Talk:Cartesian Product is not Associative

I quote from the proof:
 * "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 $\left({b, c}\right) = c$."

However, this is too strict; $A$ 'merely' needs to satisfy $a\in A, b\in B\implies (a,b)\in A$ (and something similar for $C$). There are probably pathological examples satisfying this condition, which aren't degenerate. It would be interesting to construct such examples; I however have no time momentarily, and will add this to my to-be-formed PW to-do list in my namespace (it appears even as though the sets $A$ and $C$ satisfying this can be classified... most interesting). For now, I think we'd best weaken the statement by adding something like in general. --Lord_Farin 16:35, 18 November 2011 (CST)


 * That would be interesting. I've pondered this, and not been able to penetrate it - I've seen the statement made that $\times$ is not associative, and I've seen exercises to prove it, but apart from that (and a few more-or-less coherent arguments on various forums) I haven't made headway with it. --prime mover 16:50, 18 November 2011 (CST)


 * Having given it a second thought today, the only examples are those with one of $A,B,C$ empty. This is because we would need also that $\forall c\in C:\exists c'\in C,b\in B: c = (b,c')$. This however contradicts the well-foundedness of $\in$ (i.e., that only finite sequences of sets $x_1\in x_2\in ...$ exist). I will write up a proof soon. --Lord_Farin 07:01, 1 December 2011 (CST)