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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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