[proofplan] We first record the coherence forced by the end-extension order: conditions in a filter agree below the smaller maximum. This makes the generic union $C_G$ a well-defined increasing union of closed bounded approximations. We then prove unboundedness by showing that, below every $\alpha < \kappa$, the set of conditions with maximum above $\alpha$ is dense. Finally, if $\delta$ is a [limit point](/page/Limit%20Point) of $C_G$, we trap a cofinal subset of $\delta$ inside one condition of $G$ whose maximum is above $\delta$, and the closedness of that condition gives $\delta \in C_G$. [/proofplan] [step:Use end-extension to make the generic union coherent] Let $p,q \in G$. Since $G$ is a filter, there exists $r \in G$ such that $r \leq p$ and $r \leq q$. Thus $r$ end-extends both $p$ and $q$. If $\max(p) \leq \max(q)$, then every element of $p$ lies below or equal to $\max(p) \leq \max(q)$. Since $r$ end-extends $q$, we have $q = r \cap (\max(q)+1)$, and since $p \subset r$, it follows that $p \subset q$. Here every nonempty condition has a maximum: if $s \in P_T$ is nonempty, then $s$ is a closed bounded subset of the ordinal $\kappa$, so $\sup s < \kappa$ and closedness gives $\sup s \in s$, hence $\max(s)=\sup s$. More generally, $p$ and $q$ agree on their common initial segment $p \cap (\min\{\max(p),\max(q)\}+1) = q \cap (\min\{\max(p),\max(q)\}+1)$. Therefore the family $G$ is linearly coherent by end-extension, and $C_G = \bigcup_{p \in G}p$ is the increasing union of its conditions. Since every condition $p \in P_T$ satisfies $p \subset T$, the union also satisfies $C_G \subset T$. [/step] [step:Show that conditions with maximum above any fixed ordinal are dense] Fix $\alpha < \kappa$. Define $D_\alpha := \{p \in P_T : p \neq \varnothing \text{ and } \max(p) > \alpha\}$. We prove that $D_\alpha$ is dense in $P_T$. Let $p \in P_T$. Since $p$ is bounded in $\kappa$, the ordinal $\sup p$ is below $\kappa$ when $p$ is nonempty, and we set $\sup \varnothing := 0$ if $p=\varnothing$. Because $T$ is unbounded in $\kappa$, choose $\beta \in T$ such that $\beta > \max\{\alpha,\sup p\}$. Define $q := p \cup \{\beta\}$. Then $q \subset T$, and $q$ is bounded in $\kappa$ because $\beta < \kappa$. It is closed: any increasing sequence from $q$ of length less than $\kappa$ whose supremum lies below $\kappa$ either is eventually equal to $\beta$, in which case its supremum is $\beta \in q$, or is eventually contained in $p$, in which case its supremum belongs to $p$ because $p$ is closed. Thus $q \in P_T$. Moreover $q$ end-extends $p$, because the only new point of $q$ is $\beta$, and $\beta$ is strictly above every element of $p$. Hence $q \leq p$, and $\max(q)=\beta>\alpha$. Therefore $q \in D_\alpha$, proving density. [guided] Fix an ordinal $\alpha < \kappa$. The goal is to prove that every condition can be strengthened so that its maximum is above $\alpha$. Define $D_\alpha := \{p \in P_T : p \neq \varnothing \text{ and } \max(p) > \alpha\}$. To show that $D_\alpha$ is dense, take an arbitrary condition $p \in P_T$. The condition $p$ is bounded in $\kappa$, so if $p$ is nonempty then $\sup p < \kappa$; if $p$ is empty, we use the convention $\sup \varnothing := 0$. Since $T$ is unbounded in $\kappa$, there is some $\beta \in T$ above both $\alpha$ and the old condition, namely $\beta > \max\{\alpha,\sup p\}$. Now define the extension by appending this single new point, $q := p \cup \{\beta\}$. We verify that $q$ is a valid condition. First, $q \subset T$ because $p \subset T$ and $\beta \in T$. Second, $q$ is bounded in $\kappa$ because $\beta < \kappa$ and every element of $q$ is at most $\beta$. Third, $q$ is closed. Indeed, let an increasing sequence from $q$ have length less than $\kappa$ and have supremum below $\kappa$. If the sequence contains $\beta$, then because $\beta$ is the largest element of $q$, the sequence is eventually equal to $\beta$, so its supremum is $\beta \in q$. If it does not eventually hit $\beta$, then all its terms lie in $p$, and the closedness of $p$ implies that its supremum belongs to $p \subset q$. Thus $q \in P_T$. Since $\beta$ lies strictly above every element of $p$, the condition $q$ end-extends $p$, so $q \leq p$. Finally, $\max(q)=\beta>\alpha$, so $q \in D_\alpha$. This proves that $D_\alpha$ is dense below every condition. [/guided] [/step] [step:Meet the dense sets to prove that $C_G$ is unbounded] Let $\alpha < \kappa$. Since $D_\alpha$ is dense and $G$ is generic, choose $p \in G \cap D_\alpha$. Then $\max(p) > \alpha$, and since $p \subset C_G$, there is an element of $C_G$ above $\alpha$. Therefore $C_G$ is unbounded in $\kappa$. [/step] [step:Use one condition above a limit point to prove closure] Let $\delta < \kappa$ be a limit point of $C_G$, meaning that $C_G \cap \delta$ is unbounded in $\delta$. We prove that $\delta \in C_G$. By density of $D_\delta$, choose $c \in G$ such that $\max(c)>\delta$. We claim that $c \cap \delta$ is unbounded in $\delta$. Let $\alpha < \delta$. Since $\delta$ is a limit point of $C_G$, choose $\gamma \in C_G \cap (\alpha,\delta)$. By definition of $C_G$, choose $d \in G$ such that $\gamma \in d$. Since $G$ is a filter, there is $r \in G$ with $r \leq c$ and $r \leq d$. Because $r$ end-extends $c$, we have $c = r \cap (\max(c)+1)$. Also $\gamma \in d \subset r$, and $\gamma < \delta < \max(c)$. Hence $\gamma \in r \cap (\max(c)+1)=c$. Since $\gamma \in (\alpha,\delta)$ was found inside $c$, the set $c \cap \delta$ is unbounded in $\delta$. Now $c$ is closed and $c \cap \delta$ is cofinal in $\delta$. Let $\lambda := \operatorname{cf}(\delta)$, so $\lambda < \kappa$ because $\delta < \kappa$ and $\kappa$ is regular. Choose an increasing cofinal sequence $(\gamma_i)_{i<\lambda}$ in $c \cap \delta$. Its supremum is $\delta$, and the sequence has length less than $\kappa$. Therefore the closure of $c$ under suprema of increasing sequences of length less than $\kappa$ gives $\delta = \sup_{i<\lambda}\gamma_i \in c$. Hence $\delta \in c \subset C_G$. [/step] [step:Conclude that the generic union is a club subset of $T$] We have shown that $C_G \subset T$, that $C_G$ is unbounded in $\kappa$, and that every limit point $\delta < \kappa$ of $C_G$ belongs to $C_G$. Hence $C_G$ is closed and unbounded in $\kappa$, and it is contained in $T$. This is exactly the asserted conclusion. [/step]