Lowenheim-Skolem number for satisfaction via continuous function rings

Solution 1:

This is not even close to a complete answer, but it at least shows that $\text{ZFC}\vdash\kappa>2^{\aleph_0}$, so I will record it here. Let $$\varphi\equiv\forall u\left[u^2=u\to(u=0\vee u=1)\right]$$ be the sentence expressing that there are no non-trivial idempotents. Then $\mathcal{X}\models_C\varphi$ if and only if $\mathcal{X}$ is connected. To see this, first suppose $\mathcal{X}$ is disconnected, say with a decomposition $\mathcal{X}=U\sqcup V$ for non-empty open subsets $U,V$; then the map $\alpha:\mathcal{X}\to\mathbb{R}$ defined by $x\mapsto 0$ if $x\in U$ and $x\mapsto 1$ if $x\in V$ is a non-trivial idempotent element of $C(\mathcal{X})$. Conversely, suppose $C(\mathcal{X})$ has a non-trivial idempotent $\alpha$. Since the ring $\mathbb{R}$ has no non-trivial idempotents, we have $\alpha(x)\in\{0,1\}$ for each $x\in\mathcal{X}$; since also $\alpha\neq 0,1$, we have that $\alpha^{-1}(\{0\})$ and $\alpha^{-1}(\{1\})$ are both non-empty. This gives a partition of $\mathcal{X}$ into disjoint non-trivial open subsets, whence $\mathcal{X}$ is not connected.

Now let $\psi\equiv \exists u\left[u\neq 0\wedge\forall v(u\cdot v\neq 1)\right]$. I claim that $\varphi\wedge\psi$ is $C$-satisfiable, but that $\mathcal{X}\not\models_C\varphi\wedge\psi$ for any $\mathcal{X}$ whose cardinality is smaller than $2^{\aleph_0}$. To see that it is $C$-satisfiable, note that, for example, $\mathbb{R}\models_C\varphi\wedge\psi$, where $\mathbb{R}$ has the standard topology. Indeed, $\mathbb{R}$ is connected, so $\mathbb{R}\models_C\varphi$ by the remarks above; furthermore, the function $\alpha:\mathbb{R}\to\mathbb{R}$ given by $x\mapsto 0$ when $x\leqslant 0$ and $x\mapsto x$ when $x\geqslant 0$ is a non-invertible non-zero element of $C(\mathcal{X})$, whence $\mathbb{R}\models_C\psi$ as well.

To see that $\mathcal{X}\not\models_C\varphi\wedge\psi$ for any $\mathcal{X}$ of cardinality smaller than $2^{\aleph_0}$, suppose that $|\mathcal{X}|<2^{\aleph_0}$ and that $\mathcal{X}\models_C\varphi$. Then $\mathcal{X}$ is connected by the remarks in the first paragraph, so for any $\alpha\in C(\mathcal{X})$ the subspace $\alpha(\mathcal{X})\subset\mathbb{R}$ is connected as well. Connected subsets of $\mathbb{R}$ are precisely intervals; since $|\mathcal{X}|<2^{\aleph_0}$, this forces each $\alpha(\mathcal{X})$ to be a singleton. So every element of $C(\mathcal{X})$ is a constant function, whence $C(\mathcal{X})\cong\mathbb{R}$ as rings, whence $\mathcal{X}\models_C\neg\psi$, as desired.

Solution 2:

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.

1. C. Ward Henson studied a related question in “Nonstandard hulls of Banach spaces”, for positive bounded formulas.
2. 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.
3. 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.
4. 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 $B(\mathcal Y)^\omega.$