Cardinal exponentiation below the continuum under MA

Fix $X\subseteq \kappa$. Let $P_X$ consist of pairs $(s,F)$ with $s$ a finite partial map $\omega \to 2$ and $F\subseteq \kappa \setminus X$ finite. Put $(t,H)\le (s,F)$ iff $s\subseteq t$, $F\subseteq H$, and $t^{-1}(1)\cap (\mathrm{dom}(t)\setminus \mathrm{dom}(s))\text{is disjoint from}{\bigcup }_{\alpha \in F}{A}_{\alpha }\mathrm{.}$ This is transitive: an extension never puts a new $1$ on any set already protected by the weaker condition. Conditions with the same stem $s$ are compatible, since $(s,F\cup H)$ extends both. There are only countably many finite stems, so $P_X$ is $\sigma$-centered and hence ccc.

For $\alpha \mathrm{\notin }X$, let $E_{\alpha }=\{(s,F):\alpha \in F\};$ this is dense because adjoining $\alpha$ to $F$ changes no stem. For $\alpha \in X$ and $n<\omega$, let $D_{\alpha ,n}=\{(s,F):(\mathrm{\exists }m>n)[m\in A_{\alpha }\cap \mathrm{dom}(s)\text{ and }s(m)=1]\}\mathrm{.}$ Given $(s,F)$, the set $A_{\alpha }\cap {\bigcup }_{\beta \in F}{A}_{\beta }$ is finite by almost disjointness. Since $A_{\alpha }$ is infinite, choose $m>n$ outside that finite set and $\mathrm{dom}(s)$; setting $s(m)=1$ gives an extension in $D_{\alpha ,n}$. Thus all these sets are dense. Their number is at most $\kappa \cdot {\mathrm{\aleph }}_0=\kappa$, so MA supplies a filter $G$ meeting them. Let $r_X=\{m:(\mathrm{\exists }(s,F)\in G)s(m)=1\}\mathrm{.}$ Directedness makes the stems in $G$ consistent. If $\alpha \in X$, meeting every $D_{\alpha ,n}$ makes $r_X\cap A_{\alpha }$ infinite. If $\alpha \mathrm{\notin }X$, choose $(s,F)\in G\cap E_{\alpha }$. For any $(t,H)\in G$, take $(u,K)\in G$ below both. Since $(u,K)\le (s,F)$ and $\alpha \in F$, no new $1$ of $u$ beyond $s$ lies in $A_{\alpha }$; hence $t^{-1}(1)\cap A_{\alpha }\subseteq s^{-1}(1)$. Consequently $r_X\cap A_{\alpha }\subseteq s^{-1}(1)$ is finite. Therefore $X=\{\alpha :\mathrm{\mid }{r}_X\cap A_{\alpha }\mathrm{\mid }={\mathrm{\aleph }}_0\}$.

Choosing the least real in a fixed well-order among the codes for each $X$ gives an injection $\mathcal{P}(\kappa )\hookrightarrow \mathcal{P}(\omega )$; AC is used here. Monotonicity gives $\mathfrak{c}\le 2^{\kappa }$, hence equality. If $\mathrm{cf}(\mathfrak{c})=\lambda <\mathfrak{c}$, then $\mathfrak{c}\le {\mathfrak{c}}^{\lambda }=(2^{\lambda }{)}^{\lambda }=2^{\lambda }=\mathfrak{c}$, while König gives ${\mathfrak{c}}^{\lambda }>\mathfrak{c}$, contradiction. Therefore $\mathfrak{c}$ is regular.