[proofplan] We force the desired new set by finite approximations. A condition records a finite initial piece $s \subseteq \omega$ of the new set and a finite subfamily $F \subseteq \mathcal A$ that future additions must avoid. The forcing is $\sigma$-centered, hence ccc, and Martin's axiom gives a filter meeting the dense sets that make the generic set infinite and eventually avoid every member of $\mathcal A$. The union of the finite approximations in the filter is then an infinite set almost disjoint from all old members of $\mathcal A$, so $\mathcal A$ was not maximal. [/proofplan] [step:Choose the forcing and its extension relation] Let $\lambda := |\mathcal A|+\aleph_0$. Since $\mathcal A$ is infinite and $|\mathcal A| < 2^{\aleph_0}$, we have $\lambda = |\mathcal A| < 2^{\aleph_0}$. Define a partially ordered set $\mathbb P$ as follows. Its elements are pairs $p=(s_p,F_p)$, where $s_p \in [\omega]^{<\omega}$ and $F_p \in [\mathcal A]^{<\omega}$. For $p=(s_p,F_p)$ and $q=(s_q,F_q)$ in $\mathbb P$, write $q \leq p$ if and only if \begin{align*} s_p \subseteq s_q \end{align*} \begin{align*} F_p \subseteq F_q \end{align*} \begin{align*} (s_q \setminus s_p) \cap \bigcup F_p = \varnothing \end{align*} and every element of $s_q \setminus s_p$ is larger than every element of $s_p$. If $s_p=\varnothing$, the last condition is vacuous. Thus a stronger condition may add more forbidden old sets, and may add new integers only above the current finite approximation and outside the old sets already listed in $F_p$. This relation is reflexive because $s_p \setminus s_p=\varnothing$ and $F_p \subseteq F_p$. It is transitive because, if $r \leq q \leq p$, then $F_p \subseteq F_q \subseteq F_r$, the new points in $s_q \setminus s_p$ avoid $\bigcup F_p$, and the new points in $s_r \setminus s_q$ avoid $\bigcup F_q$, hence also avoid $\bigcup F_p$. The order condition on the sizes of newly added integers is preserved for the same reason: points added from $s_q$ over $s_p$ are above $s_p$, and points added from $s_r$ over $s_q$ are above $s_q$, hence above $s_p$. Therefore $\mathbb P$ is a partially ordered set. [/step] [step:Show that the forcing is $\sigma$-centered] For each finite set $s \in [\omega]^{<\omega}$, define \begin{align*} \mathbb P_s := \{(s,F) \in \mathbb P : F \in [\mathcal A]^{<\omega}\}. \end{align*} Then \begin{align*} \mathbb P = \bigcup_{s \in [\omega]^{<\omega}} \mathbb P_s. \end{align*} The set $[\omega]^{<\omega}$ is countable, so it remains to show that each $\mathbb P_s$ is centered. Fix $s \in [\omega]^{<\omega}$ and finitely many conditions $(s,F_0),\dots,(s,F_{k-1}) \in \mathbb P_s$, where $k \in \omega$ and $k \geq 1$. Define \begin{align*} F_* := \bigcup_{i=0}^{k-1} F_i. \end{align*} Since each $F_i$ is a finite subset of $\mathcal A$, the set $F_*$ is also a finite subset of $\mathcal A$, so $(s,F_*) \in \mathbb P$. For every $i<k$, we have $s \subseteq s$, $F_i \subseteq F_*$, and no new element is added to $s$. Hence $(s,F_*) \leq (s,F_i)$ for every $i<k$. Thus every finite subset of $\mathbb P_s$ has a common extension, so $\mathbb P_s$ is centered. Therefore $\mathbb P$ is a countable union of centered subsets. In particular, $\mathbb P$ is $\sigma$-centered, and hence $\mathbb P$ satisfies the countable chain condition. [/step] [step:Define the dense sets that force infinitude and eventual avoidance] For each $A \in \mathcal A$, define \begin{align*} D_A := \{(s,F) \in \mathbb P : A \in F\}. \end{align*} We show that $D_A$ is dense in $\mathbb P$. Let $p=(s_p,F_p) \in \mathbb P$. Define $q=(s_p,F_p \cup \{A\})$. Then $q \in \mathbb P$, and $q \leq p$ because no new element is added to $s_p$ and $F_p \subseteq F_p \cup \{A\}$. Thus $D_A$ is dense. For each $n \in \omega$, define \begin{align*} E_n := \{(s,F) \in \mathbb P : \text{there exists } m \in s \text{ with } m>n\}. \end{align*} We show that $E_n$ is dense in $\mathbb P$. Let $p=(s_p,F_p) \in \mathbb P$. Since $\mathcal A$ is infinite and $F_p$ is finite, choose $B \in \mathcal A \setminus F_p$. For every $A \in F_p$, the set $B \cap A$ is finite because $\mathcal A$ is almost disjoint and $B \neq A$. Hence \begin{align*} B \cap \bigcup F_p = \bigcup_{A \in F_p} (B \cap A) \end{align*} is finite. Since $B$ is infinite, the set $B \setminus \bigcup F_p$ is infinite. Choose \begin{align*} m \in B \setminus \bigcup F_p \end{align*} with $m>n$ and with $m$ larger than every element of $s_p$. Define $q=(s_p \cup \{m\},F_p)$. Then $q \in \mathbb P$, $q \leq p$, and $q \in E_n$. Therefore $E_n$ is dense. [guided] The purpose of the dense sets is to encode the two required properties of the final set $X$. The sets $E_n$ force $X$ to contain arbitrarily large integers, and therefore to be infinite. The sets $D_A$ force each old set $A \in \mathcal A$ to enter the forbidden list at some finite stage, after which no later extension may add new points from $A$. Fix $A \in \mathcal A$. We define \begin{align*} D_A := \{(s,F) \in \mathbb P : A \in F\}. \end{align*} This set is dense because any condition can be strengthened by enlarging only its second coordinate. Indeed, if $p=(s_p,F_p) \in \mathbb P$, define $q=(s_p,F_p \cup \{A\})$. No new integer is added to $s_p$, so the avoidance requirement in the extension relation is automatically satisfied. Also $F_p \subseteq F_p \cup \{A\}$, so $q \leq p$, and by construction $q \in D_A$. Now fix $n \in \omega$ and define \begin{align*} E_n := \{(s,F) \in \mathbb P : \text{there exists } m \in s \text{ with } m>n\}. \end{align*} To prove that $E_n$ is dense, start with an arbitrary condition $p=(s_p,F_p)$. We need to add a new integer above $n$, but the extension relation only allows us to add integers outside $\bigcup F_p$. Choose $B \in \mathcal A \setminus F_p$, which is possible because $\mathcal A$ is infinite and $F_p$ is finite. For each $A \in F_p$, the intersection $B \cap A$ is finite by almost disjointness. Since $F_p$ is finite, the union \begin{align*} B \cap \bigcup F_p = \bigcup_{A \in F_p} (B \cap A) \end{align*} is finite. The set $B$ itself is infinite, so $B \setminus \bigcup F_p$ is infinite. Therefore we can choose \begin{align*} m \in B \setminus \bigcup F_p \end{align*} such that $m>n$ and $m$ is larger than every element of $s_p$. Define $q=(s_p \cup \{m\},F_p)$. The newly added point $m$ avoids $\bigcup F_p$ and lies above all old points in $s_p$, so $q \leq p$. Also $m \in s_p \cup \{m\}$ and $m>n$, so $q \in E_n$. Thus $E_n$ is dense. [/guided] [/step] [step:Apply Martin's axiom to obtain a filter meeting all dense requirements] Let \begin{align*} \mathcal D := \{D_A : A \in \mathcal A\} \cup \{E_n : n \in \omega\}. \end{align*} Then \begin{align*} |\mathcal D| \leq |\mathcal A|+\aleph_0 = \lambda. \end{align*} The poset $\mathbb P$ satisfies the countable chain condition, and $MA_\lambda$ holds by the hypothesis that $MA_\kappa$ holds for every $\kappa < 2^{\aleph_0}$ together with $\lambda < 2^{\aleph_0}$. Therefore there exists a filter $G \subseteq \mathbb P$ such that \begin{align*} G \cap D \neq \varnothing \end{align*} for every $D \in \mathcal D$. [/step] [step:Build the new set from the generic finite approximations] Define \begin{align*} X := \bigcup\{s \subseteq \omega : \text{there exists } F \in [\mathcal A]^{<\omega} \text{ with } (s,F) \in G\}. \end{align*} Then $X \subseteq \omega$. We first show that $X$ is infinite. Let $n \in \omega$. Since $G$ meets $E_n$, there exists $(s,F) \in G \cap E_n$. By the definition of $E_n$, there exists $m \in s$ with $m>n$. Since $s \subseteq X$, we have $m \in X$. Thus for every $n \in \omega$ there is $m \in X$ with $m>n$, so $X$ is infinite. [/step] [step:Show that the new set is almost disjoint from every old set] Fix $A \in \mathcal A$. Since $G$ meets $D_A$, choose a condition $p_A=(s_A,F_A) \in G \cap D_A$. Thus $A \in F_A$. We claim that \begin{align*} X \cap A \subseteq s_A. \end{align*} Let $m \in X \cap A$. By the definition of $X$, choose $q=(s_q,F_q) \in G$ such that $m \in s_q$. Since $G$ is a filter, there exists $r=(s_r,F_r) \in G$ such that $r \leq p_A$ and $r \leq q$. From $r \leq q$ and $m \in s_q$, we get $m \in s_r$. Suppose, toward a contradiction, that $m \notin s_A$. Then $m \in s_r \setminus s_A$. Since $r \leq p_A$, the extension relation gives \begin{align*} (s_r \setminus s_A) \cap \bigcup F_A = \varnothing. \end{align*} But $A \in F_A$, so $m \notin A$, contradicting $m \in X \cap A$. Therefore $m \in s_A$, proving $X \cap A \subseteq s_A$. Since $s_A$ is finite, $X \cap A$ is finite. As $A \in \mathcal A$ was arbitrary, $X$ is almost disjoint from every member of $\mathcal A$. [/step] [step:Conclude that the almost disjoint family was not maximal] The set $X$ is infinite and $X \cap A$ is finite for every $A \in \mathcal A$. In particular, $X \notin \mathcal A$, because if $X=A_0$ for some $A_0 \in \mathcal A$, then \begin{align*} X \cap A_0 = X \end{align*} would be infinite, contradicting the result just proved. Therefore $\mathcal A \cup \{X\}$ is an almost disjoint family properly extending $\mathcal A$. Hence $\mathcal A$ is not maximal almost disjoint. [/step]