Cartesian Product is not Associative

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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:

$A \times B = \left\{{\left({a, b}\right): a \in A, b \in B}\right\}$

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

Now:

Elements of $A \times \left({B \times C}\right)$ are in the form $\left({a, \left({b, c}\right)}\right)$
Elements of $\left({A \times B}\right)\times C$ are in the form $\left({\left({a, b}\right), c}\right)$.

So for $A \times \left({B \times C}\right) = \left({A \times B}\right)\times C$ we would need to have that $a = \left({a, b}\right)$ and $\left({b, c}\right) = c$.

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, $A \times \left({B \times C}\right) \ne \left({A \times B}\right) \times C$.

$\blacksquare$


Formal Proof

Assign to every set $X$ the following number $n \left({X}\right) \in \N$:

$n \left({X}\right) = \begin{cases} 0 & : X = \varnothing \\ \displaystyle 1 + \max_{Y \mathop \in X} \ n \left({Y}\right) & : \text{ otherwise} \end{cases}$

From the Axiom of Foundation:

$\forall X \in \N: n \left({X}\right) < \infty$


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

$\displaystyle n \left({a}\right) = \min_{b \mathop \in A} \ n \left({b}\right)$

Suppose that:

$\exists a' \in A, b \in B: a = \left({a', b}\right)$

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

Then it follows that:

\(\displaystyle n \left({a}\right)\) \(=\) \(\displaystyle n \left({\left\{ {\left\{ {a'}\right\}, \left\{ {a', b}\right\} }\right\} }\right)\) Definition of Ordered Pair
\(\displaystyle \) \(=\) \(\displaystyle 1 + \max \left({n \left({\left\{ {a'}\right\} }\right), n \left({\left\{ {a', b}\right\} }\right) }\right)\) Definition of $n$
\(\displaystyle n \left({\left\{ {a', b}\right\} }\right)\) \(\ge\) \(\displaystyle n \left({\left\{ {a'}\right\} }\right)\) Maximum of Subset
\(\displaystyle \) \(=\) \(\displaystyle 1 + n \left({a'}\right)\) Definition of $n$
\(\displaystyle \implies \ \ \) \(\displaystyle n \left({a}\right)\) \(\ge\) \(\displaystyle 2 + n \left({a'}\right)\)

That is:

$n \left({a'}\right) < n \left({a}\right)$

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:

$A \times \left({B \times C}\right) \ne \left({A \times B}\right) \times C$

$\blacksquare$


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, we have that:

$\size {A \times \paren {B \times C} } = \size {\paren {A \times B} \times C}$

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