi)\}$
and then claims that $\Gamma^*$ is a filter because (i) $\bot \not \in \Gamma^*$, (ii) if $\phi_1, \phi_2 \in \Gamma^*$, then $\phi_1 \wedge \phi_2 \in \Gamma^*$, and (iii) if $\phi_1 \in \Gamma^*$ and $\phi_1 \models \phi_2$, then $\phi_2 \in \Gamma^*$.
As I said, this would make sense if we were working inside Lindenbaum's algebra, so that there was a partial order in the background; I've never seen the concept of a filter defined in relation to a preorder. Is this something usual, or was Paseau just a bit sloppy in not indicating that he was working with Lindenbaum's algebra?
I've definitely seen the word "filter" used in the context of preorders to mean what it should - namely, if $\mathbb{P}$ is a preorder and $\mathbb{Q}$ is the associated partial order ($\mathbb{Q}=\mathbb{P}/\equiv_\mathbb{P}$), then a filter $F$ in $\mathbb{P}$ should be exactly the preimage of a filter in $\mathbb{Q}$ under the quotient map. It's easy to check that this is the same as $F$ being upwards closed and closed under finite meets.