For $I = \{\varnothing,\{\varnothing\}\}$ with $A_{\varnothing} = A$ and $A_{\{\varnothing\}} = B$, the map $f \mapsto (f(\varnothing), f(\{\varnothing\}))$ is a bijection $\prod_{i \in I} A_i \to A \times B$

Given: sets $A$ and $B$, the index set $I := \{\varnothing,\{\varnothing\}\}$, the family $(A_i)_{i \in I}$ above, and $P := \prod_{i \in I} A_i$.

an element of $\prod_{i \in I} A_i$ is a function with domain $I$ that takes its value at each index inside the member carried by that index (The product $\prod_{i \in I} A_i := \{\, f : I \to \bigcup_{i \in I} A_i \ \mid\ f(i) \in A_i \text{ for every } i \in I \,\}$).

$z \in A \times B$ holds if and only if $z = (a,b)$ for some $a \in A$ and some $b \in B$ (The Cartesian product $A \times B := \{\, z \in \mathcal{P}(\mathcal{P}(A \cup B)) : \exists a \in A\ \exists b \in B\ z = (a,b) \,\}$).

$(a,b) = (c,d)$ if and only if $a = c$ and $b = d$ ($(a,b) = (c,d)$ if and only if $a = c$ and $b = d$).

$f$ is injective (one-to-one) if $f(x) = f(y)$ implies $x = y$, for all $x, y \in A$ (Injection, surjection, bijection).

$f$ is surjective (onto) if for every $b \in B$ there is some $x \in A$ with $f(x) = b$ (Injection, surjection, bijection).

$f = g$ if and only if $\operatorname{dom} f = \operatorname{dom} g$ and $f(x) = g(x)$ for every $x \in \operatorname{dom} f$ (Functions $f$ and $g$ are equal if and only if $\operatorname{dom} f = \operatorname{dom} g$ and $f(x) = g(x)$ for every $x$ in that common domain).

$\{x,y\}$ is the set whose elements are exactly $x$ and $y$, and $\{x\} := \{x,x\}$ (The unordered pair $\{x,y\}$ and the singleton $\{x\} = \{x,x\}$).

There is exactly one set with no elements, written $\varnothing$ (There is exactly one set with no elements, written $\varnothing$).

A function is a relation $f$ such that $(a,b) \in f$ and $(a,c) \in f$ imply $b = c$ (A function is a relation $f$ with $(a,b) \in f$ and $(a,c) \in f$ implying $b = c$; $f : A \to B$, the value $f(a)$, domain and codomain).

An indexed family with index set $I$ is a function $A$ with $\operatorname{dom} A = I$ (An indexed family $(A_i)_{i \in I}$ is a function with domain $I$; $\{A_i : i \in I\}$ is its range).

$\operatorname{dom} R := \{\, a : \exists b\ (a,b) \in R \,\}, \qquad \operatorname{ran} R := \{\, b : \exists a\ (a,b) \in R \,\}$ (Relation, $\operatorname{dom} R$, $\operatorname{ran} R$, $\operatorname{fld} R$, and the specialisations "relation from $A$ to $B$" and "relation on $A$").

For any parameters $\bar p$ and any set $x$, there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $\varphi(z,\bar p)$ holds (The Axiom Schema of Separation: for each formula $\varphi$, $\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))$).

$z \in \bigcup_{i \in I} A_i$ holds if and only if $z \in A_i$ for some $i \in I$ ($\bigcup_{i \in I} A_i := \bigcup \{A_i : i \in I\}$, and $\bigcap_{i \in I} A_i := \bigcap \{A_i : i \in I\}$ for $I \neq \varnothing$).

$z \in a \cup b$ holds if and only if $z \in a$ or $z \in b$ ($\bigcup \varnothing = \varnothing$, $\bigcup \{a\} = a$, $\bigcup \{a,b\} = a \cup b$, $\bigcap \{a\} = a$, and $\bigcap \{a,b\} = a \cap b$).