Cardinality of Union not greater than Product

Theorem
Let $\mathcal F$ be a set of sets.

Let:
 * $\left\vert{\mathcal F}\right\vert \leq \mathbf m$

where
 * $\left\vert{\mathcal F}\right\vert$ denotes the cardinality of $\mathcal F$,
 * $\mathbf m$ is cardinal number (possibly infinite).

Let:
 * $\forall A \in \mathcal F: \left\vert{A}\right\vert \leq \mathbf n$

where
 * $\mathbf n$ is cardinal number (possibly infinite).

Then:
 * $\displaystyle \left\vert{\bigcup \mathcal F}\right\vert \leq \left\vert{\mathbf m \times \mathbf n}\right\vert = \mathbf m \mathbf n$

Proof
$\mathcal F = \varnothing$ or $\mathcal F = \{\varnothing\}$ or $\varnothing \ne \mathcal F \ne \{\varnothing\}$.

In case when $\mathcal F = \varnothing$ or $\mathcal F = \{\varnothing\}$:

In case when $\varnothing \ne \mathcal F \ne \{\varnothing\}$:
 * by Surjection iff Cardinal Inequality there exists a surjection $f: \mathbf m \to \mathcal F$ as $\left\vert{\mathbf m}\right\vert = \mathbf m$ by Cardinal of Cardinal Equal to Cardinal.

$\mathcal F$ contains non empty set $A_0$.
 * $\left\vert A_0 \right\vert > \mathbf 0$.

By assumption:
 * $\left\vert A_0 \right\vert \leq \mathbf n$.

Then:
 * $\mathbf 0 < \mathbf n$.

Hence:
 * $\left\vert \{0\} \right\vert = \mathbf 1 \leq \mathbf n$.

Define a family $\left({B_A}\right)_{A \in \mathcal F}$:
 * $B_A = \left\{{\begin{array}{ll}A& A \ne \varnothing\\ \{0\}& A = \varnothing\end{array}}\right.$

By Surjection iff Cardinal Inequality define a family $\left({g_A}\right)_{A \in \mathcal F}$ of surjections:
 * $\forall A \in \mathcal F: g_A: \mathbf n \to B_A$ is a surjection.

Define a mapping $h:\mathbf m \times \mathbf n \to \displaystyle \bigcup_{A \in \mathcal F} B_A$:
 * $\forall \alpha \in \mathbf m: \forall \beta \in \mathbf n: h \left( {\alpha, \beta} \right) = g_{f \left(\alpha\right)}\left(\beta\right)$.

We will show by definition that $h$ is a surjection.

Let $x \in \displaystyle \bigcup_{A \in \mathcal F} B_A$.

Then by definition of union:
 * $\exists A \in \mathcal F: x \in B_A$.

By definition of surjection:
 * $\exists \alpha \in \mathbf m: f \left(\alpha\right) = A$.

By definition of surjection:
 * $\exists \beta \in \mathbf n: g_A \left(\beta\right) = x$.

So:
 * $h \left({\alpha, \beta}\right) = g_{f \left(\alpha\right)}\left(\beta\right) = x$.

This ends the proof of surjection.

Hence by Surjection iff Cardinal Inequality:
 * $\displaystyle \left\vert {\bigcup_{A \in \mathcal F} B_A} \right\vert \leq \left\vert{\mathbf m \times \mathbf n}\right\vert$.

By definition of subset:
 * $\forall A \in \mathcal F: A \subseteq B_A$.

Then by Set Union Preserves Subsets:
 * $\displaystyle \bigcup \mathcal F \subseteq \bigcup_{A \in \mathcal F} B_A$.

Hence by Subset implies Cardinal Inequality:
 * $\displaystyle \left\vert {\bigcup \mathcal F} \right\vert \leq \displaystyle \left\vert {\bigcup_{A \in \mathcal F} B_A} \right\vert$

Thus the result
 * $ \left\vert {\bigcup \mathcal F} \right\vert \leq \left\vert{\mathbf m \times \mathbf n}\right\vert$.

Thus by definition of product of cardinals:
 * $\left\vert{\mathbf m \times \mathbf n}\right\vert = \mathbf m \mathbf n$.