# Cartesian products as sets of all functions.

I can't fully understand the general definition of the Cartesian products of an arbitrary collection of sets defined by an index set. The general definition states that it is simply the set of all functions from the index set to the union of the sets in the collection. Now since functions are formally defined as ordered pairs, how can the set of all functions generate a set of n-tuples, given that the collection contains n sets?

Thanks.

You mean that functions are formally defined as sets of ordered pairs. Thus, if $I=\{0,1,2\}$ and $X_0=X_1=X_2=\{a,b,c\}$, one member of $\prod\limits_{i\in I}X_i$ is the function $\{\langle 0,a\rangle,\langle 1,b\rangle,\langle 2,a\rangle\}$. You're asking, I take it, how to square this with the idea that a member of $\prod\limits_{i\in I}X_i$ 'ought' to be an ordered triple, for instance $\langle a,b,a\rangle$.

The answer depends on how we choose to define ordered n-tuple. One way, for $n>2$, is to define it as a function whose domain is $\{0,1,\dots,n-1\}$. If we do that, $\{\langle 0,a\rangle,\langle 1,b\rangle,\langle 2,a\rangle\}$ is an ordered triple.

Another approach is to define ordered triples to be ordered pairs of the form $\langle \langle a_0,a_1\rangle,a_2\rangle$, ordered $4$-tuples to be ordered pairs of the form $\langle\langle\langle a_0,a_1\rangle,a_2\rangle,a_3\rangle$, and in general to define ordered $(n+1)$-tuples to be ordered pairs of the form $\langle\pi,a\rangle$, where $\pi$ is an ordered $n$-tuple. If you adopt this definition, then the members of my little product $\prod\limits_{i\in I}X_i$ aren't actually ordered triples. However, there is a natural bijection between them and the 'real' ordered triples, given by $$\{\langle 0,x\rangle,\langle 1,y\rangle,\langle 2,z\rangle\}\leftrightarrow\langle\langle x,y\rangle, z\rangle\;,$$ by means of which any statement about the one corresponds trivially to a corresponding statement about the other.

Should ordered $n$-tuples be defined in any other reasonable way, a similar situation will obtain.

The function definition may look odd at first, especially when you're dealing only with finite index sets, but it's the only really convenient way to deal with products over infinite index sets.

• Okay, that makes perfect sense now, the bijection part eliminated all my confusion, since it answers my previous confusion about sequences, defined as a function from natural numbers to real numbers: since sequences have a specific ordering, there is nothing that prevents someone writing the outputs of a function in an unordered manner, but if there is a bijection, than it doesn't matter. – firemind May 6 '12 at 6:25

The general definition is actually more precise than that : the functions are from the index set and go into the disjoint union alright, but if $f \in \prod_{i \in I} X_i$, then $f(i) \in X_i$.

Perhaps an example would make things more precise at this point.

Suppose we just consider the simple Cartesian product. You have two sets $A_1$ and $A_2$ and want to consider their Cartesian product $A_1 \times A_2$, also written as $\prod_{i \in \{1,2\}} A_i$. Then an element of this product can be a function $f : \{1,2\} \to A_1 \cup A_2$ for which $f(1) \in A_1$, $f(2) \in A_2$. Usually, for instance when $A_1 = A_2 = \mathbb R$, it is convention to write $f = (f(1),f(2))$ to represent the function as an element of the cartesian product. But formally, it represents a function from the set $\{1,2\}$ to the set $A_1 \cup A_2$, and the element $1$ maps in $A_1$ while the element $2$ maps in $A_2$.

Hope that helps,

• Thanks for the answers, so the idea is to order the outputs of the function, taking the elements of the index set as inputs. I guess my confusion arose from the fact that a function does not embody an intrinsic ordering. In fact, I experienced a similar confusion when sequences are defined as functions, since sequences have a specific ordering, there is nothing that prevents someone writing the outputs of a function in an unordered manner. – firemind May 6 '12 at 6:23
• Yes, you're getting closer to the point. =) – Patrick Da Silva May 6 '12 at 6:24
• So, f, or actually, rather the range of f, consists of just a single element of $A_1$x$A_2$, rather then the whole product of $A_1$ with $A_2$, is that correct? – ak87 Jun 9 '17 at 15:34
• @ak87 : you sound confused. I don't see the point you're trying to make ; of course $f$ is not equal to $A_1 \times A_2$. – Patrick Da Silva Jun 9 '17 at 15:38
• thanks for clearing that up. i'm reading a book on analysis and it certainly sounded like they meant that all of the product could be represented by just the function described above... I just double checked and the book actually puts an equal sign in between the set that contains the function f described above and the product of two sets X and Y – ak87 Jun 9 '17 at 15:39

This is a common confusion.

Note that we identify $\mathbb R^n$ with $n$-tuples of real numbers, while the set itself is actually functions from $\{0,\ldots,n-1\}$ to $\mathbb R$.

We define $\prod_I X_i$ to be all the functions $f$ from $I$ into $\bigcup_I X_i$ such that $f(i)\in X_i$, and therefore $f$ is just a collection of ordered pairs. If $I$ is "simple" enough we often prefer to think of the product as tuples of some sort, but this is just because the two things are isomorphic in a very natural way.

Another way to make the distinction formal is to define an ordered pair $\langle a,b\rangle$ as before, and define an $I$-tuple as a function from $I$ into some set. Now we define the product $\prod_I X_i$ as the set of $I$-tuples. In this way we again get "different" things when we take $X_1\times X_2$ and $\prod_{i=1}^2 X_i$. The former is a collection of ordered pairs while the latter is a set of $2$-tuples.