Subset of Cartesian Product
Theorem
Let $S$ be a set of ordered pairs.
Then $S$ is the subset of the cartesian product of two sets.
Proof
Let $S$ be a set of ordered pairs.
Let $x \in S$ such that $x = \set {\set a, \set {a, b} }$ as defined in Kuratowski Formalization of Ordered Pair.
Since the elements of $S$ are sets, we can form the union $\mathbb S = \bigcup S$ of the sets in $S$.
Since $x \in S$ it follows that the elements of $x$ are elements of $\mathbb S$.
Since $\set {a, b} \in x$ it follows that $\set {a, b} \in \mathbb S$.
Now we can form the union $\mathbb S' = \bigcup \mathbb S$ of the sets in $\mathbb S$.
Since $\set {a, b} \in \mathbb S$ it follows that both $a$ and $b$ are elements of $\mathbb S' = \bigcup \bigcup S$.
Thus from the Kuratowski Formalization of Ordered Pair we have that $S$ is a subset of some $A \times B$.
We can at this stage take both $A$ and $B$ as being equal to $\bigcup \bigcup S$.
Finally, the axiom of specification is applied to construct the sets:
- $A = \set {a: \exists b: \tuple {a, b} \in S}$
and
- $B = \set {b: \exists a: \tuple {a, b} \in S}$
$A$ and $B$ are seen to be the first and second projections respectively of $S$.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 6$: Ordered Pairs