Is it the case that $A \times (B \times C) = (A \times B) \times C$ if and only if one of the factors is the empty set? I am talking about strict equality, not isomorphism. The duplicate question did not answer my question, because it never said when the associative law is true.
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The answer to a question like this depends on the precise set-theoretic definition of $A\times B$ that you use. Assuming you define $A\times B=\{(a,b):a\in A,b\in B\}$, this will depend on how you define an ordered pair $(a,b)$.
That said, for pretty much any reasonable definition, $A\times (B\times C)=(A\times B)\times C$ is true only when one of the factors is empty. In particular, this is true for any definition of $(a,b)$ such that $(a,b)$ has higher rank than $a$ and $b$. (This holds for the Kuratowski definition $(a,b)=\{\{a\},\{a,b\}\}$, for instance, whose rank is $\max(\mathrm{rank}(a),\mathrm{rank}(b))+2$.)
To prove this, first note that if $B$ are nonempty, then $A$ is uniquely determined by the set $A\times B$ since $A$ is the set of all $a$ such that $(a,b)\in A\times B$ for some $b$. Similarly, $B$ is determined by $A\times B$ if $B$ is nonempty.
So now suppose $A,B,$ and $C$ are nonempty. In order to have $A\times (B\times C)=(A\times B)\times C$, we must have $A=A\times B$. But now take an element $a\in A$ of minimal rank. Every element of $A\times B$ has rank strictly greater than the rank of $a$, so $a\not\in A\times B$. Thus $A\neq A\times B$, and so $A\times (B\times C)\neq(A\times B)\times C$.
This argument crucially uses the axiom of regularity to talk about ranks of sets. Without the axiom of regularity, there can be counterexamples. In particular, in ZF without regularity it is consistent for there to exist a set $A$ such that $A=\{A\}$. With the Kuratowski definition, we then have $$A\times A=\{(A,A)\}=\{\{\{A\},\{A,A\}\}\}=\{\{\{A\}\}\}=A$$ and so $$(A\times A)\times A=A\times A=A\times (A\times A).$$
Eric Wofsey
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