2
$\begingroup$

My textbook defined general cartesian products as: Let $\{A_\alpha| \alpha\in\mathbb{A}\}$ be a family of sets. The the cartesian product $\prod_{\alpha} A_\alpha$ is the set of all maps $c$: $\mathbb{A}\rightarrow\cup_{\alpha} A_\alpha$ having the property $\forall\alpha\in\mathbb{A}: c(\alpha)\in A_\alpha$.

Since the cartesian product $A*B\neq B*A$, I was wondering is index set $\mathbb{A}$ always has some order to make sure the general product, for example, is $A_{\alpha 1}*A_{\alpha 2}*A_{\alpha 3}*\dots$ not $A_{\alpha 2}*A_{\alpha 1}*A_{\alpha 3}*\dots$, or the general cartesian product includes both the cases $A_{\alpha 1}*A_{\alpha 2}*A_{\alpha 3}\dots$ and $A_{\alpha 2}*A_{\alpha 1}*A_{\alpha 3}*\dots$.

It somehow seems in either case the map $c(\alpha)\in A_\alpha$ is defined. Thanks!

$\endgroup$
2
  • $\begingroup$ You should write $\times$ instead of $*$. $\endgroup$ – Paul Frost Jan 20 '20 at 23:33
  • $\begingroup$ What is your definition of $A \times B$? $\endgroup$ – Paul Frost Jan 21 '20 at 0:09
1
$\begingroup$

Your book gives the most general definition possible. An useful comparison: $\Bbb R^n = \prod_{i \in \{1,\ldots,n\}} \Bbb R$ is the set of functions $v\colon \{1,\ldots, n\} \to \Bbb R$ with $v(i) \in \Bbb R$ for all $i$. We write $v(i) = v_i$ and $v = (v_1,\ldots, v_n)$. This suggests that you should think of an element $c \in \prod_{\alpha \in \mathbb{A}} A_\alpha$ as a vector with $|\mathbb{A}|$ components (even if $|\mathbb{A}|$ is infinite!), one for each factor $A_\alpha$.

As far as order go, you are correct. If $A_1 = A$ and $A_2 = B$, both $A\times B$ and $B\times A$ are equal to $\prod_{\alpha \in \{1,2\}} A_\alpha$, but choosing an order $A\times B$ or $B\times A$ to write requires a choice of ordering for the index set $\mathbb{A} = \{1,2\}$ (either $1<2$ or $2<1$).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.