Sum of Cardinals is Associative

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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