Let $I$ be a set, and let $(\kappa_i)_{i \in I}$ and $(\lambda_i)_{i \in I}$ be indexed families of cardinals. If $\kappa_i < \lambda_i$ for every $i \in I$, then \begin{align*} \sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i. \end{align*}

For every cardinal $\kappa$, there exists a cardinal $\kappa^+$ such that $\kappa < \kappa^+$ and, for every cardinal $\lambda$, if $\kappa < \lambda$, then $\kappa^+ \leq \lambda$. Equivalently, every cardinal has a least strictly larger cardinal, called its successor cardinal.

Let $\mathrm{Card}$ denote the class of cardinals, where each cardinal is represented by its initial ordinal. Define a relation $\le$ on $\mathrm{Card}$ by declaring, for $\kappa, \lambda \in \mathrm{Card}$, $\kappa \le \lambda$ iff there exists an injective map $f: \kappa \to \lambda$. Then $\le$ is a partial order on $\mathrm{Card}$: for all cardinals $\kappa, \lambda, \mu$, $\kappa \le \kappa$; if $\kappa \le \lambda$ and $\lambda \le \mu$, then $\kappa \le \mu$; and if $\kappa \le \lambda$ and $\lambda \le \kappa$, then $\kappa = \lambda$.

For every ordinal $\alpha$ and every nonzero limit ordinal $\lambda$, \begin{align*} \alpha+\lambda&=\sup\{\alpha+\beta:\beta<\lambda\},\\ \alpha\cdot\lambda&=\sup\{\alpha\cdot\beta:\beta<\lambda\}. \end{align*} Moreover, for every nonzero ordinal $\gamma$ and every nonzero limit ordinal $\lambda$, \begin{align*} \gamma^\lambda&=\sup\{\gamma^\beta:\beta<\lambda\}. \end{align*}

Let $\theta$ be an uncountable cardinal such that $H_\theta$ is infinite. Let $\mathcal{L}$ be a countable first-order language extending the language $\{\in\}$, and let $\mathcal{H}$ be an $\mathcal{L}$-structure with universe $H_\theta$ whose interpretation of $\in$ is the membership relation restricted to $H_\theta$. If $A \subset H_\theta$ is countable, then there exists a [countable set](/page/Countable%20Set) $M \subset H_\theta$ such that $A \subset M$ and the induced $\mathcal{L}$-substructure $\mathcal{M}$ on $M$ satisfies \begin{align*} \mathcal{M} \preccurlyeq \mathcal{H}. \end{align*} No transitivity of $M$ is asserted.

Let $(A,<_{A})$ and $(B,<_{B})$ be well-ordered sets. Call a subset $I \subset X$ an initial segment of a linearly ordered set $(X,<_{X})$ if, whenever $x \in I$ and $y <_{X} x$, one has $y \in I$. Exactly one of the following three alternatives holds: 1. $(A,<_{A})$ and $(B,<_{B})$ are order-isomorphic. 2. There exists a proper initial segment $J \subsetneq B$ such that $(A,<_{A})$ is order-isomorphic to $(J,<_{B}|_{J})$. 3. There exists a proper initial segment $I \subsetneq A$ such that $(B,<_{B})$ is order-isomorphic to $(I,<_{A}|_{I})$.

Let $(V_\alpha)_{\alpha \in \operatorname{Ord}}$ be the cumulative hierarchy defined by \begin{align*} V_0 &= \varnothing, \\ V_{\alpha+1} &= \mathcal{P}(V_\alpha), \\ V_\lambda &= \bigcup_{\gamma \in \lambda} V_\gamma \end{align*} for every ordinal $\alpha$ and every limit ordinal $\lambda$. Then: 1. If $\alpha$ and $\beta$ are ordinals with $\alpha \leq \beta$, then $V_\alpha \subset V_\beta$. 2. For every ordinal $\alpha$, the set $V_\alpha$ is transitive: if $x \in y$ and $y \in V_\alpha$, then $x \in V_\alpha$. 3. For every ordinal $\alpha$ and every $x \in V_\alpha$, there exists an ordinal $\gamma \in \alpha$ such that $x \subset V_\gamma$.

For every ordinal $\alpha$, $\alpha \in L$. Equivalently, $\operatorname{Ord} \subset L$, where $L = \bigcup_{\beta \in \operatorname{Ord}} L_\beta$ is the constructible universe.

Let $\alpha$ be an ordinal, and define its successor by \begin{align*} \alpha + 1 := \alpha \cup \{\alpha\}. \end{align*} Then $\alpha + 1$ is an ordinal, and $\alpha \in \alpha + 1$.

Let $A$ be a set and let $R \subset A \times A$ be a binary relation. Assume that $R$ is well-founded on $A$, meaning that every nonempty subset $B \subset A$ has an element $b_0 \in B$ such that there is no $b \in B$ with $bRb_0$. Assume also that $R$ is extensional on $A$, meaning that for all $a,a' \in A$, \begin{align*} \bigl(\forall b \in A,\ bRa \iff bRa'\bigr) \implies a=a'. \end{align*} Then there exist a transitive set $M$ and a bijection $\pi:A \to M$ such that for all $a,b \in A$, \begin{align*} aRb \iff \pi(a) \in \pi(b). \end{align*} Moreover, for every $a \in A$, \begin{align*} \pi(a)=\{\pi(b): bRa\}. \end{align*} The set $M$ and the map $\pi$ are uniquely determined by these properties.