Cartesian Product is not Associative/Proof 2

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 $B$.

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:

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

$\blacksquare$