# Cartesian Product is Anticommutative/Corollary

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## Corollary to Cartesian Product is Anticommutative

Let $S$ and $T$ be sets.

Then:

- $S \times T = T \times S \iff S = T \lor S = \O \lor T = \O$

where $S \times T$ denotes the cartesian product of $S$ and $T$.

## Proof

Suppose $S \times T = T \times S$.

Then either:

- $(1): \quad S \ne \O \land T \ne \O$ and from Cartesian Product is Anticommutative, $S = T$

or:

- $(2): \quad S = \O \lor T = \O$ and from Cartesian Product is Empty iff Factor is Empty, $S \times T = T \times S = \O$.

In either case, we see that:

- $S \times T = T \times S \implies S = T \lor S = \O \lor T = \O$

Now suppose $S = T \lor S = \O \lor T = \O$.

From Cartesian Product is Empty iff Factor is Empty, we have that:

- $S = \O \lor T = \O \implies S \times T = \O = T \times S$

Similarly:

- $S = T \land \neg \paren {S = \O \lor T = \O} \implies S \times T = T \times S$

by definition of equality.

$\blacksquare$

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 3$. Ordered pairs; cartesian product sets: Exercise $5$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $11$