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 $\map n X \in \N$:

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

From the Axiom of Foundation:

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


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

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

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:

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

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, 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