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 $11$