Let $A,B$ be two nonempty sets, then the Cartesian product of $A$ and $B$ is defined as $$A\times B:=\left\{(x,y)|\ x\in A,\ y\in B\right\}$$ Here $(x,y):=\left\{x,\left\{x,y\right\}\right\}.$

And $(x,y,z):=((x,y),z),\ A\times B\times C:=(A\times B)\times C,$ etc.

In general, there is $$((x,y),z)\neq (x,(y,z)),$$and$$(A\times B)\times C \neq A\times(B\times C).$$ So Cartesian product is not associative. I've come up with an idea of defining an "associative Cartesian product".

Definition: Let $A$ be a nonempty set. Then $\forall\ m,n\in \mathbb{N},\ \forall\ x=(x_1,\cdots,x_m)\in A^m,\ y=(y_1,\cdots,y_n)\in A^n, $ we define $$x\circledast_A y:=(x_1,x_2,\cdots,x_m,y_1,y_2,\cdots,y_n)\in A^{m+n}.$$ And $\forall C\subset A^m,\ D\subset A^n,$ we define that $$C\otimes_A D:=\left\{x\circledast_A y | \ x\in C,\ y\in D \right\}$$

I call the operation $\otimes_A$ "associative Cartesian product", because if my definition is well-defined, then we have $(C\otimes_A D)\otimes_A E=C\otimes_A (D\otimes_A E)$ for any set $C,D,E\subset A.$

And particularly, we have $\mathbb{R}^m\otimes_{\mathbb{R}} \mathbb{R}^n=\mathbb{R}^{m+n},\ $and $ (x,y)\circledast_{\mathbb{R}} z =x\circledast_{\mathbb{R}} (y,z),$ for any $x,y,z\in \mathbb{R}. $

Question: Is my definition above (in the yellow square frame) well-defined and proper ?

  • 2
    $\begingroup$ It works, after a fashion, as long as everything you want to put into your tuples are numbers. But there are subtleties that mean it's not usually done that way. See Associativity of Cartesian Product. $\endgroup$ – Henning Makholm Jun 16 '17 at 13:46

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