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Here is a statement that I can't understand:

Ordered triples, ordered quadruples, etc., may be defined as families whose index sets are unordered triples, quadruples, etc.

For now, let's stick with the ordered triples. First of all I can't understand whether the index set is a set of unordered triples like $\{ \{a, b, c\}, \{d, e, f\}\}$ or it is itself an unordered triple?

Also, I just can't imagine how having any of such sets as the index set can help me to build a set of ordered triples. Maybe some example will make things clear. I will be very grateful if you help. Thanks in advance.

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  • $\begingroup$ Where did you find this statement? Personally, I also find it difficult to parse ... $\endgroup$ – Noah Schweber May 27 '18 at 18:03
  • $\begingroup$ @NoahSchweber, Naive Set Theory of Halmos $\endgroup$ – Turkhan Badalov May 27 '18 at 18:03
  • $\begingroup$ What page? It's an entire book ... $\endgroup$ – Noah Schweber May 27 '18 at 18:03
  • $\begingroup$ @NoahSchweber, oh, sorry. Page 36 in "Families" section $\endgroup$ – Turkhan Badalov May 27 '18 at 18:05
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Yes, this is a bit unclear.

Halmos' goal is to define the notion of an arbitrary Cartesian product. This should generalize the usual Cartesian product of two sets, $A\times B$, but should "work for any number of sets." Before diving into his description, let me point out that this really is nontrivial: how should we think of the Cartesian product of "$\mathbb{R}$-many" sets?

Halmos is about to tell us, essentially, that the Cartesian product of an indexed family $\{X_i\}_{i\in I}$ of sets is just the set of all indexed families $\{x_i\}_{i\in I}$ of objects with $x_i\in X_i$. Maybe more clearly, an element of the Cartesian product is a function with domain $I$; it sends $i\in I$ to the "$i$th coordinate," which must be in $X_i$.


To motivate this, Halmos has us go back to the idea of a Cartesian product. We can rethink Cartesian products as follows:

  • Fix any two distinct sets, $a$ and $b$. We'll think of the first as "LEFT" and the second as "RIGHT."

  • Now an ordered pair has two "coordinates," a left coordinate and a right coordinate. We're going to match these up with $a$ and $b$ above: if I have a function $z$ with domain $\{a, b\}$ such that $z(a)=x\in X$ and $z(b)=y\in Y$ (here I write "$z(i)$" for Halmos's "$z_i$"), we want to think of the object $z$ as being the ordered pair $(x, y)$. Informally, $z$ says

$$\mbox{My left coordinate is $x$, and my right coordinate is $y$.}$$

Put another way:

We can think of the ordered pair $(x, y)$ as the function with domain $\{a, b\}$ (which is an unordered pair) mapping $a$ to $x$ and $b$ to $y$.

This is easiest to think about if $a=0$ and $b=1$, or something similar, but Halmos's point is that all we need is that the index set has two distinct elements. Now here's the key linguistic step Halmos makes which I think is confusing at first:

An ordered pair is an indexed set! And the indexing set is $\{a, b\}$, which is an unordered pair.

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  • $\begingroup$ are you sure you wanted to say sets in the end? "... that the Cartesian product of an indexed family $\{X_i\}_{i \in I}$ of sets is just the set of all indexed families $\{x_i\}_{i \in I}$ of sets with $x_i \in X_i$". Because as I understand, if we say the family $\{x_i\}$ of sets, its range consists of sets implying $x_i$ is a set which is, as I understand, is not said $\endgroup$ – Turkhan Badalov May 27 '18 at 18:42
  • $\begingroup$ @TurkhanBadalov My ZFC bias is showing. In formal ZFC set theory (and many other formal set theories), every object is in fact a set and everything is "built out of" the emptyset in a sense. But yes, in the context of Halmos's book that's inappropriate. Fixed! $\endgroup$ – Noah Schweber May 27 '18 at 18:51
  • $\begingroup$ Thanks for the great answer! So, to make an ordered triple, let's say, $(11, 21, 31)$ I need some unordered triple and let it be $\{a, b, c\}$, true? Then I have to have (actually I doubt this statement about having the following set, because I need to construct it additionaly, correct me if I am wrong) the family $\{X_i\}_{i \in \{a, b, c\}}$ and let it be $\{ (a,11), (b, 21), (c, 31)\}$. Then the Cartesian product of the family will be the set of all families $\{x_i\}$: $\{ \{ (a, 11), (b, 21), (c, 31)\} \}$, the element of which I interpret as the ordered triple we expected to make? $\endgroup$ – Turkhan Badalov May 27 '18 at 18:59
  • $\begingroup$ Not quite. The set $\{(a, 11), (b, 21), (c, 31)\}$ is a single ordered triple. The Cartesian product of three sets $X_a, X_b, X_c$ would be the set of all sets of the form $\{(a, x_a), (b,x_b), (c, x_c)\}$ with $x_a\in X_a, x_b\in X_b, x_c\in X_c$. So e.g. if $X_a=X_b=X_c=\mathbb{N}$, the set $\{(a, 11), (b, 21), (c, 31)\}$ would be an element of the Cartesian product, but the Cartesian product would also include other things like $\{(a, 18), (b, 467), (c, 3)\}$ (which we would think of as "$(18, 467, 3)$"). $\endgroup$ – Noah Schweber May 27 '18 at 19:02
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    $\begingroup$ @TurkhanBadalov I thought you were using my example where $X_a=X_b=X_c=\mathbb{N}$. If the $X_i$s are as you describe, then that's right. $\endgroup$ – Noah Schweber May 27 '18 at 19:26

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