Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(1) If $X$ and $Y$ are two sets, we define the Cartesian product $X \times Y$ as the set of ordered pairs $(x,y)$, such that $x \in X$ and $y \in Y$.

(2) On the other hand [Folland, Real Analysis, page 4], if $\{X_\alpha\}_{\alpha \in A}$ is infinite indexed family of sets, their Cartesian product $$ \prod_{\alpha \in A}X_\alpha $$ is defined as the set of maps $f: A \to \bigcup\limits_{\alpha \in A} X_\alpha$ such that $f(\alpha) \in X_\alpha$ for every $\alpha \in A$.

After saying this, Folland remarks:

it should be noted, and promptly forgotten, that when $A = \{1,2\}$, the previous definition of $X_1 \times X_2$ [that's (1) above] is set-theoretically different from the present definition of $\prod_1^2 X_j$ [that's (2) above]. Indeed, the latter concept depends on the mappings, which are defined in terms of the former one.

I am not grasping this remark. Specifically, here are my questions.

Question 1: How is (2) set-theoretically different from (1)? A simple illustrative example?

Question 2: If (1) is extended to infinite families, which definition would be stronger? A simple illustrative example?

Question 3: Why should this be "promptly forgotten"?

I'll probably have more questions depending on the type of answers I'll get to these.

Thanks!

share|improve this question
    
I wondered about this as well. To give some insight on your Question 2, note that it is not possible to extend (1) to uncountably infinite families. In this sense, (2) is more general, but they are essentially the same (see the bijection mentioned by William) and for most instances (I have been told) it will not hurt to think of the countably infinite case. YMMV. –  process91 Sep 14 '12 at 3:06
    
Thanks to everybody! All are such great responses, it's hard to choose an accepted answer. –  Rick Sep 14 '12 at 10:55
add comment

5 Answers 5

up vote 3 down vote accepted

Very briefly. Question 1 has been amply answered; the two definitions produce different sets, although there is a canonical bijection between them.

Question 2. There is no such thing as extending (1) to infinite families, it is a definition for pairs of sets only. Even if you want something for three sets, there are at least two ways to combine two Cartesian products, which give different (though isomorphic) results, neither of which involves ordered triples. But one can never get infinite products in any of these ways. One can interpret (2) as an alternative way to "generalize" this, in a manner that also caters for infinite products.

Question 3: We can forget the distinction between (1) and (2) in the case of a Cartesian product of two sets because of the canonical bijection between the sets produced by the two definitions, which means we can always and consistently translate back and forth between them when necessary. And we should forget the distinction because keeping track of which of the two is officially applied in which situation is a totally unproductive effort. The whole importance of giving these definitions is to have a precise model of a Cartesian product, so that its properties can be deduced from the axioms of set theory, but having more than one equivalent model does not add anything useful.

Absent question: if (2) can do all that (1) can (slightly differently but equivalently) why should we care about (1) in the first place? Because without (1) we would be in a chicken-and-egg situation when trying to formulate (2), not only because (2) produces sets of mappings, which require a Cartesian product (of two sets), but also because the notion of an indexed family of sets itself is defined in terms of mappings.

share|improve this answer
add comment
  1. An order pair and a function $f : \{1, 2\} \rightarrow X_1 \cup X_2$ are not the same thing. Depending on your definition, some people define the order pair $(a,b)$ to be the set $\{a, \{a,b\}\}$. A function is a subset of $X_1 \times X_2$ satisfing the usual property of functions. Hence set-theoretically they are different. "Indeed, the latter concept depends on the mappings, which are defined in terms of the former one." means that functions are usually defined as a subset of $X \times Y$, i.e. subset of the set of order pairs. Hence Folland is just remarking that the definition of functions uses the notion of order pairs.

  2. Folland only defines $(1)$ for finite cartesian product. So in the context of Folland's book, it doesn't make any sense to ask which definition is stronger for infinite families. According to folland, for infinite families there is only one notion of cartesian product, and that is the second one.

  3. By being "prompty forgotten", is that although order pairs and certain functions from $\{1, 2\} \rightarrow X_1 \cup X_2$ are not set theoretically the same, there is a very nice bijection $\Phi$ between the two concepts. $\Phi((x_1,x_2)) = f_{(x_1,x_2)}$ where $f_{(x_1, x_2)}$ defined by $f_{(x_1,x_2)}(1) = x_1$ and $f_{(x_1,x_2)}(2) = x_2$.

(Note that its inverse would then be : for any $f : \{0,1\} \rightarrow X_1 \cup X_2$ with the property that $f(i) \in X_i$, $f$ is map to $(f(1), f(2))$. )

share|improve this answer
add comment

Whey they're different

The usual set-theoretic construction of functions is as a relation between two sets $A$ and $B$, in which every $a \in A$ is related to exactly one $b \in B$ (but not necessarily a distinct one from any other $a' \in A$). To write it in formal symbols, $f: A \to B$ is logically equivalent to $$\Bigl(f \subseteq A \times B \Bigr) \mathbin\& \Bigl( \forall a \in A \,\exists b \in B : \bigl[ (a,b) \in f \bigr] \mathbin\& \bigl[\forall b' \in B: (a,b') \in f \implies b' = b \bigr] \Bigr) .$$ In particular, a function $f: \{ 1, 2 \} \to X \cup Y $ of the type described in your question is a two-element set $\{(1,x), (2,y)\}$ such that $x \in X$ and $y \in Y$. This is different from an element $(x,y) \in X \times Y$.

Why it doesn't matter that they're different

The reason why you should ignore the difference is because these are just two ways that you can reproduce the structure of an ordered pair in set theory. What matters is not how the ordered pairs (or any other particular structure) are built, but rather what you can do with them. For instance, the approach of category theory is to place the emphasis on the mappings of the projections $\pi_1$ and $\pi_2$ which give you the first and second entries of the tuples; and this reproduces everything you really care about for tuples, without fretting about "which set" the tuple is represented by.

Why I have a preference despite the fact that it doesn't matter that they're different

Nevertheless — I must confess a preference (purely aesthetic, mind you) for the definition in terms of functions, if you do choose to spend time contemplating definitions in terms of sets.

  • If you take the constructions very literally, $A \times B \times C \times D$ has to be interpreted as something like $A \times (B \times (C \times D))$, consisting of tuples $x = (a,(b,(c,d)))$, which is slightly ridiculous. I prefer to think of tuples as $x = (a,b,c,d) := \{(1,a), (2,b), (3,c), (4,d)\}$, which is usually what I really mean (because then $x_1 = a$, $x_2 = b$, etc. is no more than function evaluation).

  • Furthermore, an infinite product $A \times (B \times (C \times (\cdots))))$ would give rise to sets with infinitely decreasing chains of elementhood relations, which are ruled out by the usual axioms of set theory (specifically the Axiom of Foundation for ZFC or NBG). If you use functions $f: A \to \bigcup_{\alpha \in A} X_\alpha$ to define tuples, it really doesn't matter if $A$ happens to be an infinite set; the set of the tuples is still well-defined (although anyone who disbelieves in the Axiom of Choice may think that it could be empty even if none of the $X_\alpha$ are).

If you spend much time thinking about tuples in terms of sets, defining tuples as functions — and defining functions in terms of "arcs", a word which I choose almost-arbitrarily for the usual set $\{\{a\},\{a,b\}\}$ used to define ordered pairs — makes things a lot nicer.

share|improve this answer
    
"Furthermore, an infinite product $A\times(B\times(C\times(⋯))))$ would give rise to sets with infinitely decreasing chains, which are ruled out by the usual axioms of set theory". Are you referring to the axiom of foundation? What are these infinite chains? –  William Sep 14 '12 at 3:13
    
Infinite decreasing chains do exists in $ZFC$. For example $\omega^*$, or the chain in $\mathbb{Z}$. –  William Sep 14 '12 at 3:15
    
@William: $(a,b) = \{\{a\},\{a,b\}\}$; $(a,(b,c)) = \{\{a\},\{a,(b,c)\}\} = \{\{a\},\{a,\{\{b\},\{b,c\}\}\}$; $(a,(b,(c,d))) = \{\{a\},\{a,(b,c,d)\}\} = \{\{a\},\{a,\{\{b\},\{b,\{\{c\},\{c,d\}\}\}\}$, etc. The depth of the tuple scales with the number of terms in the product; infinite caretsian products lead to infinite descending chains. The two sets you give are infinite descending total orders, but what ZFC forbids is infinite descending chains for the elementhood relation, which would arise as I note above for infinite cartesian products. I've edited my response to clarify that. –  Niel de Beaudrap Sep 14 '12 at 3:17
1  
@William: the sequence of these sets has increasing depth. However, if you hypothetically took a limit element consisting of a tuple from an infinite cartesian product, that element would have an infinite descending chain: $(a,(b,(c,(d,(\cdots))) \ni \{a,(b,(c,(d,(\cdots)))\} \ni (b,(c,(d,(\cdots))) \ni \{b,(c,(d,(\cdots)))\} \ni (c,(d,(\cdots))) \ni \cdots$ –  Niel de Beaudrap Sep 14 '12 at 3:19
    
@William and Niel, you discussion is misguided, and Niel should not have mentioned $A\times(B\times(C\times(⋯))))$ in the first place. If one defines a binary operation (here Cartesian product of two sets) it can be repeated to give any $n$-ary operation, but not to an infinitary operation. Just because addition of two integers is defined does not mean we have an operation $\mathbf Z^{\mathbf N}\to\mathbf Z$ of infinitary addition. –  Marc van Leeuwen Sep 14 '12 at 6:33
show 2 more comments

Here's what I think.

Question 1: the first definition, the finite one, tells you a direct product is made of 'ordered pairs', that is, couples of elements in the sets. But the second definition, applied when $A= \{1,2 \}$, tells you the direct product is made of 'maps satisfying some conditions'. Clearly, a map and an ordered pair are not, from a set-theoretic point of view, the same thing, so here's the difference. Nonetheless, practically they are the same thing, just the pairs you obtain with your first definition are the images of all the possible maps you are considering with the second one.

Question 2: you cannot, in general, extend the first definition to infinite families, basically because your $A$ is any set and could lack a total ordering, so there is no meaningful generalization of 'ordered pair'. However, if your set $A$ actually has a total ordering, for example $A=\mathbb N$, you can give the definition of the direct product $\prod_{ i \in \mathbb N} X_i$ as the set whose elements are sequences $(x_i)$ such that $x_i \in X_i$ for all $i \in \mathbb N$. You can check yourself this definition is consistent with the 'maps' one.

Question 3: this is just my personal opinion, but it should be forgotten because when the set $A$ is finite, you clearly use the first definition, while if $A$ is infinite (and in particular with no total ordering) you use the second.

share|improve this answer
add comment

(1) defines the product as a set of ordered pairs, (2) defines it as a set of maps. Pairs aren't maps, so there's your set-theoretic difference.

I don't think the word "strength" applies to definitions. But the word "ordered" in "ordered pairs" may lead to difficulties in (1) for infinite families, since you may not have an ordering on your index set.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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