TITLE: Lowenheim-Skolem number for satisfaction via continuous function rings
QUESTION [7 upvotes]: EDIT: now asked at MO.
For $\mathcal{X}$ a topological space, let $C(\mathcal{X})$ be the ring of continuous functions $\mathcal{X}\rightarrow\mathbb{R}$. For a first-order sentence $\varphi$ in the language of rings, write $\mathcal{X}\models_{C}\varphi$ iff $C(\mathcal{X})\models\varphi$ in the usual sense. Say that $\varphi$ is C-satisfiable iff $\mathcal{X}\models_C\varphi$ for some topological space $\mathcal{X}$.
I'm generally curious about analogues of classical model-theoretic questions in this context. For example:

What is the least cardinal $\kappa$ such that, if $\varphi$ is $C$-satisfiable, then $\mathcal{X}\models_C\varphi$ for some space $\mathcal{X}$ with cardinality $<\kappa$?

The obvious Lowenheim-Skolem argument doesn't help much here. Roughly speaking, given $\mathcal{X}\models_C\varphi$ we can form a "small" elementary substructure $(\mathcal{Y}, R)$ in the usual first-order sense of (an appropriate version of) the pair $(\mathcal{X},C(\mathcal{X}))$. However, in general we might have $R\subsetneq C(\mathcal{Y})$ (this is similar to the distinction between Henkin and standard semantics for second-order logic).
(Of course number of points isn't the only cardinal invariant for which we could ask a Lowenheim-Skolem flavored question, but it seems the simplest to start with. Another natural thing to do would be to restrict attention to a "nice" class of spaces, e.g. the compact Hausdorff spaces, but again this seems like a better first step.)
One issue here is that I don't know an abstract characterization of which rings are isomorphic to some $C(\mathcal{X})$. Obviously any such ring $R$ must be the underlying ring of a commutative associative unital $\mathbb{R}$-algebra, but beyond that I don't see anything. I recall seeing a very snappy theorem addressing this, but I can't find it at the moment (I think it was more complicated than just "any commutative associative unital $\mathbb{R}$-algebra's underlying ring," but I'm not certain).

My broader interest incidentally is in what happens if we replace $\mathbb{R}$ with some other topological structure (= first-order structure + topology such that all functions are continuous and all relations are closed, as subsets of the appropriate product space). Given a topological structure $\mathcal{A}$ and a topological space $\mathcal{X}$ we can always turn the set of continuous maps from $\mathcal{X}$ to $\mathcal{A}$ into a structure of the same type as $\mathcal{A}$ by defining everything "pointwise." So for each topological structure $\mathcal{A}$ we get an analogue $\models_C^\mathcal{A}$ of the relation $\models_C$ above.
Ultimately my interest is in developing a connection between "coarse" model-theoretic properties (e.g. variants of the Lowenheim-Skolem number) of $\models_C^\mathcal{A}$ and topological algebraic properties of $\mathcal{A}$. While this question specifically asks about $\mathcal{A}=\mathbb{R}$, if anyone has information about a different ("nontrivial") $\mathcal{A}$ or class of $\mathcal{A}$s I'd be very interested.

REPLY [3 votes]: If $\mathcal Y\Vdash \phi$ for some compact Hausdorff space $\mathcal Y$ then $\mathcal X\Vdash \phi$ for some compact Hausdorff $\mathcal X$ of cardinality $\leq \beth_2.$ So a "compact Hausdorff satisfiability" version of your question has $\kappa\leq\beth_2^+.$ I don't know how to improve your bound in the general case, or whether $\beth_2^+$ is optimal for the compact Hausdorff case.
Fix a compact Hausdorff $\mathcal Y.$ We will try to find a space $\mathcal X$ of cardinality $\leq\beth_2$ that is elementarily equivalent to $\mathcal Y.$
Let $B(\mathcal Y)$ denote the set of continuous functions $\mathcal Y\to[-1,1].$ The approach will be to take a subset $A\subset B(\mathcal Y)$ of cardinality $\leq\mathfrak c$ that is “countably saturated with respect to $B(\mathcal Y)$” in the sense that for any countable $F\subseteq A,$ any type $\Lambda(f)$ over $A$ that is realized in $B(\mathcal Y)$ is also realized by a scalar multiple of an element of $A.$
Introduce variables $x_i, i<\mathfrak c.$
For each $\alpha\in\mathfrak c$ pick a 1-type $\Lambda_\alpha$ over a countable subset of $\{x_\beta:\beta<\alpha\},$ such that every such type gets hit at some point. Then build $A$ by transfinite iteration, where at step $\alpha$ we add a realization $f_\alpha\in B(\mathcal Y),$ if any exists, for the type $\Lambda_\alpha(f)$ with variables $x_\beta$ replaced by $f_\beta$ for $\beta<\alpha,$ and set $f_\alpha=0$ otherwise.
Let $P$ be the product $[-1,1]^A.$ Take $\mathcal X$ to be the image of $\phi:\mathcal Y\to P,$ where $x$ is sent to $a\mapsto a(x).$ Consider a continuous function $f:\mathcal X\to\mathbb R.$ Since $\mathcal X$ is compact, $f$ is bounded, so it is an integer multiple of a function $f/k$ with values in $[-1,1].$ I claim that $f/k$ is already a scalar multiple of a function in $A.$ We can extend $f/k$ to a function $g:P\to[-1,1].$
This $g$ factors through the projection $[-1,1]^A\to[-1,1]^{A_0}$ for some countable $A_0.$ One way to see this is that the preimage of each closed interval $[a,b]$ with $a,b\in\mathbb Q$ is a compact $G_\delta,$ which can be written as a countable intersection of finite unions of basic opens. So $g$ is continuous using the topology on $[-1,1]^A$ generated by basic opens using indices in some countable $A_0.$ But $g$ will then already be uniquely specified by some type over $A_0,$ and hence will already be in $A.$ The countable saturation, in fact just finite saturation, shows that $\mathcal Y$ is an elementary submodel of $\mathcal X.$ Note $|\mathcal Y|\leq \beth_2.$
Remarks.

C. Ward Henson studied a related question in “Nonstandard hulls of Banach spaces”, for positive bounded formulas.
For general spaces there is a ring isomorphism $C(\mathcal X)\cong C(\mathcal Y)$ if and only if the “Hewitt realcompactifications” of $\mathcal X$ and $\mathcal Y$ are homeomorphic. I’m not sure this helps directly, but it might be what you were thinking of as a characterization interesting.
Some papers (at least Henson's) talk in terms of Banach algebras. For the compact Hausdorff case we're talking about (the ring reduct of) the real Banach algebra $C(\mathcal X).$ There is a bi-interpretation of complex arithmetic with complex conjugation, so we could equivalently talk about complex Banach algebras.
My guess is that we can always take $\mathcal X$ to be a Polish space. Perhaps we can take $A$ to be a countable sequence of generic functions, i.e. an element in a comeager set of $B(\mathcal Y)^\omega.$