Infinite Cartesian product on a finite family of sets (example)

Question

everyone!

I am struggling to see how two definitions of Cartesian product (for finite and infinite cases) are equivalent practically. Imagine I have two sets, $X_{1}=\left\{ 3,4\right\} ,X_{2}=\left\{ 5,6\right\}.$ Their Cartesian product is the set of all ordered pairs : $\left\{ (3,5),(3,6),(4,5),(4,6)\right\}$ such that $x\in X_{1},y\in X_{2}.$

Now, by the definition of the infinite Cartesian product, it must be a set of maps $f:I\rightarrow\bigcup_{i\in I}X_{i}$ such that $f(i)\in X_{i}\,\forall i\in I.$ Here $I=\left\{ 1,2\right\}.$ How does this set look like? I know by definition, that the function is a subset of Cartesian product, so it must be sth like $f_{1}(1)=3, f_{2}(1)=4, f_{1}(2)=5,f_{2}(1)=6,$ so $\left\{ (1,3),(1,4),(2,5),(2,6)\right\}.$ Those two definitions does not produce the same result at all, where am I mistaken?

Answer 1

There are two problems here.

For brevity, let's denote an arbitrary function $f$ on the set $I = \{1, 2\}$ by $f = \left\langle f(1), f(2)\right\rangle.$

This is well-defined if, as in the question, we adopt the convention that a function is a set of ordered pairs.

By that convention, for any mathematical objects $a$ and $b,$ we have $\left\langle a, b\right\rangle = \{(1, a), (2, b)\}.$

In particular, for all $a \in X_1,$ $b \in X_2,$ $\left\langle a, b\right\rangle$ is the unique function $f \colon I \to \bigcup_{i \in I} X_i$ such that $f(1) = a$ and $f(2) = b$; and it is a subset of $I \times \bigcup_{i \in I}X_i.$

In this notation, the Cartesian product of $X_1$ and $X_2,$ according to the definition for arbitrary index sets, including infinite ones, is $$ \prod_{i \in I} X_i = \{\left\langle a, b\right\rangle : a \in X_1 \text{ and } b \in X_2\} = \{\left\langle 3, 5\right\rangle, \left\langle 3, 6\right\rangle, \left\langle 4, 5\right\rangle, \left\langle 4, 6\right\rangle\}. $$ Each set $\left\langle a, b\right\rangle,$ here, is a subset of $I \times \bigcup_{i \in I}X_i,$ therefore $\prod_{i \in I} X_i$ is a set of subsets of $I \times \bigcup_{i \in I}X_i.$

The set $\prod_{i \in I} X_i$ is not, as stated in the question, itself a subset of $I \times \bigcup_{i \in I}X_i$; that is the first problem.

The second problem is that the set $\prod_{i \in I} X_i,$ even when it is correctly computed, is not the same as the set $$ X_1 \times X_2 = \{(a, b) : a \in X_1 \text{ and } b \in X_2\} = \{(3, 5), (3, 6), (4, 5), (4, 6)\}. $$

However, the two sets $\prod_{i \in I} X_i$ and $X_1 \times X_2$ are, as the question puts it, "equivalent practically".

The reason for this is that functions of the form $\left\langle a, b\right\rangle$ "behave like" ordered pairs, in the sense that $$ \left\langle a, b\right\rangle = \left\langle a', b'\right\rangle \iff a = a' \text{ and } b = b' $$ for all mathematical objects $a, b, a', b'.$

When erecting a set-theoretical foundation for mathematics, it is vital to distinguish between $(a, b)$ and $\left\langle a, b\right\rangle$ - indeed, the definition of the latter depends on the former - but once the foundations have been put in place, one can "in practice" ignore the distinction.

(I'm uncomfortable with this approach myself, but it's the way things are done, and I know of no alternative.)

Answer 2

You are mis-computing the second definition.

Each map is supposed to match to an ordered pair of the Cartesian product, and is a function from a coordinate index into the pair to the value at that index. Therefore there are four maps, matched to the pairs as follows (functions are represented here as sets of ordered pairs):

(3,5) <---> { (1,3), (2,5) }
(3,6) <---> { (1,3), (2,6) }
(4,5) <---> { (1,4), (2,5) }
(4,6) <---> { (1,4), (2,6) }