TITLE: What kind of compactness does "expanding $\mathbb{R}$ by constants" have?
QUESTION [10 upvotes]: EDIT: in retrospect this question should have been split up; I've accepted Joel's answer to the first part below, and asked the second part here.

This question is crossposted at MSE; however, it has not gotten any responses there despite a bounty, so I'm asking it here.
Say that a theory $T$ in the language of ordered fields + constants is $\mathbb{R}$-satisfiable if it has a model whose ordered field part is $\mathbb{R}$, with the usual ordered field structure. I'm interested in what compactness-like properties $\mathbb{R}$-satisfiability has.
To begin with, we can easily knock off compactness itself: since $\mathbb{R}$ is Archimedean, $\mathbb{R}$-satisfiability is not compact. But things get murkier when we consider “compactness at higher cardinalities.”
For cardinals $\kappa<\lambda$, say that $\mathbb{R}$-satisfiability is $(\kappa, \lambda)$-compact if whenever $\Gamma$ is a set of sentences of cardinality $<\lambda$, and every subset of cardinality $<\kappa$ is $\mathbb{R}$-satisfiable, then $\Gamma$ is $\mathbb{R}$-satisfiable. (So usual compactness is $(\omega, \infty)$-compactness, and countable compactness is $(\omega,\omega_1$)-compactness.)
It’s easy to show that $\mathbb{R}$-satisfiability is not $(\omega_1,\omega_2)$-compact: any countable linear order embeds into $\mathbb{R}$, but $\omega_1$ does not. And it follows from Easton’s theorem that for every cardinal $\kappa$ with $cf(\kappa)>\omega$, it is consistent with ZFC that $\mathbb{R}$-satisfiability is not $(\kappa^+, \kappa^{++})$-compact (look at a theory asserting the existence of $\kappa^+$-many distinct reals assuming $\mathfrak{c}=\kappa$).
This suggests two natural questions:

Can ZFC prove that $\mathbb{R}$-satisfiability is $(\omega_{\omega+1}, \omega_{\omega+2})$-compact?

Is it consistent with ZFC that $\mathbb{R}$-satisfiability is $(\omega_2, \omega_3)$-compact?


I believe the answer to the second question should be a relatively easy "yes", while the first question should be "no" but might require some work. However, I don't immediately see how to resolve either piece.

REPLY [7 votes]: The answer to the first question is no, by an easy argument.
What I claim is that ZFC proves that
$\newcommand\R{\mathbb{R}}\R$-satisfiability is not
$(\mathfrak{c},\mathfrak{c}^+)$-compact. (This improves upon your
observation about $(\kappa^+,\kappa^{++})$.) To see this, let $T$ be the
elementary diagram of $\R$, with a constant for every real, plus
another constant, asserted to be none of them. Every
size-less-than-continuum subtheory of this theory is
$\R$-satisfiable. But for the whole theory, the only way to
interpret the $\R$-constants in an $\R$-structure is as themselves,
since the rational constants will be interpreted as themselves and
the other constants are then constrained by the type of their cuts
in the rationals. So there is nobody left for the one extra
constant.
In particular, if $2^\omega=\aleph_{\omega+1}$, then this shows
$\R$-satisfiability needn't be
$(\omega_{\omega+1},\omega_{\omega+2})$-compact.