Cardinal of Union Less than Cardinal of Cartesian Product

Theorem
Let $S$ and $T$ be sets that are equivalent to their cardinal numbers.

Let $\left|{ S }\right|$ denote the cardinal number of $S$.

Suppose $\left|{ S }\right| > 1$ and $\left|{ T }\right| > 1$.

Then:


 * $\left|{ S \cup T }\right| \le \left|{ S \times T }\right|$

Proof
Let $x_1$ and $x_2$ be distinct elements of $S$.

Let $y_1$ and $y_2$ be distinct elements of $T$.

Define the mapping $f : S \times T \to S \cup T$ as follows:


 * $f\left({x,y}\right) = \begin{cases}

y &: x = x_1 \\ x_1 &: x = x_2 \land y = y_1 \\ x &: \text {otherwise} \end{cases}$

If $x \in S$, then we have that either $x = x_1$ or $x \ne x_1$.

If $x \ne x_1$, then $f\left({x, y_2}\right) = x$ by the definition of $f$.

If $x = x_1$, then $f\left({x_2, y_1}\right) = x_1$ by the definition of $f$.

If $y \in S$, then we have that:


 * $f\left({x_1, y}\right) = y$ by the definition of $f$.

Therefore, it follows that $f : S \times T \to S \cup T$ is a surjection.

Thus, $\left|{ S \cup T }\right| \le \left|{ S \times T }\right|$ by Surjection iff Cardinal Inequality.