[proofplan] Represent cardinals by initial ordinals and use [Hartogs' theorem](/theorems/4822) to produce an ordinal too large to inject into $\kappa$. Choose the least such ordinal $\theta$; minimality implies that every smaller ordinal injects into $\kappa$. We then prove that $\theta$ is itself an initial ordinal, that it lies strictly above $\kappa$, and that every cardinal strictly above $\kappa$ must be at least $\theta$. [/proofplan] [step:Choose the least ordinal that does not inject into $\kappa$] Let $\kappa$ be a cardinal, viewed as its initial ordinal representative. By Hartogs' theorem (citing a result not yet in the wiki: Hartogs Theorem) applied to the set $\kappa$, there exists an ordinal which does not inject into $\kappa$. Let $\theta$ denote the least ordinal with the property that there is no injective map $\theta \to \kappa$. Thus, by the defining minimality of $\theta$, for every ordinal $\beta < \theta$ there exists an injective map $\beta \to \kappa$. [guided] We work with the standard convention that cardinals are represented by initial ordinals. The input cardinal $\kappa$ is therefore also a set, namely the set of all ordinals smaller than $\kappa$. Hartogs' theorem says that for every set $A$, there is an ordinal which does not inject into $A$. Applying this to the set $A := \kappa$, we obtain at least one ordinal that does not inject into $\kappa$. Since the ordinals are well-ordered by $\in$, there is a least such ordinal. Define $\theta$ to be that least ordinal. The choice of $\theta$ has two parts. First, no injective map $\theta \to \kappa$ exists. Second, every smaller ordinal does inject into $\kappa$: if $\beta < \theta$ and there were no injection $\beta \to \kappa$, then $\beta$ would be a smaller counterexample, contradicting the minimality of $\theta$. [/guided] [/step] [step:Show that $\theta$ is a cardinal] We prove that $\theta$ is an initial ordinal. Suppose, toward a contradiction, that $\theta$ is not initial. Then there exists an ordinal $\beta < \theta$ and a bijection \begin{align*} b: \theta &\to \beta. \end{align*} By minimality of $\theta$, there exists an injective map \begin{align*} j: \beta &\to \kappa. \end{align*} The composite map \begin{align*} j \circ b: \theta &\to \kappa \end{align*} is injective, since both $b$ and $j$ are injective. This contradicts the defining property of $\theta$. Hence no smaller ordinal is equipotent with $\theta$, so $\theta$ is an initial ordinal, that is, a cardinal. [guided] To show that $\theta$ is a cardinal, we must show that it is an initial ordinal: no smaller ordinal has the same cardinality as $\theta$. Assume the opposite. Then there is some ordinal $\beta < \theta$ and a bijection $b: \theta \to \beta$. Since $\beta$ is smaller than $\theta$, the minimality of $\theta$ gives an injective map $j: \beta \to \kappa$. Now compose: \begin{align*} j \circ b: \theta &\to \kappa. \end{align*} Because $b$ is bijective and $j$ is injective, the composite $j \circ b$ is injective. This produces an injection from $\theta$ into $\kappa$, contradicting the way $\theta$ was chosen. Therefore $\theta$ must be an initial ordinal, and hence a cardinal. [/guided] [/step] [step:Verify that $\theta$ is strictly larger than $\kappa$] The identity map \begin{align*} \operatorname{id}_{\kappa}: \kappa &\to \kappa \end{align*} is injective, so $\kappa$ is not an ordinal with no injection into $\kappa$. Since $\theta$ is the least ordinal with no injection into $\kappa$, we cannot have $\theta \leq \kappa$; otherwise the inclusion map $\theta \to \kappa$ would be injective. Hence $\kappa < \theta$ as ordinals. Since both $\kappa$ and $\theta$ are cardinals, this says $\kappa < \theta$ in the cardinal order. [/step] [step:Prove that no smaller cardinal lies strictly above $\kappa$] Let $\lambda$ be a cardinal such that $\kappa < \lambda$. By the meaning of strict cardinal inequality, there is an injection $\kappa \to \lambda$ and $\kappa$ is not equipotent with $\lambda$. If there were an injection $\lambda \to \kappa$, then the Schröder-Bernstein theorem (citing a result not yet in the wiki: Schröder-Bernstein Theorem) applied to the injections $\kappa \to \lambda$ and $\lambda \to \kappa$ would give a bijection between $\kappa$ and $\lambda$, contradicting $\kappa < \lambda$. Therefore no injection $\lambda \to \kappa$ exists. If $\lambda < \theta$ as ordinals, then the minimality of $\theta$ would give an injective map $\lambda \to \kappa$, contradicting the previous paragraph. Hence $\theta \leq \lambda$. Since $\lambda$ was an arbitrary cardinal strictly above $\kappa$, $\theta$ is the least cardinal strictly larger than $\kappa$. [guided] Let $\lambda$ be any cardinal with $\kappa < \lambda$. We must prove that $\theta \leq \lambda$, because this is exactly the assertion that $\theta$ is the least cardinal above $\kappa$. First we show that there cannot be an injection $\lambda \to \kappa$. Since $\kappa < \lambda$, there is an injection from $\kappa$ into $\lambda$, but $\kappa$ and $\lambda$ are not equipotent. If an injection $\lambda \to \kappa$ also existed, then the Schröder-Bernstein theorem would apply to the two injections \begin{align*} \kappa &\to \lambda, \\ \lambda &\to \kappa, \end{align*} and would produce a bijection between $\kappa$ and $\lambda$. That contradicts the strict inequality $\kappa < \lambda$. Therefore no injection $\lambda \to \kappa$ exists. Now suppose, toward a contradiction, that $\lambda < \theta$ as ordinals. Since $\lambda$ is smaller than the least ordinal that fails to inject into $\kappa$, the minimality of $\theta$ gives an injective map $\lambda \to \kappa$. This contradicts the conclusion just proved. Hence $\lambda$ cannot be below $\theta$, so $\theta \leq \lambda$. Thus every cardinal strictly larger than $\kappa$ is at least $\theta$. [/guided] [/step] [step:Define the successor cardinal] Define \begin{align*} \kappa^+ := \theta. \end{align*} The previous steps show that $\kappa^+$ is a cardinal, that $\kappa < \kappa^+$, and that for every cardinal $\lambda$, the implication $\kappa < \lambda \implies \kappa^+ \leq \lambda$ holds. Therefore $\kappa^+$ is the successor cardinal of $\kappa$. [/step]