[proofplan] We prove density of the intersection directly. Starting below an arbitrary condition $p\in P$, we recursively build a descending sequence that meets the dense [open set](/page/Open%20Set) $D_\alpha$ at the next stage after $\alpha$. At limit stages, and once at the end, $\kappa$-closure supplies lower bounds because every sequence used has length $<\kappa$. The final lower bound lies below a condition in each $D_\alpha$, so openness gives membership in every $D_\alpha$. [/proofplan] [step:Fix a condition and reduce the goal to finding a common extension] Let $p\in P$ be arbitrary. Let $\lambda<\kappa$ be a cardinal, and let $(D_\alpha)_{\alpha<\lambda}$ be a family of dense open subsets of $P$. To prove that $\bigcap_{\alpha<\lambda}D_\alpha$ is dense in $P$, it suffices to find a condition $q\in P$ such that $q\leq p$ and $q\in D_\alpha$ for every $\alpha<\lambda$. If $\lambda=0$, then $\bigcap_{\alpha<\lambda}D_\alpha=P$, so $q:=p$ works. Hence assume $\lambda>0$. [/step] [step:Build a descending sequence that meets each dense open set] We recursively define a descending sequence $(p_\xi)_{\xi\leq\lambda}$ in $P$ with $p_0=p$ and with $p_{\alpha+1}\in D_\alpha$ whenever $\alpha<\lambda$. Set $p_0:=p$. Suppose $\xi\leq\lambda$ and $p_\eta$ has been defined for all $\eta<\xi$. If $\xi=\alpha+1$ is a successor ordinal, then $p_\alpha$ is already defined. Since $D_\alpha$ is dense in $P$, there exists $r\in D_\alpha$ such that $r\leq p_\alpha$. Define $p_{\alpha+1}:=r$. If $\xi$ is a nonzero limit ordinal, then the sequence $(p_\eta)_{\eta<\xi}$ is descending and has length $\xi\leq\lambda<\kappa$. By $\kappa$-closure, it has a lower bound $r\in P$. Define $p_\xi:=r$. Then $p_\xi\leq p_\eta$ for every $\eta<\xi$. This recursion defines $(p_\xi)_{\xi\leq\lambda}$, and by construction the sequence is descending. [guided] We want to meet the dense sets one at a time while never moving above the original condition $p$. The natural object to build is a transfinite descending sequence indexed by the ordinals up to $\lambda$. Define $p_0:=p$. Now suppose that $p_\eta$ has been constructed for every $\eta<\xi$, where $\xi\leq\lambda$. First consider the successor case $\xi=\alpha+1$. The condition $p_\alpha$ is already available. Since $D_\alpha$ is dense in $P$, the definition of density gives a condition $r\in D_\alpha$ with $r\leq p_\alpha$. We set $p_{\alpha+1}:=r$. This both keeps the sequence descending and ensures that the stage immediately after $\alpha$ has entered $D_\alpha$. Now consider the limit case. Let $\xi$ be a nonzero limit ordinal. The previously constructed sequence $(p_\eta)_{\eta<\xi}$ is descending: each later condition was chosen below the earlier ones. Its length is $\xi$, and since $\xi\leq\lambda<\kappa$, this is a descending sequence of length $<\kappa$. The hypothesis that $P$ is $\kappa$-closed therefore gives a lower bound $r\in P$ such that $r\leq p_\eta$ for every $\eta<\xi$. We define $p_\xi:=r$. Thus the recursion produces a descending sequence $(p_\xi)_{\xi\leq\lambda}$ with $p_0=p$ and $p_{\alpha+1}\in D_\alpha$ for each $\alpha<\lambda$. [/guided] [/step] [step:Use the terminal lower bound to enter every dense open set] Let $q:=p_\lambda$. Since the sequence is descending and $p_0=p$, we have $q\leq p$. Fix $\alpha<\lambda$. By construction, $p_{\alpha+1}\in D_\alpha$. Since $\alpha+1\leq\lambda$, the sequence is descending, so $q=p_\lambda\leq p_{\alpha+1}$. Because $D_\alpha$ is open downward, $q\in D_\alpha$. Since $\alpha<\lambda$ was arbitrary, $q\in\bigcap_{\alpha<\lambda}D_\alpha$. Therefore, for every $p\in P$, there is $q\leq p$ with $q\in\bigcap_{\alpha<\lambda}D_\alpha$, so $\bigcap_{\alpha<\lambda}D_\alpha$ is dense in $P$. Hence $P$ is $\kappa$-distributive. [/step]