# Cartesian Product is not Associative

## Theorem

Let $A, B, C$ be non-empty sets.

Then:

- $A \times \paren {B \times C} \ne \paren {A \times B} \times C$

where $A \times B$ is the cartesian product of $A$ and $B$.

## Intuitive Proof

By definition:

- $A \times B = \set {\tuple {a, b}: a \in A, b \in B}$

that is, the set of all ordered pairs $\tuple {a, b}$ such that $a \in A$ and $b \in B$.

Now:

- Elements of $A \times \paren {B \times C}$ are in the form $\tuple {a, \tuple {b, c} }$
- Elements of $\paren {A \times B} \times C$ are in the form $\tuple {\tuple {a, b}, c}$.

So for $A \times \paren {B \times C} = \paren {A \times B} \times C$ we would need to have that $a = \tuple {a, b}$ and $\tuple {b, c} = c$.

This can not possibly be so, except perhaps in the most degenerate cases.

So from the strict perspective of the interpretation of the pure definitions:

- $A \times \paren {B \times C} \ne \paren {A \times B} \times C$

$\blacksquare$

## Formal Proof

Assign to every set $X$ the following number $\map n X \in \N$:

- $\map n X = \begin{cases} 0 & : X = \O \\ 1 + \max_{Y \mathop \in X} \map n Y & : \text{ otherwise} \end{cases}$

From the Axiom of Foundation:

- $\forall X \in \N: \map n X < \infty$

Now let $a \in A$ be such that:

- $\map n a = \min_{b \mathop \in A} \map n b$

Suppose that:

- $\exists a' \in A, b \in B: a = \tuple {a', b}$

That is, that $a$ equals the ordered pair of $a'$ and $b'$.

Then it follows that:

\(\ds \map n a\) | \(=\) | \(\ds \map n {\set {\set {a'}, \set {a', b} } }\) | Definition of Ordered Pair | |||||||||||

\(\ds \) | \(=\) | \(\ds 1 + \map \max {\map n {\set {a'} }, \map n {\set {a', b} } }\) | Definition of $n$ | |||||||||||

\(\ds \map n {\set {a', b} }\) | \(\ge\) | \(\ds \map n {\set {a'} }\) | Maximum of Subset | |||||||||||

\(\ds \) | \(=\) | \(\ds 1 + \map n {a'}\) | Definition of $n$ | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map n a\) | \(\ge\) | \(\ds 2 + \map n {a'}\) |

That is:

- $\map n {a'} < \map n a$

contradicting the assumed minimality of the latter.

Therefore:

- $a \notin A \times B$

and hence:

- $A \nsubseteq A \times B$

It follows from Equality of Cartesian Products that:

- $A \times \paren {B \times C} \ne \paren {A \times B} \times C$

$\blacksquare$

## Comment

Despite this result, the cartesian product of three sets is usually just written $A \times B \times C$ and **understood** to be the set of all ordered triples.

That is, as the set of all elements like $\tuple {a, \tuple {b, c} }$.

From Cardinality of Cartesian Product of Finite Sets, we have that:

- $\card {A \times \paren {B \times C} } = \card {\paren {A \times B} \times C}$

and so:

- $A \times \paren {B \times C} \sim \paren {A \times B} \times C$

where $\sim$ denotes set equivalence.

So it matters little whether $A \times B \times C$ is defined as being $A \times \paren {B \times C}$ or $\paren {A \times B} \times C$, and it is rare that one would even need to know.

When absolute rigour is required, the cartesian product of more than two sets can be defined using ordered $n$-tuples or, even more generally, by indexed sets.

## Also see

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): Exercise $1.2: \ 13$ - 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $1$. Sets: Exercise $8$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $1$: The Notation and Terminology of Set Theory: $\S 9 \beta$