[proofplan] We prove first that the intersection of fewer than $\kappa$ many club subsets of $\kappa$ is again club; this is the finite-support mechanism used at each stage of the diagonal construction. To show unboundedness of the diagonal intersection, we start above an arbitrary ordinal $\gamma < \kappa$ and build a countable increasing sequence whose next term lies in all clubs indexed below the previous term. The supremum of this sequence is still below $\kappa$ by regularity, and closure of the relevant clubs puts it into every $C_\alpha$ with $\alpha$ below that supremum. Finally, closure of the diagonal intersection follows by taking a limit point $\lambda$ and using points of the diagonal intersection cofinal below $\lambda$ to force $\lambda \in C_\alpha$ for each $\alpha < \lambda$. [/proofplan] [step:Show that fewer than $\kappa$ many clubs have club intersection] Let $I$ be a set of ordinals with $|I| < \kappa$, and suppose $(D_i)_{i \in I}$ is a family of closed unbounded subsets of $\kappa$. Define \begin{align*} D := \bigcap_{i \in I} D_i. \end{align*} We prove that $D$ is closed and unbounded in $\kappa$. Closure is immediate from the definition of closed subset of $\kappa$: if $\lambda < \kappa$ is a limit ordinal and $D \cap \lambda$ is unbounded in $\lambda$, then for every $i \in I$, the set $D_i \cap \lambda$ is unbounded in $\lambda$, hence $\lambda \in D_i$ by closure of $D_i$. Therefore $\lambda \in D$. For unboundedness, fix $\gamma < \kappa$. Let $\theta$ be an ordinal with $\theta < \kappa$ and let \begin{align*} e:\theta \to I \end{align*} be a map such that for every $i \in I$, the set $e^{-1}(\{i\})$ is cofinal in $\theta$; if $I$ is finite, take $\theta = \omega$, and if $I$ is infinite, take $\theta = |I|$ and use $|I| \cdot |I| = |I|$ to arrange cofinal repetition of each index. Define an increasing sequence $(\xi_\eta)_{\eta < \theta}$ of ordinals below $\kappa$ by transfinite recursion. Put $\xi_0 > \gamma$ with $\xi_0 \in D_{e(0)}$, using unboundedness of $D_{e(0)}$. At a successor stage $\eta + 1 < \theta$, choose \begin{align*} \xi_{\eta+1} \in D_{e(\eta+1)} \end{align*} with $\xi_{\eta+1} > \xi_\eta$, using unboundedness of $D_{e(\eta+1)}$. At a limit stage $\eta < \theta$, define \begin{align*} s_\eta := \sup_{\rho < \eta} \xi_\rho. \end{align*} Since $\eta < \theta < \kappa$ and $\kappa$ is regular, $s_\eta < \kappa$; choose $\xi_\eta \in D_{e(\eta)}$ with $\xi_\eta > s_\eta$, again using unboundedness. Now define \begin{align*} \xi := \sup_{\eta < \theta} \xi_\eta. \end{align*} Regularity of $\kappa$ gives $\xi < \kappa$, since $\theta < \kappa$ and every $\xi_\eta < \kappa$. Also $\xi > \gamma$. Fix $i \in I$. Because $e^{-1}(\{i\})$ is cofinal in $\theta$, the subsequence $(\xi_\eta)_{\eta \in e^{-1}(\{i\})}$ is cofinal in $\xi$ and lies in $D_i$. Since $D_i$ is closed, $\xi \in D_i$. As $i \in I$ was arbitrary, $\xi \in D$. Thus $D$ is unbounded in $\kappa$. [guided] We need one general fact: intersecting fewer than $\kappa$ many clubs still gives a club. Let $I$ be a set of indices with $|I| < \kappa$, and let $(D_i)_{i \in I}$ be closed unbounded subsets of $\kappa$. Define \begin{align*} D := \bigcap_{i \in I} D_i. \end{align*} First consider closure. Suppose $\lambda < \kappa$ is a limit ordinal and $D \cap \lambda$ is unbounded in $\lambda$. Then for every $i \in I$, because $D \subset D_i$, the set $D_i \cap \lambda$ is also unbounded in $\lambda$. Since $D_i$ is closed in $\kappa$, this implies $\lambda \in D_i$. This holds for every $i \in I$, so $\lambda \in D$. The main point is unboundedness. Fix a starting ordinal $\gamma < \kappa$. We want to find some point of $D$ above $\gamma$, meaning a single ordinal that lies in every $D_i$. To do this, we build an increasing sequence that visits each club cofinally often. Choose an ordinal $\theta < \kappa$ and a map \begin{align*} e:\theta \to I \end{align*} such that each index $i \in I$ appears cofinally often: $e^{-1}(\{i\})$ is cofinal in $\theta$. If $I$ is finite, we may take $\theta = \omega$, since $\kappa$ is uncountable and hence $\omega < \kappa$. If $I$ is infinite, take $\theta = |I|$ and use the cardinal identity $|I| \cdot |I| = |I|$ to list all indices cofinally often. We now recursively construct an increasing sequence $(\xi_\eta)_{\eta < \theta}$ below $\kappa$. At the initial stage, choose $\xi_0 > \gamma$ with $\xi_0 \in D_{e(0)}$, possible because $D_{e(0)}$ is unbounded in $\kappa$. At a successor stage $\eta+1$, choose \begin{align*} \xi_{\eta+1} \in D_{e(\eta+1)} \end{align*} with $\xi_{\eta+1} > \xi_\eta$. At a limit stage $\eta < \theta$, define \begin{align*} s_\eta := \sup_{\rho < \eta} \xi_\rho. \end{align*} Since $\eta < \theta < \kappa$ and $\kappa$ is regular, the supremum of this $<\kappa$-length sequence of ordinals below $\kappa$ is still below $\kappa$, so $s_\eta < \kappa$. Then choose $\xi_\eta \in D_{e(\eta)}$ above $s_\eta$. Finally set \begin{align*} \xi := \sup_{\eta < \theta} \xi_\eta. \end{align*} Again regularity gives $\xi < \kappa$, and the construction gives $\xi > \gamma$. Fix any $i \in I$. Since $i$ occurs cofinally often in the indexing map $e$, the ordinals $\xi_\eta$ with $e(\eta)=i$ form a cofinal subset of $\xi$, and each of those ordinals lies in $D_i$. Because $D_i$ is closed, $\xi \in D_i$. This is true for every $i \in I$, hence $\xi \in D$. Therefore $D$ is unbounded in $\kappa$. [/guided] [/step] [step:Build a point of the diagonal intersection above an arbitrary ordinal] Fix $\gamma < \kappa$. We recursively define an increasing sequence $(\beta_n)_{n \in \omega}$ of ordinals below $\kappa$. Choose $\beta_0 < \kappa$ with $\beta_0 > \gamma$. Suppose $\beta_n$ has been defined. Define \begin{align*} I_n := \{\alpha < \kappa : \alpha \leq \beta_n\}. \end{align*} Since $\beta_n < \kappa$, we have $|I_n| < \kappa$. By the previous step, \begin{align*} E_n := \bigcap_{\alpha \in I_n} C_\alpha \end{align*} is club in $\kappa$. Hence $E_n$ is unbounded in $\kappa$, so choose $\beta_{n+1} \in E_n$ with $\beta_{n+1} > \beta_n$. Define \begin{align*} \beta := \sup_{n \in \omega} \beta_n. \end{align*} Because $\kappa$ is regular and uncountable, $\omega < \kappa$, so $\beta < \kappa$. Also $\beta > \gamma$. We claim that $\beta \in \Delta_{\alpha < \kappa} C_\alpha$. Let $\alpha < \beta$. Since $\beta = \sup_{n \in \omega} \beta_n$, there exists $n \in \omega$ such that $\alpha \leq \beta_n$. For every $m \geq n+1$, the construction gives $\beta_m \in C_\alpha$, because $\alpha \in I_{m-1}$. Thus the tail $(\beta_m)_{m \geq n+1}$ is cofinal in $\beta$ and lies in $C_\alpha$. Since $C_\alpha$ is closed, $\beta \in C_\alpha$. As $\alpha < \beta$ was arbitrary, $\beta \in \Delta_{\alpha < \kappa} C_\alpha$. Therefore for every $\gamma < \kappa$ there exists $\beta \in \Delta_{\alpha < \kappa} C_\alpha$ with $\beta > \gamma$, so the diagonal intersection is unbounded in $\kappa$. [guided] Fix an arbitrary ordinal $\gamma < \kappa$. To prove unboundedness, we must produce an element of the diagonal intersection above $\gamma$. The diagonal condition at a point $\beta$ asks for membership in every $C_\alpha$ with $\alpha < \beta$, so we build $\beta$ as the limit of a sequence that gradually meets all clubs whose indices have appeared below earlier stages. Choose $\beta_0 < \kappa$ with $\beta_0 > \gamma$. Suppose $\beta_n$ is already defined. Define the set of indices currently visible below $\beta_n$ by \begin{align*} I_n := \{\alpha < \kappa : \alpha \leq \beta_n\}. \end{align*} Because $\beta_n < \kappa$, this set has cardinality $|I_n| < \kappa$. The previous step applies to the family $(C_\alpha)_{\alpha \in I_n}$, so \begin{align*} E_n := \bigcap_{\alpha \in I_n} C_\alpha \end{align*} is closed and unbounded in $\kappa$. In particular, $E_n$ is unbounded, and we may choose $\beta_{n+1} \in E_n$ with $\beta_{n+1} > \beta_n$. Now define the candidate diagonal point by \begin{align*} \beta := \sup_{n \in \omega} \beta_n. \end{align*} The sequence has length $\omega$, and $\omega < \kappa$ because $\kappa$ is uncountable. Since $\kappa$ is regular, the supremum of countably many ordinals below $\kappa$ is still below $\kappa$, so $\beta < \kappa$. Also $\beta > \gamma$. We verify the diagonal condition. Let $\alpha < \beta$. Since $\beta$ is the supremum of the increasing sequence $(\beta_n)_{n \in \omega}$, there is some $n \in \omega$ with $\alpha \leq \beta_n$. For every $m \geq n+1$, we have $\alpha \leq \beta_n < \beta_{m-1}$ or at least $\alpha \leq \beta_{m-1}$, so $\alpha \in I_{m-1}$. By construction, \begin{align*} \beta_m \in E_{m-1} = \bigcap_{\xi \in I_{m-1}} C_\xi, \end{align*} and hence $\beta_m \in C_\alpha$. Thus all sufficiently late terms of the sequence lie in $C_\alpha$, and those late terms are cofinal in $\beta$. Since $C_\alpha$ is closed in $\kappa$, it contains the supremum of this cofinal sequence, namely $\beta$. Therefore $\beta \in C_\alpha$. Because the argument works for every $\alpha < \beta$, we have \begin{align*} \beta \in \Delta_{\alpha < \kappa} C_\alpha. \end{align*} Since $\beta > \gamma$, and $\gamma < \kappa$ was arbitrary, the diagonal intersection is unbounded in $\kappa$. [/guided] [/step] [step:Prove that the diagonal intersection is closed] Let \begin{align*} D := \Delta_{\alpha < \kappa} C_\alpha. \end{align*} Let $\lambda < \kappa$ be a limit ordinal such that $D \cap \lambda$ is unbounded in $\lambda$. We prove $\lambda \in D$. Fix $\alpha < \lambda$. Since $D \cap \lambda$ is unbounded in $\lambda$, the set \begin{align*} D \cap \lambda \cap (\alpha,\lambda) \end{align*} is unbounded in $\lambda$. If $\delta \in D \cap \lambda$ and $\delta > \alpha$, then the definition of $D$ gives $\delta \in C_\alpha$. Hence $C_\alpha \cap \lambda$ is unbounded in $\lambda$. Since $C_\alpha$ is closed, $\lambda \in C_\alpha$. As this holds for every $\alpha < \lambda$, we have \begin{align*} \lambda \in \Delta_{\alpha < \kappa} C_\alpha = D. \end{align*} Thus $D$ is closed in $\kappa$. Together with the unboundedness proved above, $D$ is club in $\kappa$. [guided] Let \begin{align*} D := \Delta_{\alpha < \kappa} C_\alpha. \end{align*} To prove that $D$ is closed, take a limit ordinal $\lambda < \kappa$ such that $D \cap \lambda$ is unbounded in $\lambda$. We must show that $\lambda \in D$, which means proving $\lambda \in C_\alpha$ for every $\alpha < \lambda$. Fix one such $\alpha < \lambda$. Because $D \cap \lambda$ is unbounded in $\lambda$, there are points of $D$ arbitrarily close to $\lambda$ from below, and in particular there are unboundedly many $\delta \in D \cap \lambda$ with $\delta > \alpha$. For each such $\delta$, the diagonal definition of $D$ applies: since $\alpha < \delta$ and $\delta \in D$, we get $\delta \in C_\alpha$. Therefore $C_\alpha \cap \lambda$ is unbounded in $\lambda$. Since $C_\alpha$ is closed in $\kappa$, it contains the limit of any unbounded subset of itself below $\lambda$, so $\lambda \in C_\alpha$. This holds for every $\alpha < \lambda$, hence \begin{align*} \lambda \in \Delta_{\alpha < \kappa} C_\alpha = D. \end{align*} Thus $D$ is closed. The previous step showed that $D$ is unbounded, so $D$ is closed and unbounded in $\kappa$. [/guided] [/step]