General Cartesian Product
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1.8. General Cartesian Product#
In this section, we extend the definition of Cartesian product to an arbitrary number of sets.
(Cartesian product)
Let \(\{ A_i \}_{i \in I}\) be a family of sets. Then the Cartesian product \(\prod_{i \in I} A_i\) or \(\prod A_i\) is defined to be the set consisting of all functions \(f : I \to \cup_{i \in I}A_i\) such that \(x_i = f(i) \in A_i\) for each \(i \in I\).
In other words, the function \(f\) chooses an element \(x_i\) from the set \(A_i\) for each index \(i \in I\).
The general definition of the Cartesian product allows the index set to be finite, countably infinite as well as uncountably infinite.
Note that we didn’t require \(A_i\) to be non-empty. This is discussed below.
(Choice function)
A member function \(f\) of the Cartesian product \(\prod A_i\) is called a choice function and often denoted by \((x_i)_{i \in I}\) or simply by \((x_i)\).
For a family \(\{A_i\}_{i \in I}\), if any of the \(A_i\) is empty, then the Cartesian product \(\prod A_i\) is empty.
This follows from the definition of the Cartesian product as a choice function \(f\) must choose an element from each \(A_i\). If an \(A_i\) is empty, a choice function cannot choose any element from it, hence the choice function cannot exist.
If the family of sets \(\{A_i\}_{i \in I}\) satisfies \(A_i = A \Forall i \in I\), then \(\prod_{i \in I} A_i\) is written as \(A^I\).
i.e. \(A^I\) is the set of all functions from \(I\) to \(A\).
1.8.1. Examples#
(Binary functions on the real line)
Let \(A = \{0, 1\}\). \(A^{\RR}\) is a set of all functions on \(\RR\) which can take only one of the two values \(0\) or \(1\).
(Binary sequences)
Let \(A = \{0, 1\}\). \(A^{\Nat}\) is a set of all sequences of \(0\)s and \(1\)s.
(Real sequences)
\(\RR^{\Nat}\) is a set of all real sequences. It is also denoted as \(\RR^{\infty}\).
(Real valued functions on the real line )
\(\RR^\RR\) is a set of all functions from \(\RR\) to \(\RR\).
1.8.2. Axiom of choice#
If a Cartesian product is non-empty, then each \(A_i\) must be non-empty. We can therefore ask: If each \(A_i\) is non-empty, is then the Cartesian product \(\prod A_i\) nonempty? An affirmative answer cannot be proven within the usual axioms of set theory. This requires us to introduce the axiom of choice.
(Axiom of choice)
If \(\{A_i\}_{i \in I}\) is a nonempty family of sets such that \(A_i\) is nonempty for each \(i \in I\), then the Cartesian product \(\prod A_i\) is nonempty.
This means that if every member of a family of sets is non empty, then it is possible to pick one element from each of the members.
Another way to state the axiom of choice is:
(Axiom of choice (disjoint sets formulation))
If \(\{A_i\}_{i \in I}\) is a nonempty family of pairwise disjoint sets such that \(A_i \neq \EmptySet \) for each \(i \in I\), then there exists a set \(E \subseteq \cup_{i \in I} A_i\) such that \(E \cap A_i\) consists of precisely one element for each \(i \in I\).