Strong compactness, fine measures and infinitary logic

Assume filter extension and fix $\lambda \ge \kappa$. For $a\in P_{\kappa }(\lambda )$, let $C_a$ be the cone of $x$ containing $a$. The sets containing some $C_a$ form a proper filter: every cone contains $a$ itself, and $C_a\cap C_b=C_{a\cup b}$. An intersection of $\eta <\kappa$ such filter members contains the cone above the union of their witnessing $a_{\xi }$. F3 selects those witnesses; regularity makes their union have size below $\kappa$ (bound all sizes below one cardinal below $\kappa$ and use the infinite-cardinal product bound). Thus the filter is $\kappa$-complete. Extend it by F1. Since $C_{\alpha }$ belongs to it for each $\alpha <\lambda$, the extension is fine.

Assume the fine-measure assertion and fix a theory $T$ as in the $L_{\kappa ,\kappa }$ assertion. Choose a cardinal $\lambda \ge \kappa$ and an injection $e:T\to \lambda$. Push a fine measure on $P_{\kappa }(\lambda )$ forward by $x\mapsto e^{-1}[x]$ to get a proper $\kappa$-complete ultrafilter $W$ on $P_{\kappa }(T)$. Inverse images preserve all intersections and complements, proving these laws. For every $t\in T$, the inverse image of the cone of small subtheories containing $t$ contains the $e(t)$-cone, so $W$ is fine.

Each small subtheory $a$ has a set model in the common signature. For each $a$ take the least rank ${\rho }_a$ of such a model code. This is uniquely defined by set satisfaction, and Replacement collects the ranks. Separation in $V_{{\rho }_a+1}$ gives the nonempty set of rank-${\rho }_a$ model codes satisfying $a$; F3 chooses one for each $a$. This bounds the choices before AC and does not choose from classes of models. Take their set ultraproduct by $W$. For each $t\in T$ the coordinate truth set contains the cone of $a$ containing $t$, hence is in $W$. F2 makes $t$ true in the ultraproduct. Thus $T$ has a model. The $L_{\kappa ,\omega }$ assertion follows by restriction of syntax.

Assume the $L_{\kappa ,\omega }$ assertion and let $F$ be a proper $\kappa$-complete filter on a set $I$. Introduce one constant $d$ and one unary relation $R_X$ for every subset $X$ of $I$; abbreviate $R_X(d)$ by $r_X$. Include $r_I$, not $r_{\mathrm{\varnothing }}$, the equivalences $r_{I\setminus X}\leftrightarrow \mathrm{\neg }{r}_X$, and $r_{{\bigcap }_{\xi <\eta }{X}_{\xi }}\leftrightarrow {\bigwedge }_{\xi <\eta }{r}_{X_{\xi }}$ for every $\eta <\kappa$ and every such sequence of subsets. Also include $r_X$ for each $X\in F$. This is a set theory in $L_{\kappa ,\omega }$, using no quantifier blocks. For any subtheory of size below $\kappa$, the fewer-than-$\kappa$ $F$-members explicitly required have a nonempty intersection by completeness and properness. Take $i$ in it and interpret the predicates on a one-element carrier by $r_X$ true exactly when $i\in X$. Every identity axiom is then true, even those mentioning other subsets, and all the chosen $F$-requirements hold. Empty collections use $I$, which is nonempty since $F$ is proper.

Compactness yields a full model. Define $U$ to be the subsets $X$ for which $r_X$ is true there. The top and bottom axioms give properness. The intersection identities give $\kappa$-completeness; their binary instances also give upward closure, since $X\subseteq Y$ implies $X\cap Y=X$ and $r_X$ forces $r_Y$. Complement identities decide exactly one of $X$ and its complement. Thus $U$ is a $\kappa$-complete ultrafilter, and the $F$-requirements give $F\subseteq U$. This proves filter extension, closing the cycle of equivalences.