Let $\varepsilon_0$ denote the least ordinal satisfying $\omega^{\varepsilon_0}=\varepsilon_0$. For every nonzero ordinal $\alpha<\varepsilon_0$, there exist a unique integer $k\in\mathbb N$, unique positive integers $n_1,\dots,n_k\in\mathbb N$, and unique ordinals $\alpha_1,\dots,\alpha_k<\varepsilon_0$ such that \begin{align*} \alpha=\omega^{\alpha_1}n_1+\omega^{\alpha_2}n_2+\cdots+\omega^{\alpha_k}n_k \end{align*} and \begin{align*} \alpha_1>\alpha_2>\cdots>\alpha_k\geq 0. \end{align*} Here ordinal addition, multiplication, and exponentiation are used.

Assume the Axiom of Foundation. For every set $x$, one has $x \notin x$.

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*}

For all ordinals $\alpha$, $\beta$, and $\gamma$ with $\beta \leq \gamma$, the following inequalities hold: \begin{align*} \alpha + \beta &\leq \alpha + \gamma,\\ \alpha \cdot \beta &\leq \alpha \cdot \gamma. \end{align*} Moreover, if $\alpha > 0$, then \begin{align*} \alpha^\beta \leq \alpha^\gamma. \end{align*}

For all ordinals $\alpha$, $\beta$, and $\gamma$, \begin{align*} (\alpha \cdot \beta) \cdot \gamma = \alpha \cdot (\beta \cdot \gamma). \end{align*}

Let $A$ be a set such that every element of $A$ is an ordinal. Then the membership relation $\in$ well-orders $A$: for all $\alpha,\beta \in A$, exactly one of $\alpha = \beta$, $\alpha \in \beta$, or $\beta \in \alpha$ holds, and every nonempty subset $B \subset A$ has an $\in$-least element.

Let $A$ be a set such that every element of $A$ is an ordinal. Then $\bigcup A$ is an ordinal. Moreover, for every $\alpha \in A$ one has $\alpha \subseteq \bigcup A$, and if $\gamma$ is an ordinal such that $\alpha \subseteq \gamma$ for every $\alpha \in A$, then $\bigcup A \subseteq \gamma$.

Let $\approx$ be the relation on sets defined by $A \approx B$ if and only if there exists a bijection $f: A \to B$. Then $\approx$ is reflexive, symmetric, and transitive: for every set $A$ one has $A \approx A$; for all sets $A$ and $B$, if $A \approx B$, then $B \approx A$; and for all sets $A$, $B$, and $C$, if $A \approx B$ and $B \approx C$, then $A \approx C$.

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