Cardinal Product Distributes over Cardinal Sum

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Theorem

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.


Then:

$\mathbf a \paren {\mathbf b + \mathbf c} = \mathbf a \mathbf b + \mathbf a \mathbf c$

where:

$\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.
$\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$.


Proof

Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.

Let $B$ and $C$ be pairwise disjoint, that is:

$B \cap C = \O$

Then we can define:

$B \sqcup C := B \cup C$

where $B \sqcup C$ denotes the disjoint union of $B$ and $C$.


Then we have:

$\mathbf b + \mathbf c = \card {B \sqcup C} = \card {B \cup C}$


Then:

\(\ds \paren {A \times B} \cap \paren {A \times C}\) \(=\) \(\ds A \times \paren {B \cap C}\) Cartesian Product of Intersections/Corollary 1
\(\ds \) \(=\) \(\ds A \times \O\) as $B \cap C = \O$
\(\ds \) \(=\) \(\ds \O\) Cartesian Product is Empty iff Factor is Empty


Then:

\(\ds \card {\paren {A \times B} \cup \paren {A \times C} }\) \(=\) \(\ds \card {A \times B} + \card {A \times C}\) as $A \times B$ and $A \times C$ are disjoint from above
\(\ds \) \(=\) \(\ds \mathbf a \mathbf b + \mathbf a \mathbf c\) Definition of Product of Cardinals


Then:

\(\ds \card {\paren {A \times B} \cup \paren {A \times C} }\) \(=\) \(\ds \card {A \times \paren {B \cup C} }\) Cartesian Product Distributes over Union
\(\ds \) \(=\) \(\ds \mathbf a \paren {\mathbf b + \mathbf c}\) Definition of Product of Cardinals and Definition of Sum of Cardinals

$\blacksquare$


Sources

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