[proofplan] We prove separately that $C \cap D$ is [unbounded](/page/Unbounded%20Set) in $\kappa$ and [closed](/page/Closed%20Set) in $\kappa$, which is exactly what it means to be a [club](/page/Club%20Set) subset of $\kappa$. For unboundedness, starting above an arbitrary ordinal $\alpha < \kappa$, we build a strictly increasing countable sequence by alternating between points of $C$ and points of $D$. Regularity of $\kappa$ keeps the supremum of this countable sequence below $\kappa$, while closedness of both clubs puts that supremum in both $C$ and $D$. Closedness of the intersection follows because any limit point of $C \cap D$ is simultaneously a limit point of $C$ and of $D$. [/proofplan] [step:Build an alternating sequence above an arbitrary ordinal] Let $\omega$ denote the first infinite ordinal, used here as the index set of the nonnegative integers. Fix an ordinal $\alpha < \kappa$. Since $C$ is unbounded in $\kappa$, choose $c_0 \in C$ such that $\alpha < c_0 < \kappa$. Since $D$ is unbounded in $\kappa$, choose $d_0 \in D$ such that $c_0 < d_0 < \kappa$. Inductively, suppose that $c_n \in C$ and $d_n \in D$ have been chosen for some $n \in \omega$. Using unboundedness of $C$ and $D$ again, choose ordinals $c_{n+1} \in C$ and $d_{n+1} \in D$ such that \begin{align*} d_n < c_{n+1} < d_{n+1} < \kappa. \end{align*} Thus we obtain maps \begin{align*} c: \omega &\to C, & n &\mapsto c_n,\\ d: \omega &\to D, & n &\mapsto d_n \end{align*} satisfying \begin{align*} \alpha < c_0 < d_0 < c_1 < d_1 < c_2 < d_2 < \cdots < \kappa. \end{align*} [guided] We start above an arbitrary ordinal $\alpha < \kappa$ because unboundedness of $C \cap D$ means that we must eventually find some ordinal $\lambda \in C \cap D$ with $\alpha < \lambda < \kappa$. Let $\omega$ denote the first infinite ordinal, used here as the index set of the nonnegative integers. Since $C$ is unbounded in $\kappa$, there is an element $c_0 \in C$ satisfying $\alpha < c_0 < \kappa$. Since $D$ is unbounded in $\kappa$, there is an element $d_0 \in D$ satisfying $c_0 < d_0 < \kappa$. We then repeat the same two choices. If $c_n \in C$ and $d_n \in D$ have already been chosen for some $n \in \omega$, unboundedness of $C$ gives $c_{n+1} \in C$ with $d_n < c_{n+1} < \kappa$, and unboundedness of $D$ gives $d_{n+1} \in D$ with $c_{n+1} < d_{n+1} < \kappa$. This defines maps \begin{align*} c: \omega &\to C, & n &\mapsto c_n,\\ d: \omega &\to D, & n &\mapsto d_n \end{align*} with \begin{align*} \alpha < c_0 < d_0 < c_1 < d_1 < c_2 < d_2 < \cdots < \kappa. \end{align*} The point of alternating is that the eventual supremum will be approached from inside $C$ and also from inside $D$, so closedness of both sets can be used. [/guided] [/step] [step:Take the supremum and use regularity to keep it below $\kappa$] Define the ordinal \begin{align*} \lambda := \sup \{d_n : n \in \omega\}. \end{align*} The set $\{d_n : n \in \omega\}$ is countable. Since $\kappa$ is regular and uncountable, every countable subset of $\kappa$ is bounded below $\kappa$. Hence $\lambda < \kappa$. Also $c_0 > \alpha$, so \begin{align*} \alpha < \lambda < \kappa. \end{align*} [guided] Define \begin{align*} \lambda := \sup \{d_n : n \in \omega\}. \end{align*} This supremum is an ordinal. We must check that it still lies below $\kappa$, because the definition of club subsets of $\kappa$ concerns ordinals below $\kappa$. The set $\{d_n : n \in \omega\}$ is countable by the definition of $\omega$. Since $\kappa$ is regular and uncountable, its cofinality is $\kappa$, so no countable subset of $\kappa$ is cofinal in $\kappa$. Therefore the [countable set](/page/Countable%20Set) $\{d_n : n \in \omega\}$ is bounded in $\kappa$, and its supremum satisfies $\lambda < \kappa$. Moreover, $\alpha < c_0 < d_0 \leq \lambda$, so $\alpha < \lambda$. Hence \begin{align*} \alpha < \lambda < \kappa. \end{align*} This gives a candidate point above $\alpha$; the remaining work is to show that $\lambda$ belongs to both clubs. [/guided] [/step] [step:Use closedness of both clubs to put the supremum in the intersection] The set $\{d_n : n \in \omega\}$ is cofinal in the limit ordinal $\lambda$ by definition of $\lambda$. Since $\lambda < \kappa$, each $d_n$ lies in $D$, and $D$ is closed in $\kappa$, the closure criterion for clubs gives $\lambda \in D$. The set $\{c_n : n \in \omega\}$ is also cofinal in $\lambda$. Indeed, if $\xi < \lambda$, then by the definition of $\lambda$ there exists $n \in \omega$ such that $\xi < d_n$, and then \begin{align*} \xi < d_n < c_{n+1} < \lambda. \end{align*} Since each $c_n$ lies in $C$ and $C$ is closed in $\kappa$, it follows that $\lambda \in C$. Therefore $\lambda \in C \cap D$. Because the original ordinal $\alpha < \kappa$ was arbitrary and we have found $\lambda \in C \cap D$ with $\alpha < \lambda < \kappa$, the set $C \cap D$ is unbounded in $\kappa$. [guided] We first show that $\lambda$ belongs to $D$. By definition, \begin{align*} \lambda = \sup \{d_n : n \in \omega\}. \end{align*} Thus the set $\{d_n : n \in \omega\}$ is unbounded in $\lambda$. Since every $d_n$ lies in $D$, the set $D \cap \lambda$ is unbounded in $\lambda$. The set $D$ is closed in $\kappa$, so whenever a limit ordinal below $\kappa$ is approached cofinally by points of $D$, that limit ordinal belongs to $D$. Since $\lambda < \kappa$, we conclude that $\lambda \in D$. Next we show that $\lambda$ belongs to $C$. The sequence $(c_n)_{n=0}^{\infty}$ is not the sequence used to define $\lambda$, so we verify cofinality explicitly. Let $\xi < \lambda$. Since $\lambda$ is the supremum of the $d_n$, there exists $n \in \omega$ such that $\xi < d_n$. The alternating construction gives \begin{align*} \xi < d_n < c_{n+1} < \lambda. \end{align*} Thus the points $c_n \in C$ are also cofinal in $\lambda$. Since $C$ is closed in $\kappa$ and $\lambda < \kappa$, this implies $\lambda \in C$. We have proved $\lambda \in C$ and $\lambda \in D$, hence $\lambda \in C \cap D$. Since $\alpha < \lambda < \kappa$ and $\alpha < \kappa$ was arbitrary, $C \cap D$ is unbounded in $\kappa$. [/guided] [/step] [step:Show that the intersection is closed] Let $\mu < \kappa$ be a limit ordinal such that $(C \cap D) \cap \mu$ is unbounded in $\mu$. Since \begin{align*} (C \cap D) \cap \mu \subset C \cap \mu \end{align*} and $(C \cap D) \cap \mu$ is unbounded in $\mu$, the set $C \cap \mu$ is unbounded in $\mu$. Because $C$ is closed in $\kappa$, this gives $\mu \in C$. Similarly, \begin{align*} (C \cap D) \cap \mu \subset D \cap \mu, \end{align*} so $D \cap \mu$ is unbounded in $\mu$. Since $D$ is closed in $\kappa$, we have $\mu \in D$. Therefore $\mu \in C \cap D$. Thus $C \cap D$ is closed in $\kappa$. Combining this with unboundedness, $C \cap D$ is a closed unbounded subset of $\kappa$. [/step]