[proofplan] Closure follows directly from the definition of a [closed unbounded set](/page/Closed%20Unbounded%20Set): a limit point of the intersection is a limit point of every club in the family. For unboundedness, we start above an arbitrary ordinal $\alpha<\kappa$ and construct an increasing sequence of length $\mu \cdot \omega$, repeatedly visiting every club $C_i$, where $\omega$ denotes the first infinite ordinal. [Regularity](/page/Regular%20Cardinal) of $\kappa$ keeps the final supremum below $\kappa$, and because each $C_i$ is visited cofinally often in the construction, closedness puts the final supremum in every $C_i$. [/proofplan] [step:Show that the intersection is closed in $\kappa$] Let \begin{align*} C := \bigcap_{i<\mu} C_i. \end{align*} We prove that $C$ is closed in $\kappa$. Let $\lambda<\kappa$ be a limit ordinal such that \begin{align*} \sup(C \cap \lambda)=\lambda. \end{align*} For each $i<\mu$, we have $C \subset C_i$, so \begin{align*} C \cap \lambda \subset C_i \cap \lambda. \end{align*} Hence $\sup(C_i \cap \lambda)=\lambda$. Since $C_i$ is closed in $\kappa$, it follows that $\lambda \in C_i$. This holds for every $i<\mu$, so $\lambda \in C$. Therefore $C$ is closed in $\kappa$. [/step] [step:Handle the empty family case] If $\mu=0$, then \begin{align*} \bigcap_{i<0} C_i=\kappa. \end{align*} The set $\kappa$ is closed and unbounded in itself, so the theorem holds in this case. For the rest of the proof assume $\mu>0$. [/step] [step:Construct an increasing sequence that visits every club cofinally often] Fix an ordinal $\alpha<\kappa$. We will find an element of $C$ above $\alpha$. Let $\omega$ denote the first infinite ordinal. Let $\theta$ be the ordinal product \begin{align*} \theta := \mu \cdot \omega. \end{align*} Since $\mu<\kappa$ and $\kappa$ is uncountable, we have $|\theta|<\kappa$. Define a function \begin{align*} j:\theta &\to \mu \end{align*} as follows: for each $\xi<\theta$, write uniquely $\xi=\mu n+i$ with $n<\omega$ and $i<\mu$, and set $j(\xi):=i$. We recursively define a strictly increasing sequence $(\alpha_\xi)_{\xi<\theta}$ of ordinals below $\kappa$. Suppose $\xi<\theta$ and $\alpha_\eta<\kappa$ has been defined for all $\eta<\xi$. Define \begin{align*} \gamma_\xi := \sup\bigl(\{\alpha+1\}\cup\{\alpha_\eta+1:\eta<\xi\}\bigr). \end{align*} The set whose supremum defines $\gamma_\xi$ has cardinality less than $\kappa$, and all its elements are below $\kappa$; since $\kappa$ is regular, $\gamma_\xi<\kappa$. Because $C_{j(\xi)}$ is unbounded in $\kappa$, choose \begin{align*} \alpha_\xi \in C_{j(\xi)} \end{align*} with $\gamma_\xi\leq \alpha_\xi<\kappa$. Then $(\alpha_\xi)_{\xi<\theta}$ is strictly increasing, every term lies below $\kappa$, and $\alpha<\alpha_0$ because $\gamma_0=\alpha+1$. [guided] Fix $\alpha<\kappa$. To prove unboundedness, it is enough to produce some ordinal in the intersection above $\alpha$. The construction must do more than choose one point from each club. If we choose just one point from $C_i$, the final supremum need not be a limit point of $C_i$. Therefore we arrange that each $C_i$ is visited infinitely often and cofinally in the construction. Let $\omega$ denote the first infinite ordinal, and let \begin{align*} \theta := \mu \cdot \omega. \end{align*} Because $\mu<\kappa$ and $\kappa$ is uncountable, the ordinal $\theta$ has cardinality less than $\kappa$. For each $\xi<\theta$, write $\xi=\mu n+i$ with $n<\omega$ and $i<\mu$, and define \begin{align*} j:\theta &\to \mu,\\ \xi &\mapsto i. \end{align*} Thus $j$ cycles through all indices $i<\mu$ once in each block of length $\mu$. We now build an increasing sequence $(\alpha_\xi)_{\xi<\theta}$. At stage $\xi$, assume all earlier $\alpha_\eta$ with $\eta<\xi$ have been chosen below $\kappa$. Define \begin{align*} \gamma_\xi := \sup\bigl(\{\alpha+1\}\cup\{\alpha_\eta+1:\eta<\xi\}\bigr). \end{align*} The set inside this supremum has size less than $\kappa$, because $\xi<\theta$ and $|\theta|<\kappa$. Since $\kappa$ is regular, the supremum of fewer than $\kappa$ ordinals below $\kappa$ is still below $\kappa$. Hence $\gamma_\xi<\kappa$. Now use unboundedness of the club $C_{j(\xi)}$: there exists \begin{align*} \alpha_\xi \in C_{j(\xi)} \end{align*} such that $\gamma_\xi\leq \alpha_\xi<\kappa$. The definition of $\gamma_\xi$ ensures that $\alpha_\eta<\alpha_\xi$ whenever $\eta<\xi$, so the sequence is strictly increasing. It also ensures $\alpha<\alpha_0$, since at the first stage $\gamma_0=\alpha+1$. [/guided] [/step] [step:Take the final supremum and keep it below $\kappa$] Define \begin{align*} \lambda := \sup_{\xi<\theta} \alpha_\xi. \end{align*} Since $|\theta|<\kappa$, every $\alpha_\xi<\kappa$, and $\kappa$ is regular, we have $\lambda<\kappa$. Also $\alpha<\lambda$, because $\alpha_0>\alpha$ and $\alpha_0\leq \lambda$. [/step] [step:Use cofinal visits and closedness to place the supremum in every club] Fix $i<\mu$. For each $n<\omega$, define \begin{align*} \xi_n := \mu n+i. \end{align*} Then $\xi_n<\theta$, $j(\xi_n)=i$, and therefore \begin{align*} \alpha_{\xi_n}\in C_i. \end{align*} The sequence $(\alpha_{\xi_n})_{n<\omega}$ is increasing and cofinal in $\lambda$, because the stages $\xi_n$ are cofinal in $\theta$ and $(\alpha_\xi)_{\xi<\theta}$ is increasing. Hence \begin{align*} \sup_{n<\omega}\alpha_{\xi_n}=\lambda. \end{align*} Thus $\lambda$ is a limit point of $C_i$. Since $C_i$ is closed in $\kappa$, $\lambda\in C_i$. Because $i<\mu$ was arbitrary, $\lambda\in\bigcap_{i<\mu}C_i=C$. Since $\lambda>\alpha$, we have found an element of $C$ above the arbitrary ordinal $\alpha<\kappa$. Therefore $C$ is unbounded in $\kappa$. Together with closure from the first step, $C$ is club in $\kappa$. [/step]