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Section 5.11 Products of sets

An indexed set is really just another name for a map

\(U\colon A \to \Xi\text{;}\) we typically abuse notation and write \(U_a\) instead of \(U(a)\text{,}\) and we write \((U_a)_{a\in A}\) for the map \(U\text{.}\) The union of the indexed set \((U_a)_{a\in A}\) will mean the union of the set \(\{U_a : a\in A\}\text{:}\)

\begin{equation*} \bigcup_{a\in A}U_a\coloneq\bigcup\{U_a : a\in A\}\text{.} \end{equation*}

It may seem a bit silly to belabour this point, but the purpose will, we hope, become clear.

For any indexed set \((U_a)_{a\in A}\text{,}\) the product is the set

\begin{equation*} \prod_{a\in A} U_a \coloneq \left\{x \in \Map\left(A,\bigcup_{a\in A}U_a\right) : (\forall a\in A)\ x(a)\in U_a\right\}\text{.} \end{equation*}

An element of \(\prod_{a \in A} U_a\) is thus a map \(x \colon A \to \bigcup_{a \in A} U_a\) such that for every \(a \in A\text{,}\) one has \(x(a) \in U_a\text{.}\) One usually writes \(x_a\) instead of \(x(a)\text{,}\) and one often writes \(x = (x_a)_{a\in A}\text{.}\)

When \(A = \{1,2\}\text{,}\) an indexed set consists of two sets \(U_1\) and \(U_2\text{.}\) The product \(\prod_{a \in \{1,2\}} U_a\) thus consists of pairs \((x_1, x_2)\) with \(x_1 \in U_1\) and \(x_2 \in U_2\text{.}\) The assignment \((x_1,x_2) \mapsto \angs{x_1,x_2}\) is a bijection \(\prod_{a \in \{1,2\}} U_a \to U_1 \times U_2\text{.}\) Most mathematicians are happy to pretend as if there is no difference between these sets; indeed, there is no interesting difference!

More generally, we think of the product \(\prod_{a\in A} U_a\) as the set of ordered “\(A\)-tuples”.

For every \(b\in A\text{,}\) there is an attached map

\begin{equation*} \pi_b \colon \prod_{a\in A}U_a \to U_b \end{equation*}

given by the assignment \(x \mapsto x_b\text{,}\) called the projection onto the \(b\)-th factor.

For every set \(S\text{,}\) every indexed set \(\{U_a\}_{a\in A}\text{,}\) and every indexed set \(\left\{f_a\colon\fromto{S}{U_a}\right\}\) of maps, there exists a unique map

\begin{equation*} f\colon S \to \prod_{a\in A}U_a \end{equation*}

such that \(\pi_a\circ f=f_a\text{.}\) Indeed, the map \(f\) is given by the assignment \(s \mapsto (f_a(x))_{a\in A}\text{.}\)