TITLE: Does this logic have the downward Skolem-Löwenheim theorem?
QUESTION [9 upvotes]: Let $\mathcal L_Q$ denote the logic obtained from adding the quantifier $\newcommand{\almost}{\forall^\infty}\almost$ to the usual first-order logic, where the semantic interpretation of $\almost x\varphi$ is "All but finitely many $x$ satisfy $\varphi$", or formally:  $$M\models\almost x\varphi(x)\iff\Big|\{m\in M\mid M\not\models\varphi[m]\}\Big|\text{ is finite}.$$
It's not very hard to show that this logic is not compact$^*$, and does not satisfy the upward Skolem-Löwenheim theorem (e.g. the order $(\Bbb N,\leq)$ has a categorical axiomatization). But what about its downward counterpart?
According to Lindström theorem either compactness fails, or the downward Skolem-Löwenheim theorem should fail. One fails, what about the other?

(*) Please don't discuss the failure of the compactness theorem for $\mathcal L_Q$ here before June 17th, 2013. I gave that part as a homework assignment to my students - some of whom are reading this site.

REPLY [3 votes]: There's a cute nuke which works here:
Given a structure $\mathfrak{A}$ of arbitrary cardinality in a language $\Sigma$, fix $\theta$ large enough that $\Sigma,\mathfrak{A}\in H_\theta$ and $H_\theta$ satisfies "enough set theory." Now by downward Lowenheim-Skolem for first-order logic, let $M$ be an elementary submodel of $H_\theta$ with $\mathfrak{A},\Sigma\in M$, $\Sigma\subseteq M$, and $\vert M\vert=\aleph_0+\vert\Sigma\vert$. Set $\mathfrak{B}=\mathfrak{A}\cap M$; since finiteness is absolute to $M$, it's not hard to show that $\mathfrak{B}\preccurlyeq_{\mathsf{FOL\forall^\infty}}\mathfrak{A}$ as desired.
And of course this functions as a general recipe for proving appropriate downward Lowenheim-Skolem properties for "concrete" logics.