Let $A,A',B,B'$ be sets. If there exist bijections $f:A\to A'$ and $g:B\to B'$, then there exist bijections \begin{align*} A\sqcup B &\to A'\sqcup B',\\ A\times B &\to A'\times B',\\ A^B &\to (A')^{B'}. \end{align*} Equivalently, if $|A|=|A'|$ and $|B|=|B'|$, then \begin{align*} |A\sqcup B|&=|A'\sqcup B'|,\\ |A\times B|&=|A'\times B'|,\\ |A^B|&=|(A')^{B'}|. \end{align*}

Let $\kappa$ be a regular uncountable cardinal. If $C \subset \kappa$ and $D \subset \kappa$ are closed unbounded subsets of $\kappa$, then $C \cap D$ is a closed unbounded subset of $\kappa$.

Let $\theta$ be a cardinal such that $\omega_1 \in H_\theta$. Let $M \preccurlyeq (H_\theta,\in)$ be a countable elementary submodel with $\omega_1 \in M$. Then $M$ is countable as a set in the ambient universe, but \begin{align*} (M,\in) \models \text{``there is no surjection from } \omega \text{ onto } \omega_1\text{''}. \end{align*} Equivalently, there is no $f \in M$ such that $M$ regards $f$ as a surjective map $f:\omega \to \omega_1$.

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*}

Let $A$ and $B$ be sets equipped with well-orders $<_A$ and $<_B$, respectively. If \begin{align*} f: A &\to B \end{align*} and \begin{align*} g: A &\to B \end{align*} are order isomorphisms from $(A, <_A)$ onto $(B, <_B)$, then $f = g$. Equivalently, between two well-ordered sets there is at most one order isomorphism.

Let $\kappa$ be a regular uncountable cardinal, and let $(C_\alpha)_{\alpha < \kappa}$ be a sequence such that each $C_\alpha \subset \kappa$ is closed and unbounded in $\kappa$. Define the diagonal intersection \begin{align*} \Delta_{\alpha < \kappa} C_\alpha := \{\beta < \kappa : \beta \in C_\alpha \text{ for every } \alpha < \beta\}. \end{align*} Then $\Delta_{\alpha < \kappa} C_\alpha$ is closed and unbounded in $\kappa$.

Assume the [Axiom of Choice](/page/Axiom%20of%20Choice). For all cardinals $\kappa$ and $\lambda$, if at least one of $\kappa$ and $\lambda$ is infinite, then \begin{align*} \kappa + \lambda = \max\{\kappa,\lambda\}. \end{align*} Moreover, if $\kappa \neq 0$ and $\lambda \neq 0$, then \begin{align*} \kappa \cdot \lambda = \max\{\kappa,\lambda\}. \end{align*}

In ZF set theory, the following implications hold. First, the [Axiom of Choice](/page/Axiom%20of%20Choice) implies the Axiom of Dependent Choice: if $X$ is a nonempty set and $R \subset X \times X$ is a serial binary relation, meaning that for every $x \in X$ there exists $y \in X$ such that $(x,y) \in R$, then there exists a sequence $f: \mathbb{N} \to X$ such that $(f(n), f(n+1)) \in R$ for every $n \in \mathbb{N}$. Second, the Axiom of Dependent Choice implies the Axiom of Countable Choice: for every sequence $(X_n)_{n \in \mathbb{N}}$ of nonempty sets, there exists a function $c: \mathbb{N} \to \bigcup_{n \in \mathbb{N}} X_n$ such that $c(n) \in X_n$ for every $n \in \mathbb{N}$.

In ZF, assume the following surjection splitting principle: for every pair of sets $X$ and $Y$ and every surjective map $f: X \to Y$, there exists a map $r: Y \to X$ such that $f \circ r = \operatorname{id}_Y$. Then the [Axiom of Choice](/page/Axiom%20of%20Choice) holds; that is, for every set $A$ whose elements are nonempty sets, there exists a map $c: A \to \bigcup A$ such that $c(x) \in x$ for every $x \in A$.

For every $n\in\omega$ and every indexed family of sets $(a_i)_{i\in n}$, the set $\{a_i : i\in n\}$ exists. In particular, for all sets $a_0,\dots,a_{n-1}$, the set $\{a_0,\dots,a_{n-1}\}$ exists.