[proofplan] We first verify inside $M$ that the finite-support Cohen forcing $\operatorname{Add}(\omega,\kappa)^M$ has the countable chain condition. The only combinatorial input is the $\Delta$-system argument for finite domains: an uncountable family of finite partial functions contains two conditions with the same values on the common root, hence two compatible conditions. Once ccc is known, we prove directly that a ccc forcing cannot collapse a ground-model cardinal, because each coordinate of a name for a function has only countably many possible values below any fixed condition. The same countable-possibilities argument rules out new cofinal maps of smaller length into an uncountable regular cardinal. [/proofplan] [step:Verify that finite-support Cohen forcing is ccc inside $M$] In $M$, the forcing notion $\mathbb{P}$ is the set of finite partial functions \begin{align*} p: \operatorname{dom}(p) \to 2 \end{align*} where $\operatorname{dom}(p)$ is a finite subset of $\kappa \times \omega$, ordered by reverse inclusion: $q \leq p$ iff $q \supseteq p$. We prove that $M \models ``\mathbb{P}$ is ccc". Let $A \in M$ and suppose \begin{align*} M \models ``A \subseteq \mathbb{P} \text{ is uncountable}". \end{align*} Working in $M$, apply the $\Delta$-system lemma to the family \begin{align*} \{\operatorname{dom}(p) : p \in A\}. \end{align*} There is an uncountable set $A_0 \subseteq A$ and a finite set $r \subseteq \kappa \times \omega$ such that \begin{align*} \operatorname{dom}(p) \cap \operatorname{dom}(q) = r \end{align*} for all distinct $p,q \in A_0$. Since $r$ is finite, there are only finitely many functions from $r$ to $2$. Hence, thinning $A_0$ inside $M$, there is an uncountable set $A_1 \subseteq A_0$ such that \begin{align*} p|_r = q|_r \end{align*} for all $p,q \in A_1$. Choose distinct $p,q \in A_1$. The functions $p$ and $q$ agree on the intersection of their domains, because that intersection is $r$ and their restrictions to $r$ are equal. Therefore the union \begin{align*} s: \operatorname{dom}(p) \cup \operatorname{dom}(q) \to 2 \end{align*} defined by $s(z)=p(z)$ for $z \in \operatorname{dom}(p)$ and $s(z)=q(z)$ for $z \in \operatorname{dom}(q)$ is a well-defined finite partial function. Thus $s \in \mathbb{P}$ and $s \leq p,q$, so $p$ and $q$ are compatible. Hence $A$ is not an antichain. Since every uncountable subset of $\mathbb{P}$ in $M$ contains two compatible conditions, $M \models ``\mathbb{P}$ has the countable chain condition". [/step] [step:Bound the possible values of each coordinate of a name] Let $p_0 \in \mathbb{P}$, let $\mu,\lambda \in M$ be ordinals, and let $\dot f \in M$ be a $\mathbb{P}$-name such that \begin{align*} p_0 \Vdash_{\mathbb{P}} ``\dot f:\mu \to \lambda". \end{align*} For each $\alpha \in \mu$, define in $M$ the set of possible values below $p_0$ by \begin{align*} B_\alpha = \{\beta \in \lambda : \text{there exists } q \leq p_0 \text{ such that } q \Vdash_{\mathbb{P}} \dot f(\alpha)=\beta\}. \end{align*} We claim that $M$ satisfies that $B_\alpha$ is countable for every $\alpha \in \mu$. Fix $\alpha \in \mu$. For each $\beta \in B_\alpha$, choose a condition $q_\beta \leq p_0$ such that \begin{align*} q_\beta \Vdash_{\mathbb{P}} \dot f(\alpha)=\beta. \end{align*} If $\beta,\gamma \in B_\alpha$ and $\beta \neq \gamma$, then $q_\beta$ and $q_\gamma$ are incompatible; otherwise a common extension would force both $\dot f(\alpha)=\beta$ and $\dot f(\alpha)=\gamma$, contradicting $\beta \neq \gamma$. Thus \begin{align*} \{q_\beta : \beta \in B_\alpha\} \end{align*} is an antichain in $\mathbb{P}$. Since $M \models ``\mathbb{P}$ is ccc", this antichain is countable in $M$. Therefore $B_\alpha$ is countable in $M$. [guided] The purpose of this step is to isolate the one place where ccc controls names. A name for a function may be complicated, but ccc says that below a fixed condition, a single input cannot be forced to take uncountably many different values. Fix an ordinal $\alpha \in \mu$. We define the set \begin{align*} B_\alpha = \{\beta \in \lambda : \text{there exists } q \leq p_0 \text{ such that } q \Vdash_{\mathbb{P}} \dot f(\alpha)=\beta\}. \end{align*} Thus $B_\alpha$ is the set of all values that some extension of $p_0$ can still force for the $\alpha$-coordinate of $\dot f$. We prove that $B_\alpha$ is countable in $M$. For each $\beta \in B_\alpha$, choose a condition $q_\beta \leq p_0$ with \begin{align*} q_\beta \Vdash_{\mathbb{P}} \dot f(\alpha)=\beta. \end{align*} Now suppose $\beta,\gamma \in B_\alpha$ and $\beta \neq \gamma$. If $q_\beta$ and $q_\gamma$ were compatible, there would be a condition $r \leq q_\beta,q_\gamma$. Since forcing is monotone downward, $r$ would force both statements \begin{align*} \dot f(\alpha)=\beta \end{align*} and \begin{align*} \dot f(\alpha)=\gamma. \end{align*} A condition cannot force two distinct ordinal values for the same term, so this is impossible. Hence $q_\beta$ and $q_\gamma$ are incompatible whenever $\beta \neq \gamma$. Therefore \begin{align*} \{q_\beta : \beta \in B_\alpha\} \end{align*} is an antichain in $\mathbb{P}$. Since the previous step proved that $M$ satisfies the countable chain condition for $\mathbb{P}$, every antichain in $\mathbb{P}$ belonging to $M$ is countable in $M$. It follows that $B_\alpha$ is countable in $M$. [/guided] [/step] [step:Preserve finite cardinals and $\omega$ separately] Let $\lambda \in M$ be a cardinal of $M$ with $M \models \lambda \leq \omega$. We prove first that $\lambda$ remains a cardinal in $M[G]$. If $M \models \lambda < \omega$, then $\lambda$ is a finite ordinal. For every ordinal $\mu \in M$ with $M \models \mu < \lambda$, the assertion that there is a surjection from $\mu$ onto $\lambda$ is a finite first-order statement about the ordinals below $\lambda$, hence is absolute between the transitive models $M$ and $M[G]$. Since $M$ satisfies that no such surjection exists, $M[G]$ also satisfies that no such surjection exists. Thus every finite cardinal of $M$ remains a cardinal in $M[G]$. Now suppose $M \models \lambda = \omega$. If $\omega$ were not a cardinal in $M[G]$, then some condition $p_0 \in G$ would force that there are an ordinal $\mu \in M$ with $M \models \mu < \omega$ and a $\mathbb{P}$-name $\dot f \in M$ such that \begin{align*} p_0 \Vdash_{\mathbb{P}} ``\dot f:\mu \to \omega \text{ is surjective}". \end{align*} But $M \models \mu < \omega$ means that $\mu$ is finite, and forcing preserves the finite ordinal $\mu$ and its set of coordinates. In every forcing extension, the range of a function with finite domain $\mu$ has cardinality at most the finite ordinal $\mu$, so it cannot be all of $\omega$. Thus no condition can force such a surjection, and $\omega^M$ remains a cardinal in $M[G]$. [/step] [step:Preserve uncountable ground-model cardinals] Let $\lambda \in M$ be a cardinal of $M$ with $M \models \lambda > \omega$. We prove that $\lambda$ is still a cardinal in $M[G]$. Suppose toward a contradiction that some condition $p_0 \in G$ forces that $\lambda$ is not a cardinal. Then there are an ordinal $\mu \in M$ with $M \models \mu < \lambda$ and a $\mathbb{P}$-name $\dot f \in M$ such that \begin{align*} p_0 \Vdash_{\mathbb{P}} ``\dot f:\mu \to \lambda \text{ is surjective}". \end{align*} For each $\alpha \in \mu$, let $B_\alpha \in M$ be the [countable set](/page/Countable%20Set) of possible values of $\dot f(\alpha)$ below $p_0$ defined in the previous step. Define \begin{align*} B = \bigcup_{\alpha \in \mu} B_\alpha. \end{align*} In $M$, the set $B$ has cardinality at most $|\mu| \cdot \omega$. Since $M \models ``\lambda$ is an uncountable cardinal" and $M \models \mu < \lambda$, the cardinal $|\mu|^M$ is strictly below $\lambda$, and cardinal arithmetic in $M$ gives \begin{align*} M \models |\mu| \cdot \omega = \max\{|\mu|,\omega\} < \lambda. \end{align*} Hence \begin{align*} M \models |B| < \lambda. \end{align*} Therefore there exists $\beta \in \lambda \setminus B$ in $M$. Because $p_0$ forces that $\dot f$ is surjective onto $\lambda$, there is an extension $q \leq p_0$ and an ordinal $\alpha \in \mu$ such that \begin{align*} q \Vdash_{\mathbb{P}} \dot f(\alpha)=\beta. \end{align*} By the definition of $B_\alpha$, this implies $\beta \in B_\alpha \subseteq B$, contradicting $\beta \notin B$. Therefore no such name for a surjection exists, and every uncountable cardinal of $M$ remains a cardinal in $M[G]$. [/step] [step:Preserve regularity of uncountable regular cardinals] Let $\lambda \in M$ and suppose \begin{align*} M \models ``\lambda \text{ is an uncountable regular cardinal}". \end{align*} We prove that $M[G] \models ``\lambda$ is regular". Suppose toward a contradiction that some $p_0 \in G$ forces that $\lambda$ is singular. Then there are an ordinal $\mu \in M$ with $M \models \mu < \lambda$ and a $\mathbb{P}$-name $\dot f \in M$ such that \begin{align*} p_0 \Vdash_{\mathbb{P}} ``\dot f:\mu \to \lambda \text{ is cofinal}". \end{align*} For each $\alpha \in \mu$, define $B_\alpha$ as in the coordinate-bounding step, and set \begin{align*} B = \bigcup_{\alpha \in \mu} B_\alpha. \end{align*} As before, $M$ satisfies $|B| \leq |\mu| \cdot \omega < \lambda$. Since $\lambda$ is regular in $M$, every subset of $\lambda$ of cardinality less than $\lambda$ is bounded in $\lambda$. Therefore there is an ordinal $\delta \in \lambda$ such that \begin{align*} B \subseteq \delta. \end{align*} Because $p_0$ forces that $\dot f$ is cofinal in $\lambda$, there are $q \leq p_0$ and $\alpha \in \mu$ such that \begin{align*} q \Vdash_{\mathbb{P}} \dot f(\alpha)>\delta. \end{align*} Strengthening $q$ if necessary, choose $r \leq q$ and $\beta \in \lambda$ such that \begin{align*} r \Vdash_{\mathbb{P}} \dot f(\alpha)=\beta. \end{align*} Then $r \leq p_0$, so $\beta \in B_\alpha \subseteq B \subseteq \delta$. This contradicts the fact that $r$ also forces $\dot f(\alpha)>\delta$. Hence $\lambda$ remains regular in $M[G]$. [/step] [step:Identify the first two uncountable cardinals in the extension] By the cardinal-preservation step, $\omega_1^M$ and $\omega_2^M$ remain cardinals in $M[G]$. Since forcing does not add ordinals, every ordinal of $M[G]$ is already an ordinal of $M$. If $\alpha < \omega_1^M$, then $M$ has a bijection from $\omega$ onto $\alpha$, and the same bijection belongs to $M[G]$. Thus no ordinal below $\omega_1^M$ is uncountable in $M[G]$, so \begin{align*} \omega_1^{M[G]} = \omega_1^M. \end{align*} If $\omega_1^M \leq \alpha < \omega_2^M$, then $M$ has a bijection from $\omega_1^M$ onto $\alpha$. Since $\omega_1^M$ remains a cardinal in $M[G]$, the ordinal $\alpha$ has cardinality at most $\omega_1^{M[G]}$ in $M[G]$. Therefore no cardinal strictly between $\omega_1^M$ and $\omega_2^M$ appears in the extension, and \begin{align*} \omega_2^{M[G]} = \omega_2^M. \end{align*} This proves the stated preservation conclusions for the Cohen continuum calculation. [/step]