Does $\mathsf{ZFC}$ prove that the field of real numbers has one of these compactness properties?

Suppose $\mathcal{A}$ is a $\Sigma$-structure and $\kappa<\lambda$ are infinite cardinals. Say that $\mathcal{A}$-satisfiability is $(\kappa,\lambda)$-compact iff for every theory $T$ in an expansion $\Sigma'$ of $\Sigma$ by constant symbols with $\vert T\vert<\lambda$, if every subtheory of $T$ of size $<\kappa$ has a model whose $\Sigma$-reduct is $\mathcal{A}$ then $T$ has a model whose $\Sigma$-reduct is $\mathcal{A}$.

I'm specifically interested in $\mathcal{R}=(\mathbb{R};+,\times)$. Combining my own observations with those of Joel David Hamkins, Farmer S, and Harry West, what I know so far roughly amounts to the following:

• $\mathsf{ZFC}$ proves that $\mathcal{R}$-satisfiability is not $(\omega_1,\omega_2)$-compact, $(\omega_2,\omega_3)$-compact, $(2^\omega,(2^{\omega})^+)$-compact, or $((2^\omega)^+,(2^\omega)^{++})$-compact.

• If $\kappa$ is measurable then $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact, and if $\kappa$ is strongly compact then $\mathcal{R}$-satisfiability is $(\kappa,\lambda$)-compact for every $\lambda>\kappa$.

At this point I'd like to know what positive results are guaranteed. Here's one which seems particularly interesting to me at the moment:

Does $\mathsf{ZFC}$ prove that there is a $\kappa>(2^{\omega})^+$ such that $\mathcal{R}$-satisfiability is $(\kappa,\kappa^{++})$-compact?

This sort of "gap-$2$" compactness seems rather hard to get, as far as I can tell, but at the same time I can't immediately extract any strength from it as a hypothesis. I suspect I'm missing something rather easy here.

(Edit: This replaces an earlier attempt I had in the opposite direction, which was bogus, as pointed out by @HarryWest, and to which some comments below refer.)

(Edit 2: It now gets weak compactness as a consistency strength lower bound.)

(Edit 3: A part initially missing at the end of the Subcase 2.2 argument has been filled in.)

(Edit 4: 2 observations specifically on $(\kappa,\kappa^{++})$-compactness added at the end.)

Remark: It looks like @HarryWest has already answered the original question. Below I work more directly with $\mathcal{R}$-satisfiability and extract some more strength. I don't know to what extent Harry's method would also lead to the following.

Remark 2: The reader unfamiliar with the Dodd-Jensen core model $K^{\mathrm{DJ}}$, but familiar with $0^\sharp$, should replace "there is an inner model with a measurable cardinal" in the claim below with "$0^\sharp$ exists", and replace $K^{\mathrm{DJ}}$ with $L$ throughout; this results in a weaker result, but the proof is essentially identical, and it suffices for Corollaries 1 and 3.

Claim. Assume ZFC. Let $\kappa$ be an uncountable cardinal such that $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact. Then there is a weakly inaccessible cardinal $\mu\leq\kappa$, and either:

• there is an inner model with a measurable cardinal, or

• letting $K=K^{\mathrm{DJ}}$ be the Dodd-Jensen core model below a measurable, $K\models$"$\kappa$ is weakly compact", and if $2^\gamma\leq\kappa$ for all $\gamma<\kappa$ then $\kappa$ is weakly compact.

Corollary 1. ZFC + "There is an uncountable cardinal $\kappa$ such that $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact" is equiconsistent with ZFC + "There is a weakly compact cardinal".

Corollary 2. Assume ZFC + GCH + "there is no inner model with a measurable cardinal". Let $\kappa$ be an uncountable cardinal. Then $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact iff $\kappa$ is weakly compact.

Corollary 3. Assume ZFC + GCH + "$0^\sharp$ does not exist". Let $\kappa$ be an uncountable cardinal. Then $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact iff $\kappa$ is weakly compact.

(Note that the converse direction of Corollaries 2 and 3, i.e. if $\kappa$ is weakly compact then $\mathcal{R}$-satisfiability is $(\kappa,\kappa^+)$-compact, holds in general.)

Proof of Claim: Fix $\kappa$. Except for the proof that there is a weakly inaccessible cardinal $\leq\kappa$, we may assume there is no inner model with a measurable cardinal, so write $K=K^{\mathrm{DJ}}$. (As mentioned in Remark 2, the reader unfamiliar with $K$ but familiar with $0^\#$ should just assume $0^\#$ does not exist, in which case $K=L$.) I will formally assume this, but it is easy to drop this assumption and still get the weakly inaccessible $\mu\leq\kappa$.

The plan is to find a reasonable elementary substructure $\bar{\mathcal{H}}$ of some fragment of $V$ and an elementary embedding $j:\bar{\mathcal{H}}\to M$ with a critical point $\mu$, and use this to get the desired conclusions.

If there is $\gamma<\kappa$ such that $2^\gamma\geq\kappa$ then letting $\gamma$ be least such, fix a sequence $\left<A_\alpha\right>_{\alpha<\kappa}$ of pairwise distinct subsets of $\gamma$. Otherwise let $A_\alpha=\emptyset$ for all $\alpha<\kappa$. Let $\mathcal{H}=(\mathcal{H}_{\kappa^+},\vec{A},<^*)$, where $<^*$ is a wellorder of $\mathcal{H}_{\kappa^+}$ (the set of all sets hereditarily of cardinality $\leq\kappa$).

We will build a theory $T$, of size $\kappa$, such that every sub-theory of size ${<\kappa}$ is $\mathcal{R}$-realizable. Basically, we want $T$ to describe a model which contains a version of $\mathcal{H}$ as an element, together with the statement "$\kappa$ is not a cardinal". The theory $T$ will use primary constants $\dot{\vec{\alpha}}$ for certain sequences $\vec{\alpha}$ of ordinals ${<\kappa^+}$. The sequences $\vec{\alpha}$ used we call the relevant sequences, and which are considered relevant depends on the following cases. If (i) $\gamma^{\omega}\leq\kappa$ for all $\gamma<\kappa$ then there are only $\kappa$-many $\omega$-sequences $\vec{\alpha}\in{^\omega}\kappa$ such that $\vec{\alpha}$ is bounded in $\kappa$ (and of course if $\mathrm{cof}(\kappa)>\omega$, "bounded in $\kappa$" can be struck out), and in this case these sequences $\vec{\alpha}$ are the relevant ones. Suppose instead (ii) there is $\gamma<\kappa$ such that $\gamma^{\omega}>\kappa$. If (ii.1) $\kappa^{+K}<\kappa^+$ then the relevant sequences are just the finite tuples $\vec{\alpha}\in\kappa^{<\omega}$. Suppose instead (ii.2) $\kappa^{+K}=\kappa^+$. In this case we need to be a little more careful. Fix from now on an ordinal $\eta\in(\kappa,\kappa^+)$ of cofinality $\kappa$ and such that letting $\bar{\mathcal{H}}=\mathrm{Hull}^{\mathcal{H}}(\eta)$, we have $\eta=\bar{\mathcal{H}}\cap\eta$. Then the relevant sequences are the finite tuples $\vec{\alpha}\in\eta^{<\omega}$. (Also for the proof that there is a weakly inaccessible cardinal $\mu\leq\kappa$, i.e. without assuming $K$ exists, in case (ii), proceed as in (ii.1).)

Using the process in the answer here: https://mathoverflow.net/questions/394526/is-this-compactness-property-for-satisfiability-on-mathbbr-consistent, we then also augment the theory with some secondary constants, so as to allow us to talk about subsets of $\omega$ coded by reals, and to make arithmetic statements about those coded sets using (infinitely many) statements in $T$. We will skip the details of those extra constants, and just directly make arithmetic statements about the coded subsets of $\omega$.

For each relevant $\vec{\alpha}$, let $t_{\vec{\alpha}}$ be the full theory of the single parameter $\vec{\alpha}$ in the structure $\mathcal{H}$ (note this includes predicates for $\vec{A}$ and $<^*$). Add the following statements to $T$ (they mostly refer to theories coded by reals):

• $\dot{\vec{\alpha}}$ codes a consistent complete theory $u_{\dot{\vec{\alpha}}}$ in the language of set theory augmented with symbols $\hat{\vec{\beta}},\hat{\mathcal{H}},\hat{f},\hat{\kappa},\hat{\xi}$,

• $u_{\dot{\vec{\alpha}}}$ contains the formula "$\hat{\mathcal{H}}=(\mathcal{J},<',\vec{A}')$ is a structure with transitive universe $\hat{\mathcal{J}}$, $<'$ is a wellorder of $\mathcal{J}$, and $\hat{\kappa}$ is the largest cardinal of $\mathcal{J}$",

• $u_{\dot{\vec{\alpha}}}$ contains the formula "$V=L(\hat{\mathcal{H}},\hat{f})$ and $\hat{\xi}<\hat{\kappa}$ and $\hat{f}:\hat{\xi}\to\hat{\kappa}$ is a surjection",

• the model determined by $u_{\dot{\vec{\alpha}}}$ has standard $\omega$,

and for each formula $\varphi[\vec{\alpha}]\in t_{\vec{\alpha}}$, add the statement

• $u_{\dot{\vec{\alpha}}}$ contains the formula "$\hat{\mathcal{H}}\models\varphi[\dot{\vec{\beta}}]$"

to $T$. Moreover, if $\vec{\gamma}$ is also relevant and $\mathrm{rg}(\vec{\alpha})\subseteq\mathrm{rg}(\vec{\gamma})$ and in the $\omega$-sequence case, $\vec{\alpha}$ itself is easily computed from $\vec{\gamma}$ (say there is a recursive function $i:\omega\to\omega$ such that $\vec{\alpha}_n=\vec{\gamma}_{i(n)}$), then we add the formula

• $u_{\dot{\vec{\alpha}}}$ is the theory induced by $u_{\dot{\vec{\gamma}}}$ (according to how $\vec{\alpha}$ is computed from $\vec{\gamma}$)

to $T$.

Now let $S\subseteq T$ be a sub-theory of size ${<\kappa}$. We find an $\mathcal{R}$-realization of $S$. Let $C$ be the set of relevant sequences $\vec{\alpha}$ used in $S$; so $C$ has size ${<\kappa}$, and $C\subseteq\mathcal{H}$. Let $H=\mathrm{Hull}^{\mathcal{H}}(C)$; that is, the structure whose universe is the set of all elements of $\mathcal{H}_{\kappa^+}$ definable over $\mathcal{H}$ from parameters in $C$ (using the predicates of $\mathcal{H}$), and with predicates being the restrictions of those of $\mathcal{H}$. So $H\preccurlyeq\mathcal{H}$. Let $\bar{H}$ be the transitive collapse of $H$. Let $\pi:\bar{H}\to H$ be the uncollapse map. Let $\bar{\kappa}=\pi^{-1}(\kappa)$ etc. By enlarging $S,C$ if necessary, we may assume that $\bar{\kappa}$ has cardinality ${\xi<\bar{\kappa}}$. So let $f:\xi\to\bar{\kappa}$ be a surjection. Let $N=L_\beta(\bar{H},f)$ for some ordinal $\beta>0$.

Now for each $\vec{\alpha}\in C$, let $u_{\vec{\alpha}}$ be the theory in $N$ of the parameters $\bar{\vec{\alpha}},\bar{H},f,\xi,\bar{\kappa}$, where bars denote preimage under $\pi$. Then note that by interpreting $\dot{\vec{\alpha}}$ as the real naturally coding $u_{\vec{\alpha}}$, we get an $\mathcal{R}$-realization of $S$.

By $(\kappa,\kappa^+)$-compactness, we can fix an $\mathcal{R}$-realization $\mathcal{R}^+$ of $T$. Let $u_{\vec{\alpha}}$ be the theory coded by $\dot{\vec{\alpha}}^{\mathcal{R}^+}$. Note that we can define a term model $M$, pointwise definable from constants $\widetilde{\vec{\alpha}}$ (for relevant sequences $\vec{\alpha}$ as before) and constants $\widetilde{\mathcal{H}},\widetilde{f},\widetilde{\xi},\widetilde{\kappa}$, and such that $u_{\vec{\alpha}}$ is just the theory in $M$ of $\widetilde{\vec{\alpha}}^M,\widetilde{\mathcal{H}}^M,\widetilde{f}^M,\widetilde{\xi}^M,\widetilde{\kappa}^M$ (the super-$M$ denotes the interpretation of a constant in $M$).

Let $\bar{\mathcal{H}}=\mathrm{Hull}^{\mathcal{H}}(C)$ where $C$ is the set of all relevant sequences (the hull is uncollapsed); note that $\bar{\mathcal{H}}$ is in fact transitive (so actually it's the same as the collapsed hull) so $\bar{\mathcal{H}}\preccurlyeq\mathcal{H}$ (here $\bar{\mathcal{H}}$ is a structure with predicates induced by those of $\mathcal{H}$) (and note that in case (ii.2), what we defined as $\bar{\mathcal{H}}$ earlier is the same as the model we have just now defined, and so in this case $\mathrm{OR}\cap\bar{\mathcal{H}}=\eta$). We have $\kappa+1\subseteq\bar{\mathcal{H}}$. (But $\kappa^+\not\subseteq\bar{\mathcal{H}}$, since $\bar{\mathcal{H}}$ has cardinality $\kappa$.) Note that there is an elementary embedding $j:\bar{\mathcal{H}}\to\widetilde{\mathcal{H}}^M$ given by setting $j(\vec{\alpha})=\widetilde{\vec{\alpha}}^M$ for each relevant $\vec{\alpha}$.

We have $j(\kappa)=\kappa^M$. Since $\kappa$ is a cardinal in $V$, but $\kappa^M$ is not a cardinal in $M$ (though it is of course a cardinal in $\widetilde{\mathcal{H}}^M$), $j$ has a critical point $\mu\leq\kappa$ (here $M$ might be illfounded). So $\mu$ is a regular cardinal in $\bar{\mathcal{H}}$, and hence also regular in $V$. Note that $\mu>\omega$, as $M$ has standard $\omega$, as this held for each sub-model coded by $u_{\vec{\alpha}}$. Note that $\mu$ is not a successor cardinal (if $\mu=\gamma^+$ then $$\gamma^+=\mu<j(\mu)=j(\gamma^+)=j(\gamma)^{+M}=\gamma^{+M},$$ so $\mu$ is collapsed in $M$, a contradiction). So $\mu$ is weakly inaccessible, in particular giving the existence of a weakly inaccessible $\leq\kappa$ (we didn't yet use $K$). So from now on we do assume we have $K$, i.e. there is no inner model with a measurable.

Case 1: $\gamma^{\omega}\leq\kappa$ for all $\gamma<\kappa$.

So the relvant sequences are the bounded-in-$\kappa$ $\omega$-sequences $\vec{\alpha}$. We consider two subcases.

Subcase 1.1: $\mu<\kappa$.

In this case there is an inner model with a measurable cardinal. For if not, then the Dodd-Jensen core model $K$ (below an inner model with a measurable cardinal) exists. So there is no transitive proper class $K'$ and non-trivial elementary $k:K\to K'$. And $K\models GCH$, and $\mathcal{P}(\mu)\cap K\subseteq\bar{\mathcal{H}}$.

Let $\delta$ be some "ordinal" of $M$ such that $\beta<\delta<j(\mu)$ for all $\beta<\mu$ (it doesn't matter whether $M$ is wellfounded; but we will use that $M$ has wellfounded $\omega$). Let $U$ be the $K$-ultrafilter derived from $j$ with seed $\delta$. Then $\mathrm{Ult}(K,U)$ is wellfounded. For suppose not, and let $\left<f_n,X_n\right>_{n<\omega}\subseteq K$ be such that $f_n:\mu\to\mathrm{OR}$ and $X_n\subseteq\mu$ and $X_n\in U$ and $f_{n+1}(\alpha)<f_n(\alpha)$ for all $\alpha\in X_n$, for all $n<\omega$. Let $\vec{X}=\left<X_n\right>_{n<\omega}$. Because $K|\mu^{+K}$ is definable over $\mathcal{H}_\kappa$ and satisfies "$V=\mathrm{HOD}$", each $X_n$ is specified by some ordinal $\alpha<\mu^+$, so there is a relevant $\vec{\alpha}$ such that $\left<X_n\right>_{n<\omega}$ is defined from $\vec{\alpha}$. Let $X=\bigcap_{n<\omega}X_n$. So $X\in\bar{\mathcal{H}}$, and (as $M$ has wellfounded $\omega$) $X\in U$, and in particular $X\neq\emptyset$. But then letting $\beta\in X$, we have $f_{n+1}(\beta)<f_n(\beta)$ for all $n<\omega$, a contradiction.

So $K'=\mathrm{Ult}(K,U)$ is wellfounded, but then the ultrapower map $k:K\to K'$ is non-trivial, a contradiction.

Subcase 1.2: $\mu=\kappa$.

So all $\omega$-sequences $\subseteq\kappa$ are relevant, and it easily follows that $M$ is closed under $\omega$-sequences, and hence wellfounded. We claim that $K\models$"$\kappa$ is weakly compact". For since $\kappa$ is weakly inaccessible, it is inaccessible in $K$. Suppose $\kappa$ is not weakly compact in $K$. We have $K_\kappa=V_\kappa^K\subseteq\bar{\mathcal{H}}$ and $V_{\kappa+1}^K\subseteq\mathcal{H}$, and so $\bar{\mathcal{H}}\models$"$\kappa$ is not weakly compact in $K$". Let $A\in V_{\kappa+1}^K$ with $A\in\bar{\mathcal{H}}$ and $\varphi$ be a $\Pi_1$ formula such that $(V_{\kappa+1}^K,A)\models\varphi$ but there is no $\kappa'<\kappa$ such that $(V_{\kappa'+1}^K,A\cap V_{\kappa'}^K)\models\varphi$. So the same holds for $j(\kappa)$ in $M$. Since $M$ is wellfounded, we have $\kappa\in M$, and can apply the statement at $\kappa'=\kappa$ in $M$, which implies $(V_{\kappa+1}^{K^M},j(A)\cap V_{\kappa}^{K^M})\models\neg\varphi$, and so in fact $(V_{\kappa+1}^{K^M},A)\models\neg\varphi$. But $M$ is closed under $\omega$-sequences, which implies that $K^M$ is iterable, and so $K^M|\kappa^{+K^M}$ is a segment of $K$ so $V_{\kappa+1}^{K^M}\subseteq K^M$, and since $\neg\varphi$ is $\Sigma_1$, therefore $(V_{\kappa+1}^K,A)\models\neg\varphi$, a contradiction.

Now suppose $2^\gamma\leq\kappa$ for all $\gamma<\kappa$. We first claim that $\kappa$ is actually inaccessible. For suppose not, and let $\gamma<\kappa$ be least such that $2^\gamma\geq\kappa$; by our assumption then, $2^\gamma=\kappa$. Thus, we chose $\vec{A}$ to enumerate all subsets of $\gamma$, in a one-to-one fashion. But then $j(\vec{A})$ properly extends $\vec{A}$ with new subsets of $\gamma$, a contradiction. Now one can show that $\kappa$ is weakly compact via an argument just like that used for $K$ above, but now in $V$ (and it is easier).

Case 2: Otherwise (there is $\gamma<\kappa$ such that $\gamma^{\omega}>\kappa$).

This case will be dealt with similarly to the previous one, but it is a little subtler. The relevant sequences are the finite tuples $\vec{\alpha}\in\eta^{<\omega}$, where $\eta=\kappa$ in case (ii.1).

Subcase 2.1: $\mu<\kappa$.

Define $\delta,U$ as in Subcase 1.1. We claim again that $\mathrm{Ult}(K,U)$ is wellfounded, which is again enough. Suppose not and let $\left<f_n,X_n\right>_{n<\omega}\subseteq K$ be as before. We have $X_n\subseteq\mu$. Let $\alpha_n$ be the rank of $X_n$ in the $K$-constructibility order. Then $\alpha_n<\mu^{+K}\leq\mu^+\leq\kappa$. By covering for $K$, there is a set $\mathcal{X}\in K$, of cardinality $\leq\aleph_1^V$, such that $X_n\in\mathcal{X}$ for each $n$, and $Y\subseteq\mu$ for each $Y\in\mathcal{X}$. Since $\aleph_2^V<\mu$, the usual argument shows that $Z=\{Y\in\mathcal{X}\bigm|Y\in U\}$ is in $K$. (That is, consider $\{Y\in j(\mathcal{X})\bigm|\delta\in Y\}\in M$, and use the agreement between $M$ and $\bar{\mathcal{H}}$ below $\mu$.) But then also since $\aleph_2^V<\mu$, it follows that $\bigcap Z\neq\emptyset$. Therefore $\bigcap_{n<\omega}X_n\neq\emptyset$, so now we can argue for contradiction like in Subcase 2.1.

Subcase 2.2: $\mu=\kappa$.

In this case we show that $\kappa$ is weakly compact in $K$.

Suppose (ii.2) holds, so $\kappa^{+K}=\kappa^+$ and $\bar{\mathcal{H}}=\mathrm{Hull}^{\mathcal{H}}(\eta)$, and $\mathrm{cof}(\eta)=\kappa$. Suppose $\kappa$ is not weakly compact in $K$. Then $K|\eta=K^{\bar{\mathcal{H}}}\models$"$\kappa$ is not weakly compact". So fix a counterexample $A,\varphi\in K|\eta$.

Define $U,\delta$ as before. We claim that $\mathrm{Ult}(K|\eta,U)$ is wellfounded. So suppose otherwise and let $\left<f_n,X_n\right>_{n<\omega}$ be a counterexample $\subseteq K|\eta$. Let $\alpha_n$ be the rank of $X_n$ in the order of constructibility of $K$. Let $\mathcal{X}=\{\alpha_n\bigm|n<\omega\}$. So by covering, there is $\mathcal{Y}\subset\mathrm{OR}$ such that $\mathcal{Y}\in K$ and $\mathcal{X}\subseteq\mathcal{Y}$ and $\mathcal{Y}$ has cardinality $\leq\aleph_1^V$. It suffices to see there is such a $\mathcal{Y}\in K|\eta$, as then we can argue as before. Since $\kappa$ is the largest cardinal of $K|\eta$, there are cofinally many $\beta<\eta$ such that $K|\beta$ projects to $\kappa$, which means here that there is a surjection $f:\kappa\to K|\beta$ which is definable without parameters over $K|\beta$. Fix such a $\beta,f$ with $X_n\in K|\beta$ for all $n$; this exists because $\mathrm{cof}(\eta)=\kappa$ and $\kappa$ is weakly inaccessible. So $f\in K|(\beta+1)\subseteq K|\eta$. Let $\mathcal{X}'$ be the set of all $\alpha<\kappa$ such that $f(\alpha)\in\mathcal{X}'$ and for no $\alpha'<\alpha$ is $f(\alpha')=f(\alpha)$. So $\mathcal{X}'\subset\kappa$ and $\mathcal{X}'$ is countable. So by covering, there is $\mathcal{Y}'\in K$ with $|\mathcal{Y}'|\leq\aleph_1^V$ and $\mathcal{X}'\subseteq\mathcal{Y}'$. Since $\kappa$ is weakly inaccessible, $\mathcal{Y}'\in K|\kappa$. But then $\mathcal{Y}=f``\mathcal{Y}'$ covers $\mathcal{X}$ and $\mathcal{Y}\in K|(\beta+1)\subseteq K|\eta$, as desired.

Also, $N=\mathrm{Ult}(K|\eta,U)$ is iterable. For it suffices to see that all countable elementary substructures of $N$ are iterable. For this, given $\left<f_n\right>_{n<\omega}\subseteq K$, it suffices to see that $\bar{N}=\mathrm{Hull}^N(\{[f_n]\bigm|n<\omega\})$ is iterable (where $f_n$ represents $[f_n]\in N$). But we can find sets $X_n\in K|\eta$ such that the $\Sigma_n$-elementary theory of $(f_0(\alpha),\ldots,f_n(\alpha))$ in $K|\eta$ is independent of $\alpha\in X_n$, and then arguing as before, $\bigcap_{n<\omega}X_n\neq\emptyset$, so letting $\alpha\in\bigcap_{n<\omega}X_n$, $\bar{N}$ is isomorphic to $\mathrm{Hull}^{K|\eta}(\{f_n(\alpha)\bigm|n<\omega\})$, and is therefore iterable.

By the iterability, $N|\kappa^{+N}=K|\kappa^{+N}$, and so the failure of weak compactness now leads to a contradiction like before (applying the ultrapower map $i^{K_\eta}_U:K|\eta\to N$).

Finally if (ii.1) holds, it is almost the same as for (ii.2), but easier: note that $\kappa^{+K}<\eta=\mathrm{OR}\cap\bar{\mathcal{H}}$ in this case, so taking any (small) covering set $\mathcal{Y}\in K$ for a countable set $\mathcal{X}$, we get $\mathcal{Y}\in\bar{\mathcal{H}}$, which is again enough.

Regarding specifically $(\kappa,\kappa^{++})$-compactness of $\mathcal{R}$-satisfiability:

Observation 1: Assuming ZFC + $\kappa$ is uncountable + $\mathcal{R}$-satisfiability is $(\kappa,\kappa^{++})$-compact, we get an inner model with a measurable, by arguing much as above, but allowing relevant sequences to take values anywhere ${<\kappa^+}$; this leads to a model $\bar{\mathcal{H}}\preccurlyeq\mathcal{H}$ with $\kappa^+\subseteq\bar{\mathcal{H}}$ and an embedding $j:\bar{\mathcal{H}}\to\widetilde{\mathcal{H}}^M$ at the end, and $j(\kappa)>\kappa$. This then gives an inner model with a measurable like when $\mu<\kappa$ in the preceding proof.

Observation 2: In the converse direction: Assume ZFC + GCH + $\kappa$ is $\kappa^+$-supercompact. Let $G$ be generic over $V$ for adding a $\kappa^{++}$-seqence of Cohen reals with the finite support product. Then $V[G]\models$"$2^{\aleph_0}=\kappa^{++}$ and $\mathcal{R}$-satisfiability is $(\kappa,\kappa^{++})$-compact".

This can be shown using a small variant of the argument in Update 2 of my answer to https://mathoverflow.net/questions/394526/is-this-compactness-property-for-satisfiability-on-mathbbr-consistent, which does the analogous thing for $(\kappa,\kappa^+)$-compactness from weak compactness.

No: working in a model of ZFC+GCH+no inaccessibles, $\mathcal R$-satisfiability is not $(\kappa,\kappa^+)$-compact for any uncountable $\kappa.$ $\DeclareMathOperator{\bool}{bool} \DeclareMathOperator{\true}{true} \DeclareMathOperator{\false}{false} \newcommand{\Mp}{M_{\mathrm{prop}}}$

I will just argue a simpler illustrative statement, which should make the above claim seem plausible.

The logic I’ll use is many-sorted first order with equality. Let $\Sigma$ be the signature with two sorts $\bool,0,$ with bool constants for $\false,\true,$ and sort $0$ constants $c^0_i$ for each element $i\in \omega,$ and no other functions or relations. Let $\mathcal O$ be the $\Sigma$-structure where the bool sort is interpreted as $\{\false,\true\},$ and sort $0$ is interpreted as $\omega,$ with constants $c^0_i$ interpreted as $i.$ I claim that $\mathcal O$-satisfiability is not $(\beth_2,\beth_2^+)$-compact.

Start by defining an extension $\mathcal V$ of $\mathcal O$ over a signature $\Sigma_{\mathcal V}\supset\Sigma$ with new sorts $1,2$ interpreted as sets $\mathcal P(\omega),\mathcal P(\mathcal P(\omega)).$ The signature $\Sigma_{\mathcal V}$ also has constants $c^i_j$ of sort $i$ for each (external) element $j\in\mathcal P^i(\omega),$ the two membership relations $\in^1$ of type $0\times 1$ and $\in^2$ of type $1\times 2,$ and two Skolem functions witnessing disequality: $u^0_1$ of type $1\times 1\to 0$ and $u^1_2$ of type $2\times 2\to 1.$ We need to pick such functions to define the model $\mathcal V,$ though all I really need is the two axioms:

• U1: $(\forall a^1,b^1)((u^0_1(a^1,b^1)\in^1 a^1 \iff u^0_1(a^1,b^1)\in^1 b^1)\implies a^1=b^1)$
• U2: $(\forall a^2,b^2)((u^1_2(a^2,b^2)\in^2 a^2 \iff u^1_2(a^2,b^2)\in^2 b^2)\implies a^2=b^2)$

Add to the true theorems of this structure a constant $t^2$ of sort $2$ with axioms $t^2\neq c^2_i$ for each $i\in \mathcal P(\mathcal P(\omega)).$ This produces a theory $T$ of signature $\Sigma_T.$

Add bool constants $I^{\bool}[\phi]$ for each formula $\phi$ in the signature of $T,$ and sort $0$ constants $I^0[s^0]$ for each term $s^0$ of sort $0,$ and add the following "interpretation axiom scheme":

• "$I^{\bool}[\phi]=\true\iff \phi$" for each $\phi$
• "$I^0[s^0]=s^0$" for each $s^0$

Add all logical consequences. Take the subtheory $T_0$ that restricts the signature to constants of sort $0$ and $\bool,$ and keep the theorems that can be expressed in this signature. $T_0$ is the theory we will use to disprove $(\beth_2,\beth_2^+)$-compactness. Note that its signature is an extension of $\Sigma$ by constants, as required.

Any subtheory $T’$ of $T_0$ of cardinality less than $\mathcal P(\mathcal P(\omega))$ has a model whose $\Sigma$-retract is $\mathcal O.$ Just pick a value $i$ of $t^2$ such that no theorem in $T’$ uses $c^2_i.$ Then $\mathcal V$ can be turned into a model of $T’$ by swapping the symbols $t^2$ and $c^2_i$ and interpreting the $I^{\bool}[\phi]$ and $I^0[s^0]$ constants in the unique manner consistent with the "interpretation axiom scheme".

Suppose for contradiction that $T_0$ itself has a model $M$ with $\Sigma$-reduct $\mathcal O.$ Let $\Mp$ be the set of statements $\phi$ in the language of $\Sigma_T$ such that $M\models I^{\bool}[\phi]=\true.$ This is a complete propositional theory extending the underlying propositional theory of $T.$ Define:

\begin{align*} z&=\{y\in\mathcal P(\omega) : \Mp \vdash c^1_y \in^2 t^2 \} \in\mathcal P(\mathcal P(\omega))\\ t^1&=u^1_2(t^2,c^2_z)\text{ (just abbreviating the term)}\\ y&=\{x\in\omega : \Mp \vdash c^0_x \in^1 t^1 \} \in\mathcal P(\omega)\\ t^0&=u^0_1(t^1,c^1_y)\\ x&=(\text{$M$’s interpretation of $I^0[t^0]$})\in\omega \end{align*}

$M \Vdash I^0[t^0]=c^0_x,$ and $T_0\vdash I^0[t^0]=c^0_x\implies I^{\bool}[t^0=c^0_x]=\true.$ So $\Mp \vdash t^0=c^0_x.$

And $\Mp\vdash t^1=c^1_y$:

• by substitution into axiom (U1), $T\vdash (t^0\in^1 t^1 \iff t^0\in^1 c^1_y)\implies t^1=c^1_y$
• hence $\Mp\vdash (t^0\in^1 t^1 \iff t^0\in^1 c^1_y) \implies t^1=c^1_y$
• we know $\Mp \vdash t^0=c^0_x$ (see previous paragraph)
• also, $\Mp \vdash c^0_x\in^1 t^1 \iff c^0_x\in^1 c^1_y$ by construction of $y$
• substitution rules using terms in $T$ are theorems of $\Mp$
• hence $\Mp\vdash t^0\in^1 t^1 \iff t^0\in^1 c^1_y$
• hence $\Mp\vdash t^1=c^1_y$

And $\Mp\vdash t^2=c^2_z$:

• by substitution into axiom (U2), $T\vdash (t^1\in^2 t^2 \iff t^1\in^2 c^2_z)\implies t^2=c^2_z$
• hence $\Mp\vdash (t^1\in^2 t^2 \iff t^1\in^2 c^2_z) \implies t^2=c^2_z$
• we know $\Mp \vdash t^1=c^1_y$ (see previous paragraph)
• also, $\Mp \vdash c^1_y\in^2 t^2 \iff c^1_y\in^2 c^2_z$ by construction of $z$
• substitution rules using terms in $T$ are theorems of $\Mp$
• hence $\Mp\vdash t^1\in^2 t^2 \iff t^1\in^2 c^2_z$
• hence $\Mp\vdash t^2=c^2_z$

But $\Mp\vdash t^2\neq c^2_z,$ a contradiction.