I have some trouble trying to understand Halmos' explanations about generalized Cartesian products. I've read these entries already:

Is his generalization the same as a Cartesian product of several sets? As here: Multiple products

What puzzles me is $\{a,b\}$. Where does it come from, what does it do? I know that it is supposed to be arbitrary. How does it identify distinct elements of $X \times Y$? Does some function map each $a \in A$ to an element in $X$? Or is it always the same $a$?

What does he mean by

the set $Z$ of all families $z$, indexed by $\{a,b\}$, such that $z_a \in X$ and $z_b \in Y$

? What does does an element of $Z$, namely one family $z$, consist of? By family, does he mean an indexing function or the indexed elements? This comes to a very important question for me: Does family mean a function or the set of all indexed elements?

I will try to explain how I understand it thus far: $Z$ contains elements of the form $\{(a,x),(b,y)\}$. The choice of $x$ depends on $a$, so an $x$ in a $z \in Z$ equals the $z_a$ of $X$. Now, we want to map these pairs to $X \times Y$, so we have a function $f$ that takes $z \in Z$ and connects it to the corresponding ordered pair $(z_a, z_b)$, where $z_a \in X$. $X \times Y$ is not the same as $Z$; they only, effectively, contain equal elements.

He goes on:

If $\{X_i\}$ is a family of sets ($i \in I$), the Cartesian product of the family is, by definition, the set of all families $\{x_i\}$ with $x_i \in X_i$ for each $i$ in $I$.

How I understand it: $\{X_i\}$ contains, as elements, several sets, each single one a set $X$ identified by an index/element of $I$. Then, we go over all indeces $i$ in $I$, take from each set $X$, namely $X_i$, an element $x$ and add it to current set of the newly created sets/families. So, the lower index $i$ does not refer to an element of $X_i$ indexed by $i$, but rather calls all elements of $X_i$ $x_i$. We end up with a set, defined as Cartesian product, of sets, in which each element (set) contains ordered elements, each from a different $X_i$. How does the ordering occur?

$\prod_i X_i$, the Cartesian product, becomes equal to $X^I$, if all $X_i$ are the same set $X$. $X^I$ means all possible combinations of mapping from $I$ to $X$. We randomly take an elements $x$ from some $X_i$, put it in a set/family with other randomly obtained $x$, put the entire set into a set denoted $X^I$, and continue doing so, till we have no more options. Then, each set in the new $X^I$ contains a mapping of some $i$ to some $x$. Since we exhausted the options, they are indeed all combinations.

Where did I go wrong? Someone needs to spoonfeed my the answer; I am completely confused by now.


A few simple steps: A family of elements of $A$ that is indexed by $I$ is just a function $f\colon I\to A$, however, usually writing $f_i$ instead of $f(i)$. Thus a "family $z$, indexed by $\{a,b\}$ such that $z_a\in X$ and $z_b\in Y$" is nothing but a function $z\colon \{a,b\}\to X\cup Y$ with $z(a)\in X$ and $z(b)\in Y$. Indeed, $a$ and $b$ are arbitrary (but distinct). Some authors use a fixed choice $a=\emptyset$ and $b=\{\emptyset\}$, but $a=\text{horse}$ and $b=\text{apple}$ would fulfill the same purpose if applicable. After seing this, $Z$ is the set of all such families (but still with fixed choices of $a$ and $b$). Indeed, this is essentially the same as $X\times Y$, as the author continues, if one identifies such a family $z$ with the ordered pair $(z(a),z(b))=(z_a,z_b)$. There is no ordering in $z$ unless we say that $a$ "precedes" $b$. Yet the identification with $X\times Y$ depends on the fact that we associate $a$ with $X$ and $b$ with $Y$, so this makes $a$ somewhat the first element and $b$ the second, if we want to identify $Z$with $X\times Y$ because $X$ is the first and $Y$ the second factor in this artesian product. We can also identify $Z$ with $Y\times X$ (then having ipso facto $b$ "preceeding" $a$). (Now what if $X=Y$?).

Note that in general, a family of sets $\{A_i\}_{i\in I}$ indexed by some index set $I$ is not really ordered unless $I$ is ordered. But nevertheless we know what it means that $\{B_i\}_{i\in I}$ is ordered "in the same way" even though neither is actually ordered. That is, we associate $A_i$ with $B_i$ for the same index $i$ (and not permuting the elements of $I$ on the way).

  • $\begingroup$ Now I think I know what I got wrong: $z$ does not, at once, go from a set $\{a,b\}$ but from the individual elements. That is actually obvious, I guess. $z_a$ is the same as $z(a)$? That clears it up. $a$ and $b$ are used to distinguish $X$ and $Y$. Then, $z(a)$, each time, refers to a different element in $X \cup Y$ or $X$. So, $Z$ is a set of functions (the families)? And it can be interpreted as a set of pairs, though the elements as such are not actually pairs? $\endgroup$ – Edgar Feb 23 '13 at 17:23

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