I have seen that being used like in n-ary Cartesian product.

n-ary Cartesian product

But we know that Cartesian product is not associative so $(A \times B) \times C \neq A \times (B \times C)$ and therefore $A \times B \times C$ is not well-defined. (Like the cross product, we need to have the parentheses.)


My main concern is how all of these things can be defined (listed below)? Are all of them definable at all?

As long as we are dealing with only two sets, $A \times B$ can only mean one thing.

But ambiguities arise when the number of sets involved increases:

(For compactness, I will use the $AB$ notation instead of $A\times B$)

3 sets: $$(AB)C; \quad ABC; \quad A(BC)$$

4 sets: $$ABCD$$ $$(AB)CD; \quad A(BC)D; \quad AB(CD)$$ $$(ABC)D; \quad A(BCD)$$ $$((AB)C)D; \quad A((BC)D)$$ $$(A(BC))D; \quad A(B(CD))$$

And so on.

Up to now, It has been made clear that:

  • Sets like $X_1 \cdots X_n$ are accepted to have elements of the form $(x_1,\dots,x_n)$ -- Like $\mathbb{R}^n$.
  • Sets that have clearly shown the order of operation, have elements of the form similar to the parentheses placement. For example $(x_1,((x_2,x_3),x_4)) \in A((BC)D)$. (Am I right?)

But again some of the items of the above list remain ambiguous. Like $(ABC)D$.

Can we define this?

Can we set some rule for defining even more complicated combinations?

  • 2
    $\begingroup$ See Associativity of cartesian product. $\endgroup$ Apr 11, 2018 at 7:24
  • $\begingroup$ @MauroALLEGRANZA Oh, seems here we have a duplicate thought I think the way I have asked it differs. I'll see if I can find my answer there. Thank you. $\endgroup$
    – AHB
    Apr 11, 2018 at 7:29
  • $\begingroup$ There is an issue about convention: we have to decide if $A \times B \times C$ is $(A \times B) \times C$ instead of $A \times (B \times C)$. $\endgroup$ Apr 11, 2018 at 7:34
  • $\begingroup$ And there an issue about "practicality": in many context, the convention adopted is immaterial: $\mathbb R^n$. $\endgroup$ Apr 11, 2018 at 7:34
  • 1
    $\begingroup$ It is true $(A \times B)\times C \ne A \times (B \times C)$ but there is a natural correspondence among their elements. In math, when you are faced with objects with natural correspondences/isomorphisms and so, you typically treat them as the same thing. In fact, you can go the axiomatic way and declare an n-ary Cartesian product is something that follows certain axioms and has certain universal properties. Different ways of defining them (say the order you pick) just provides concrete models to realize the axioms. $\endgroup$ Apr 11, 2018 at 8:27

1 Answer 1


In general we do not have $\langle\langle a,b\rangle,c\rangle=\langle a,\langle b,c\rangle\rangle$ so definitions like $A\times B\times C:=(A\times B)\times C$ or $A\times B\times C:=A\times(B\times C)$ are not canonical.

A nice solution for that is defining $X_1\times\cdots\times X_n$ as the set of functions $f:\{1,\dots,n\}\to\bigcup_{i=1}^nX_i$ that satisfy $f(i)\in X_i$ for every $i\in\{1,\dots,n\}$.

If $f$ is an element of that set and $f(i)=x_i\in X_i$ for $i=1,\dots,n$ then $\langle x_1,\dots,x_n\rangle$ can be looked at as a nice notation for that function, and nothing more than that.

This also works for other index sets:

$$\prod_{i\in I}X_i=\{f\mid f\text{ is a function }I\to\bigcup_{i\in I}X_i\text{ s.t. }\forall i\in I[f(i)\in X_i]\}$$

The projection $p_i:\prod_{i\in I}X_i\to X_i$ is prescribed by $f\mapsto f(i)$.

  • 2
    $\begingroup$ It might be worth mentioning that there's an obvious bijection between the versions of $A\times B\times C $, and in many cases that bijection preserves further structure (group, topology, etc.). So that is another reason why in practice the distinction doesn't matter. $\endgroup$
    – Mark S.
    Apr 11, 2018 at 10:24
  • $\begingroup$ @MarkS. see the edit. $\endgroup$
    – AHB
    Apr 11, 2018 at 14:40

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