# Cartesian Product of Sets is Set

## Theorem

Let $V$ be a basic universe.

Let $A$ and $B$ be sets in $V$.

Then $A \times B$ is also a set.

## Proof

Let $A$ and $B$ be sets in $V$.

Because $V$ is a basic universe, the basic universe axioms apply.

Hence by the axiom of pairing $\set {A, B}$ is a set.

Then by the axiom of unions $\displaystyle \bigcup \set {A, B}$ is also a set.

We have that $A \cup B = \displaystyle \bigcup \set {A, B}$.

By the axiom of powers $\powerset {A \cup B}$ is a set.

Therefore, also by the axiom of powers, so is $\powerset {\powerset {A \cup B} }$.

It remains to be shown that $A \times B$ is a subclass of $\powerset {\powerset {A \cup B} }$.

Let $x \in A \times B$.

Then $x = \tuple {a, b}$ for some $a \in A$ and $b \in B$.

By definition of ordered pair:

- $x = \set {\set a, \set {a, b} }$

From Set is Subset of Union, we have that $A \subseteq A \cup B$ and $B \subseteq A \cup B$.

Hence by definition of subset:

- $a \in A \cup B$

and:

- $b \in A \cup B$

Thus:

- $\set a \subseteq A \cup B$

and:

- $\set {a, b} \subseteq A \cup B$

Thus by definition of power set:

- $\set a \in \powerset {A \cup B}$

and:

- $\set {a, b} \in \powerset {A \cup B}$

Hence:

- $\set {\set a, \set {a, b} } \subseteq \powerset {A \cup B}$

That is:

- $\tuple {a, b} \subseteq \powerset {A \cup B}$

and so:

- $x \subseteq \powerset {A \cup B}$

which means:

- $x \in \powerset {\powerset {A \cup B} }$

Hence by definition of subclass:

- $A \times B \subseteq \powerset {\powerset {A \cup B} }$

We have that $\powerset {\powerset {A \cup B} }$ is a set.

By the axiom of swelledness it follows that $A \times B$ is also a set.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 7$ Cartesian products: Theorem $7.1$