Cartesian Product Preserves Cardinality

Theorem

Let $R$, $S$, and $T$ be sets.

Suppose that $S$ is equivalent to $T$.

Then:

$R \times S \sim R \times T$

$S \times R \sim T \times R$

Since $S$ and $T$ are equivalent, there exists a bijection $f: S \to T$.

Let $g: T \to S$ be the inverse of $f$; its existence is assured by Bijection iff Left and Right Inverse.

Define $\hat f: R \times S \to R \times T$ by:

$\map {\hat f} {r, s} := \tuple {r, \map f s}$

Next, define $\hat g: R \times T \to R \times S$ by:

$\map {\hat g} {r, t} := \tuple {r, \map g t}$

Then, for all $\tuple {r, s} \in R \times S$:

$\ds \hat g \circ \map {\hat f} {r, s}$

$=$

$\ds \map {\hat g} {r, \map f s}$

$\ds $

$=$

$\ds \tuple {r, \map g {\map f s} }$

$\ds $

$=$

$\ds \tuple {r, s}$

$g$ is the inverse of $f$

Similarly, for all $\tuple {r, t} \in R \times T$:

$\ds \hat f \circ \map {\hat g} {r, s}$

$=$

$\ds \map {\hat f} {r, \map g s}$

$\ds $

$=$

$\ds \tuple {r, \map f {\map g s} }$

$\ds $

$=$

$\ds \tuple {r, s}$

$g$ is the inverse of $f$

Thus, it follows that $\hat g$ is a inverse for $\hat f$.

Hence $R \times S \sim R \times T$, by definition of set equivalence.

Mutatis mutandis, the other relation also follows.

$\blacksquare$