Cartesian products as sets of all functions.

Solution 1:

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.

Solution 2:

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.

Solution 3:

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,