[proofplan] We take a descending sequence in the final iteration and build a single lower bound. The support of the lower bound is the countable union of the supports of the sequence, so countable support is preserved. At each coordinate, the partial lower bound already constructed forces that the coordinate names form a descending sequence in the next iterand; countable closure of that iterand supplies a name for a coordinatewise lower bound. Finally, we verify that the assembled condition is below every member of the original sequence and prove preservation of $\omega_1$ directly from countable closure. [/proofplan] [step:Collect the coordinates that can matter] Let $(p_n)_{n\in\mathbb N}$ be a sequence of conditions in $P_\kappa$ such that $p_{n+1}\leq_{P_\kappa} p_n$ for every $n\in\mathbb N$. For each $n\in\mathbb N$, let \begin{align*} S_n:=\operatorname{supp}(p_n)\subset \kappa \end{align*} be the countable support of $p_n$. Define \begin{align*} S:=\bigcup_{n\in\mathbb N}S_n. \end{align*} Since $S$ is a [countable union of countable sets](/theorems/755), $S$ is countable. We shall define a condition $p\in P_\kappa$ with $\operatorname{supp}(p)\subset S$. For every $\alpha<\kappa$ outside $S$, we put $p(\alpha)=1_{\dot Q_\alpha}$, the canonical $P_\alpha$-name for the top condition of $\dot Q_\alpha$. It remains to define $p(\alpha)$ for $\alpha\in S$. [/step] [step:Choose each coordinate by countable closure of the next iterand] We define $p(\alpha)$ by transfinite recursion on $\alpha<\kappa$, maintaining the induction invariant that $p\restriction\alpha$ is a condition in $P_\alpha$ and satisfies \begin{align*} p\restriction\alpha\leq_{P_\alpha}p_n\restriction\alpha \end{align*} for every $n\in\mathbb N$. At a limit ordinal $\lambda\leq\kappa$, the function $p\restriction\lambda$ has support contained in $S\cap\lambda$, which is countable, and each earlier coordinate has already been chosen so that the iteration conditionhood and order requirements hold below $\lambda$; hence $p\restriction\lambda\in P_\lambda$ and the invariant holds at $\lambda$. Now assume that $p\restriction\alpha$ has been constructed with this invariant. If $\alpha\notin S$, define $p(\alpha):=1_{\dot Q_\alpha}$. Then \begin{align*} p\restriction\alpha\Vdash_{P_\alpha} ``p(\alpha)\leq_{\dot Q_\alpha} p_n(\alpha)'' \end{align*} for every $n\in\mathbb N$, because $p_n(\alpha)=1_{\dot Q_\alpha}$ whenever $\alpha\notin S$. Now suppose $\alpha\in S$. For every $n\in\mathbb N$, since $p\restriction\alpha\leq p_n\restriction\alpha$, the condition $p\restriction\alpha$ forces that $p_n(\alpha)$ is a condition in $\dot Q_\alpha$. Also, if $m\geq n$, then $p_m\leq p_n$ in $P_\kappa$, so by the definition of the iteration order, \begin{align*} p_m\restriction\alpha\Vdash_{P_\alpha} ``p_m(\alpha)\leq_{\dot Q_\alpha}p_n(\alpha)''. \end{align*} Because $p\restriction\alpha\leq p_m\restriction\alpha$, it follows that \begin{align*} p\restriction\alpha\Vdash_{P_\alpha} ``p_m(\alpha)\leq_{\dot Q_\alpha}p_n(\alpha)'' \end{align*} whenever $m\geq n$. Thus $p\restriction\alpha$ forces that $(p_n(\alpha))_{n\in\mathbb N}$ is a descending sequence in $\dot Q_\alpha$. Since \begin{align*} 1_{P_\alpha}\Vdash_{P_\alpha} ``\dot Q_\alpha \text{ is countably closed}'', \end{align*} the condition $p\restriction\alpha$ forces that this descending sequence has a lower bound in $\dot Q_\alpha$. By the forcing maximum principle, choose a $P_\alpha$-name $p(\alpha)$ such that \begin{align*} p\restriction\alpha\Vdash_{P_\alpha} ``p(\alpha)\in \dot Q_\alpha \text{ and } p(\alpha)\leq_{\dot Q_\alpha}p_n(\alpha)\text{ for every }n\in\mathbb N''. \end{align*} [guided] The recursion works because the order in an iteration is checked coordinate by coordinate, but the statement at coordinate $\alpha$ is evaluated over the earlier forcing $P_\alpha$. So before choosing $p(\alpha)$, we must already know that the initial segment $p\restriction\alpha$ is below every $p_n\restriction\alpha$. Fix $\alpha<\kappa$ and assume that $p\restriction\alpha\in P_\alpha$ has been constructed with \begin{align*} p\restriction\alpha\leq_{P_\alpha}p_n\restriction\alpha \end{align*} for all $n\in\mathbb N$. If $\alpha\notin S$, then no original condition has a nontrivial coordinate at $\alpha$. By the definition of $S$, this means $p_n(\alpha)=1_{\dot Q_\alpha}$ for every $n\in\mathbb N$. We therefore set $p(\alpha)=1_{\dot Q_\alpha}$, and the desired inequality at this coordinate is immediate. Now let $\alpha\in S$. The goal is to find one name $p(\alpha)$ forced to be below all the coordinate names $p_n(\alpha)$. First we verify that those names really form a descending sequence as seen from $p\restriction\alpha$. Since $p\restriction\alpha\leq p_n\restriction\alpha$, every statement forced by $p_n\restriction\alpha$ is also forced by $p\restriction\alpha$ after strengthening. In particular, $p\restriction\alpha$ forces that $p_n(\alpha)$ is a condition in $\dot Q_\alpha$. Next take $m\geq n$. Because the original sequence is descending in $P_\kappa$, we have $p_m\leq p_n$. The definition of the countable-support iteration order says that this implies \begin{align*} p_m\restriction\alpha\Vdash_{P_\alpha} ``p_m(\alpha)\leq_{\dot Q_\alpha}p_n(\alpha)''. \end{align*} Since $p\restriction\alpha\leq p_m\restriction\alpha$, the same forced inequality is also forced by $p\restriction\alpha$: \begin{align*} p\restriction\alpha\Vdash_{P_\alpha} ``p_m(\alpha)\leq_{\dot Q_\alpha}p_n(\alpha)''. \end{align*} Therefore $p\restriction\alpha$ forces that $(p_n(\alpha))_{n\in\mathbb N}$ is a descending sequence in $\dot Q_\alpha$. The hypothesis on the iteration gives \begin{align*} 1_{P_\alpha}\Vdash_{P_\alpha} ``\dot Q_\alpha \text{ is countably closed}''. \end{align*} Since $p\restriction\alpha\leq 1_{P_\alpha}$, the same countable-closure statement is available below $p\restriction\alpha$. Hence $p\restriction\alpha$ forces that there is a condition of $\dot Q_\alpha$ below every $p_n(\alpha)$. By the forcing maximum principle, which turns a forced existential statement into a single name witnessing it, choose a $P_\alpha$-name $p(\alpha)$ satisfying \begin{align*} p\restriction\alpha\Vdash_{P_\alpha} ``p(\alpha)\in \dot Q_\alpha \text{ and } p(\alpha)\leq_{\dot Q_\alpha}p_n(\alpha)\text{ for every }n\in\mathbb N''. \end{align*} This is exactly the coordinate needed to continue the recursion. [/guided] [/step] [step:Assemble the recursive choices into a condition] Define the function $p$ with domain $\kappa$ by the recursion above. Its support satisfies \begin{align*} \operatorname{supp}(p)\subset S, \end{align*} because $p(\alpha)=1_{\dot Q_\alpha}$ for every $\alpha\notin S$. Since $S$ is countable, $p$ has countable support. For every $\alpha<\kappa$, the recursive construction gives \begin{align*} p\restriction\alpha\Vdash_{P_\alpha} ``p(\alpha)\in\dot Q_\alpha''. \end{align*} Together with countable support, this is exactly the conditionhood requirement in the countable-support iteration. Hence $p\in P_\kappa$. [/step] [step:Verify that the assembled condition is a common lower bound] Fix $n\in\mathbb N$. We prove that $p\leq_{P_\kappa}p_n$. By the definition of the order on a countable-support iteration, it is enough to show that for every $\alpha<\kappa$, \begin{align*} p\restriction\alpha\Vdash_{P_\alpha} ``p(\alpha)\leq_{\dot Q_\alpha}p_n(\alpha)''. \end{align*} This was arranged in the recursive construction. If $\alpha\notin S$, both coordinates are the top condition $1_{\dot Q_\alpha}$. If $\alpha\in S$, the chosen name $p(\alpha)$ was forced by $p\restriction\alpha$ to be below $p_n(\alpha)$. Therefore $p\leq p_n$ for every $n\in\mathbb N$. Since every descending $\omega$-sequence in $P_\kappa$ has a lower bound, $P_\kappa$ is countably closed. [/step] [step:Derive preservation of $\omega_1$ from countable closure] It remains to show that $P_\kappa$ preserves $\omega_1$. Let $\dot f$ be a $P_\kappa$-name and let $q\in P_\kappa$ force \begin{align*} q\Vdash_{P_\kappa} ``\dot f:\omega\to \omega_1''. \end{align*} We construct a descending sequence $(r_n)_{n\in\mathbb N}$ below $q$ and a sequence $(\beta_n)_{n\in\mathbb N}$ of ordinals below $\omega_1$ such that \begin{align*} r_n\Vdash_{P_\kappa} ``\dot f(n)=\beta_n'' \end{align*} for every $n\in\mathbb N$. This is possible by the forcing decision property, applied at stage $n$ below the already chosen condition. By countable closure of $P_\kappa$, choose $p\in P_\kappa$ such that $p\leq r_n$ for every $n\in\mathbb N$. Define \begin{align*} \beta:=\sup\{\beta_n:n\in\mathbb N\}. \end{align*} Since $\omega_1$ is the first uncountable ordinal and $\{\beta_n:n\in\mathbb N\}$ is countable, we have $\beta<\omega_1$. Moreover, $p$ forces \begin{align*} \dot f(n)=\beta_n<\beta+1<\omega_1 \end{align*} for every $n\in\mathbb N$. Hence \begin{align*} p\Vdash_{P_\kappa} ``\operatorname{range}(\dot f)\subset \beta+1''. \end{align*} Thus no condition can force that a name $\dot f:\omega\to\omega_1$ is cofinal in $\omega_1$. Therefore forcing with $P_\kappa$ adds no new countable cofinal sequence in $\omega_1$, and so $P_\kappa$ preserves $\omega_1$. [/step]