Cardinality of Union not greater than Product

Theorem
Let $\FF$ be a set of sets.

Let:
 * $\size \FF \le \mathbf m$

where
 * $\size \FF$ denotes the cardinality of $\FF$
 * $\mathbf m$ is cardinal number (possibly infinite).

Let:
 * $\forall A \in \FF: \size A \le \mathbf n$

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

Then:
 * $\ds \size {\bigcup \FF} \le \size {\mathbf m \times \mathbf n} = \mathbf m \mathbf n$

Proof
$\FF = \O$ or $\FF = \set \O$ or $\O \ne \FF \ne \set \O$.

In case when $\FF = \O$ or $\FF = \set \O$:

In case when $\O \ne \FF \ne \{\O\}$:
 * by Surjection iff Cardinal Inequality there exists a surjection $f: \mathbf m \to \FF$ as $\size {\mathbf m} = \mathbf m$ by Cardinal of Cardinal Equal to Cardinal.

$\FF$ contains non empty set $A_0$.
 * $\size {A_0} > \mathbf 0$.

By assumption:
 * $\size {A_0} \le \mathbf n$.

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

Hence:
 * $\size {\set 0} = \mathbf 1 \le \mathbf n$.

Define a family $\family {B_A}_{A \mathop \in \FF}$:
 * $B_A = \begin{cases} A & A \ne \O \\ \set 0 & A = \O \end {cases}$

By Surjection iff Cardinal Inequality define a family $\family {g_A}_{A \mathop \in \FF}$ of surjections:


 * $\forall A \in \FF: g_A: \mathbf n \to B_A$ is a surjection.

Define a mapping $h:\mathbf m \times \mathbf n \to \ds \bigcup_{A \mathop \in \FF} B_A$:
 * $\forall \alpha \in \mathbf m: \forall \beta \in \mathbf n: \map h {\alpha, \beta} = \map {g_{\map f \alpha} } \beta$

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

Let $x \in \ds \bigcup_{A \mathop \in \FF} B_A$.

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

By definition of surjection:
 * $\exists \alpha \in \mathbf m: \map f \alpha = A$.

By definition of surjection:
 * $\exists \beta \in \mathbf n: \map {g_A} \beta = x$

So:
 * $\map h {\alpha, \beta} = \map {g_{\map f \alpha} } \beta = x$

This ends the proof of surjection.

Hence by Surjection iff Cardinal Inequality:
 * $\ds \size {\bigcup_{A \mathop \in \FF} B_A} \le \size {\mathbf m \times \mathbf n}$

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

Then by Set Union Preserves Subsets:
 * $\ds \bigcup \FF \subseteq \bigcup_{A \mathop \in \FF} B_A$.

Hence by Subset implies Cardinal Inequality:
 * $\ds \size {\bigcup \FF} \le \ds \size {\bigcup_{A \mathop \in \FF} B_A}$

Thus the result:
 * $\size {\bigcup \FF} \le \size {\mathbf m \times \mathbf n}$

Thus by definition of product of cardinals:
 * $\size {\mathbf m \times \mathbf n} = \mathbf m \mathbf n$