Cartesian Product is not Associative

Theorem
Let $$A, B, C$$ be sets.

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

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

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

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 $$\left({a, \left({b, c}\right)}\right)$$.

From Cardinality of Cartesian Product, we have that:
 * $$\left|{A \times \left({B \times C}\right)}\right| \sim \left|{\left({A \times B}\right)\times C}\right|$$

and so:
 * $$A \times \left({B \times C}\right) \sim \left({A \times B}\right)\times C$$

where $$\sim$$ denotes set equivalence.

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