[proofplan] We argue by contradiction. If $\kappa^+$ had cofinality $\mu < \kappa^+$, then, because $\kappa^+$ is the least cardinal strictly larger than $\kappa$, one would have $\mu \leq \kappa$. A cofinal family of length $\mu$ in the initial ordinal $\kappa^+$ writes $\kappa^+$ as the union of $\mu$ many ordinals, each of cardinality at most $\kappa$. The [Axiom of Choice](/page/Axiom%20of%20Choice) gives the cardinal arithmetic bound $\mu \cdot \kappa = \kappa$, contradicting that $\kappa^+$ has cardinality strictly larger than $\kappa$. [/proofplan] [step:Assume a smaller cofinality and choose a cofinal sequence] Let $\kappa$ be an infinite cardinal, and regard $\kappa^+$ as the initial ordinal of cardinality $\kappa^+$. Suppose, toward a contradiction, that $\kappa^+$ is singular. Define \begin{align*} \mu := \operatorname{cf}(\kappa^+). \end{align*} Then $\mu < \kappa^+$. Since $\mu$ is a cardinal and $\kappa^+$ is the least cardinal strictly larger than $\kappa$, it follows that $\mu \leq \kappa$. By the definition of cofinality, there exists a strictly increasing cofinal map \begin{align*} \lambda: \mu &\to \kappa^+ \\ i &\mapsto \lambda_i \end{align*} such that \begin{align*} \sup_{i < \mu} \lambda_i = \kappa^+. \end{align*} For each $i < \mu$, the ordinal $\lambda_i$ satisfies $\lambda_i < \kappa^+$. Since $\kappa^+$ is the initial ordinal of the successor cardinal of $\kappa$, every ordinal below $\kappa^+$ has cardinality at most $\kappa$. Hence \begin{align*} |\lambda_i| \leq \kappa \end{align*} for every $i < \mu$. [/step] [step:Convert cofinality into a union representation of $\kappa^+$] Define the set \begin{align*} U := \bigcup_{i < \mu} \lambda_i. \end{align*} Because each $\lambda_i$ is an ordinal and the sequence $(\lambda_i)_{i<\mu}$ is cofinal in $\kappa^+$, we have \begin{align*} U = \kappa^+. \end{align*} Indeed, the inclusion $U \subseteq \kappa^+$ follows from $\lambda_i < \kappa^+$ for all $i < \mu$. Conversely, let $\alpha \in \kappa^+$. Since $\sup_{i<\mu}\lambda_i = \kappa^+$, there exists $i < \mu$ with $\alpha < \lambda_i$. Because ordinals are transitive, $\alpha \in \lambda_i \subseteq U$. Thus $\kappa^+ \subseteq U$, proving equality. [/step] [step:Bound the cardinality of the union by $\kappa$] For each $i < \mu$, choose an injection \begin{align*} f_i: \lambda_i &\to \kappa. \end{align*} Such injections exist because $|\lambda_i| \leq \kappa$. Define \begin{align*} F: U &\to \mu \times \kappa \\ \alpha &\mapsto (i_\alpha, f_{i_\alpha}(\alpha)), \end{align*} where $i_\alpha$ denotes the least $i < \mu$ such that $\alpha \in \lambda_i$. The index $i_\alpha$ exists because $U = \bigcup_{i<\mu}\lambda_i$, and it is unique by leastness. The map $F$ is injective: if $F(\alpha) = F(\beta)$, then $i_\alpha = i_\beta = i$ and $f_i(\alpha) = f_i(\beta)$; since $f_i$ is injective, $\alpha = \beta$. Therefore \begin{align*} |U| \leq |\mu \times \kappa|. \end{align*} Since $\mu \leq \kappa$ and $\kappa$ is infinite, cardinal arithmetic under the Axiom of Choice gives \begin{align*} |\mu \times \kappa| \leq |\kappa \times \kappa| = \kappa. \end{align*} Thus \begin{align*} |U| \leq \kappa. \end{align*} [/step] [step:Derive the contradiction and conclude regularity] From the union representation proved above, $U = \kappa^+$. Hence \begin{align*} \kappa^+ = |U| \leq \kappa. \end{align*} This contradicts the defining property of the successor cardinal $\kappa^+$, namely $\kappa < \kappa^+$. Therefore the assumption $\operatorname{cf}(\kappa^+) < \kappa^+$ is false. Hence \begin{align*} \operatorname{cf}(\kappa^+) = \kappa^+, \end{align*} so $\kappa^+$ is regular. [/step]