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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.

Turkhan Badalov
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1 Answers1

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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.

Noah Schweber
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  • 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 – Turkhan Badalov May 27 '18 at 18:42
  • @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! – Noah Schweber May 27 '18 at 18:51
  • 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? – Turkhan Badalov May 27 '18 at 18:59
  • 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 "$(18, 467, 3)$"). – Noah Schweber May 27 '18 at 19:02
  • Oh, as $X_i$ is a set by itself the family ${X_i}_{i \in {a, b, c}}$ should look like ${(a, {11}), (b, {21}), (c, {31})}$, assuming I only want to obtain a single ordered triple $(11, 21, 31)$?. And is it true, that I also need to construct the sets $X_i$ (for instance by axiom of specification or writing them on the paper to make sure about their existence) ? – Turkhan Badalov May 27 '18 at 19:08
  • @TurkhanBadalov No, that's still wrong. There are three things floating around here: (1) The family ${X_i}{i\in{a, b, c}}$. If we ignore details for a moment, we can think of this as essentially the three-element set ${X_a,X_b,X_c}$. (2) The Cartesian product $Q=\prod{i\in{a,b,c}}X_i$ (not quite in Halmos's notation). This is a set of three-element sets. $Q$ probably has many more than three elements; e.g. if each $X_i$ has two elements, $Q$ will have $2^3=8$ elements. (3) A single "ordered triple" in the Cartesian product $Q$ is a three-element set of the type described above. – Noah Schweber May 27 '18 at 19:12
  • As to the last sentence of your comment, yes, you need to have the sets $X_i$ "in hand" before you can talk about their Cartesian product. – Noah Schweber May 27 '18 at 19:13
  • Before reading further after ${X_a, X_b, X_c}$ I want to say that I can't agree with this. Becuase, as I understand, Halmos defines a family as a function which in its turn is a set of ordered pairs. That is why I write ${(a, {11}), (b, {21}), (c, {31})}$ and not ${ {11}, {21}, {31} }$. – Turkhan Badalov May 27 '18 at 19:16
  • @TurkhanBadalov Yes, that's why I said "if we ignore details for a moment." My point is just that the indexed family of $X_i$s is (generally) much "smaller" than their Cartesian product. If you prefer, I'm saying that ${(a, X_a), (b, X_b), (c, X_c)}$ has only three elements. The point is to distinguish it from the Cartesian product, and to distinguish both of those from a single ordered triple (= element of the Cartesian product). – Noah Schweber May 27 '18 at 19:17
  • So is it still wrong to write the family ${X_i}_{i\in{a, b, c}}$ as ${(a, {11}), (b, {21}), (c, {31})}$ where $X_a ={11}$, $X_b = {21}$, $X_c = {31}$? (2) I understood what you say, that if at least one of $X_i$ has more than one elements in it like $X_a = {11, 10}$ then in the Cartesian product there will be more than one three-element sets, being exact $2 * 1 * 1 = 2$, assuming $X_b$ and $X_c$ both have one element. – Turkhan Badalov May 27 '18 at 19:24
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    @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. – Noah Schweber May 27 '18 at 19:26