Equivalence of Definitions of Ordered Pair

Theorem
The following definitions of an ordered pair are equivalent:

Equality of Ordered Pairs
From Equality of Ordered Pairs, we have that:
 * $\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} = \left\{{\left\{{c}\right\}, \left\{{c, d}\right\}}\right\} \iff a = c, b = d$

hence showing that the Kuratowski formalization fulfils the requirement for equality.

Existence of Cartesian Product
Let $A$ and $B$ be non-empty sets.

Let $a \in A$ and $b \in B$.

Then $\left\{{a}\right\} \subseteq A$ and $\left\{{b}\right\} \subseteq B$.

Therefore $\left\{{a, b}\right\} \subseteq A \cup B$ where $\cup$ denotes union.

Because $\left\{{a}\right\} \subseteq A \cup B$, it follows that:
 * $\left\{{a}\right\}, \left\{{a, b}\right\} \in \mathcal P \left({A \cup B}\right)$

where $\mathcal P \left({A \cup B}\right)$ is the power set of $A \cup B$.

Thus:
 * $\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} \subseteq \mathcal P \left({A \cup B}\right)$

and so by definition of power set:
 * $\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} \in \mathcal P \left({\mathcal P \left({A \cup B}\right)}\right)$

Applying the axiom of specification and the axiom of extension, the unique set $A \times B$ is created which consists exactly of ordered pairs $\left({a, b}\right)$ such that $a \in A$ and $b \in B$.

Thus it has been demonstrated that the cartesian product $A \times B$ exists and is non-empty.

Finally, in Subset of Cartesian Product it is demonstrated that every set of ordered pairs is a subset of the cartesian product of two sets.