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
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 8$: Theorem $8.6$