Sum of Cardinals is Associative

Theorem

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

Then:

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

where $\mathbf a + \mathbf b$ denotes the sum 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 $A, B, C$ be pairwise disjoint, that is:

$A \cap B = \O$

$B \cap C = \O$

$A \cap C = \O$

Then we can define:

$A \sqcup B := A \cup B$

$B \sqcup C := B \cup C$

$A \sqcup C := A \cup C$

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

Then we have:

$\mathbf a + \mathbf b = \card {A \sqcup B} = \card {A \cup B}$

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

Then:

\(\ds \paren {A \cup B} \cap C\)

\(=\)

\(\ds \paren {A \cap C} \cup \paren {B \cap C}\)

Intersection Distributes over Union

\(\ds \)

\(=\)

\(\ds \O \cup \O\)

as $A \cap C = \O$ and $B \cap C = \O$

\(\ds \)

\(=\)

\(\ds \O\)

Union with Empty Set

Then:

\(\ds \card {\paren {A \cup B} \cup C}\)

\(=\)

\(\ds \card {A \cup B} + \card C\)

as $\paren {A \cup B} \cap C = \O$ from above

\(\ds \)

\(=\)

\(\ds \paren {\mathbf a + \mathbf b} + \mathbf c\)

Definition of Sum of Cardinals

Similarly:

\(\ds A \cap \paren {B \cup C}\)

\(=\)

\(\ds \paren {A \cap B} \cup \paren {A \cap C}\)

Intersection Distributes over Union

\(\ds \)

\(=\)

\(\ds \O \cup \O\)

as $A \cap B = \O$ and $A \cap C = \O$

\(\ds \)

\(=\)

\(\ds \O\)

Union with Empty Set

Then:

\(\ds \card {A \cup \paren {B \cup C} }\)

\(=\)

\(\ds \card A + \card {B \cup C}\)

as $A \cap \paren {B \cup C} = \O$ from above

\(\ds \)

\(=\)

\(\ds \mathbf a + \paren {\mathbf b + \mathbf c}\)

Definition of Sum of Cardinals

Finally note that from Union is Associative:

$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 8$: Theorem $8.5: \ (2)$