Distinction between Finite Cartesian Product and Order Pair

The beginning of a measure theory book I am read introduces Cartesian products. It contains the following statements that confuse me:

If $$\{X_\alpha\}_{\alpha \in A}$$ is an indexed family of sets, their Cartesian product $$\prod_{\alpha \in A}X_\alpha$$ is the set of all maps $$f: A \rightarrow \bigcup_{\alpha\in A}X_\alpha$$ s.t. $$f(\alpha) \in X_{\alpha} \, \forall \alpha \in A$$. It should be noted, and then promptly forgotten, that when $$A = \{1,2\}$$, the definition of $$X_1 \times X_2$$ (the set of all ordered pairs where the first term comes from $$X_1$$ and the second from $$X_2$$) is set theoretically different from the present definition $$\prod^2_{i=1}X_i$$. Indeed the latter concept depends on mappings, which are defined in terms of the former.

I'm not understanding what distinction the book is trying to make about Cartesian products and the set of ordered pairs. Is it just that Cartesian products is defined in terms of functions?

A general element of $$X_1\times X_2$$ is an ordered pair $$(x_1,x_2)$$, whereas a general element of $$\prod_{i=1}^2 X_i$$ is a function from $$\{1,2\}$$ to $$X_1\cup X_2$$ where $$f(1)\in X_1$$ and $$f(2)\in X_2$$. That is, the former contains $$(x_1,x_2)$$ while the latter contains $$\{(1,x_1),(2,x_2)\}$$ Therefore, the elements of $$X_1\times X_2$$ and $$\prod_{i=1}^2 X_i$$ are "cosmetically" different, but there is a bijective correspondence between these two sets, where one of the two above objects is mapped to the other. This is the reason that the point being made should be "promptly forgotten;" the two constructions capture the same concept in different ways, so can be used interchangeably in any application.
It should be noted, then promptly forgotten, that the ordered pair notation $$(a,b)$$ is actually shorthand for a particular set, usually chosen to be $$\{\{a\},\{a,b\}\}$$.
• Some authors (e.g. K. Kunen) define $(a,b)=\{a, \{a,b\}\}.$ With Kuratowski's definition $(a,b)=\{\{a\},\{a,b\}\}$ we can show that $(a,b)=(c,d)\implies (a=c \land b=d)$ without using the Axiom of Foundation (a.k.a. Regularity).... I am not sure this can be shown without Foundation using the def'n of $(a,b)$ favored by Kunen unless $a$ or $c$ is assumed to belong to the class of well-founded sets.............+1 Oct 1, 2018 at 22:14