Properties of Products

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Partial Products and Associativity

To discuss whether the product of sets is associative, we must first define partial products.

Definition 1 Let a family \((A_i)_{i\in I}\) and its product \(\prod_{i\in I} A_i\) be given. Then for a subset \(J\subseteq I\) of the index set, we call \(\prod_{j\in J} A_j\) a partial product.

Let a partial product \(\prod_{j\in J}A_j\) of \(\prod_{i\in I}A_i\) be given. Then for any \(F\in\prod_{i\in I}A_i\),

\[f\circ\id_J=\biggl(F\circ\Delta_J, J, \bigcup_{j\in J} A_j\biggr)\]

is a new function, and for each \(j\) it satisfies \((f\circ\id_J)(j)=f(j)\in A_j\). Thus \(F\circ\Delta_J\) is an element of \(\prod_{j\in J}A_j\).

By the preceding paragraph, \(F\mapsto F\circ\Delta_J\) defines a function from \(\prod_{i\in I}A_i\) to \(\prod_{j\in J}A_j\). We denote this by \(\pr_J\), borrowing the notation for projection functions. Then for \(K\subseteq J\subseteq I\), the composition of the \(J\)-th projection from \(\prod_{i\in I}A_i\) to the partial product \(\prod_{j\in J}A_j\) and the \(K\)-th projection from \(\prod_{j\in J}A_j\) to its partial product \(\prod_{k\in K}A_k\)

\[\prod_{i\in I}A_i\longrightarrow \prod_{j\in J}A_j\longrightarrow \prod_{k\in K}A_k\]

is simply the \(K\)-th projection \(\pr_K\) from \(\prod_{i\in I}A_i\) to its partial product \(\prod_{k\in K}A_k\). This follows from \(\Delta_K=\Delta_J\circ\Delta_K\).

Proposition 2 Consider a family \((A_i)_{i\in I}\) whose every component is non-empty, and let \(J\subseteq I\). If \(g:J\rightarrow\bigcup_{i\in I} A_i\) satisfies \(g(j)\in A_j\), then there exists an extension \(f:I\rightarrow\bigcup_{i\in I} A_i\) of \(g\) such that \(f(i)\in A_i\).

Proof

Let \(g=(G,J,\bigcup A_i)\). For each \(i\in I\setminus J\), since \(A_i\) is non-empty we can choose an element \(x_i\in A_i\). Now define

\[F=G\cup\biggl(\bigcup_{i\in I\setminus J}\{(i, x_i)\}\biggr)\]

and set \(f=(F,I,\bigcup A_i)\). This yields the desired result.

Proposition 3 Let a family \((A_i)_{i\in I}\) with non-empty index set \(I\) be given. If \((J_k)_{k\in K}\) is a partition of \(I\), then the function \(f\mapsto (\pr_{J_k}(f))_{k\in K}\) from \(\prod_{i\in I}A_i\) to \(\prod_{k\in K}\left(\prod_{j\in J_k}A_j\right)\) is also a bijection.

Proof

Since \((J_k)_{k\in K}\) is a partition, the functions \(f_k:J_k\rightarrow \bigcup_{i\in I} A_i\) form a family of functions with pairwise disjoint domains; hence by §Sum of Sets, ⁋Proposition 2 we obtain a bijection.

The proof above is concise, but the following argument using the universal property is equally elegant.

Proof of Proposition 3

For notational cleanliness we adopt the following uniform convention:

• The \(k\)-th projection of the product \(\prod_{k\in K}\left(\prod_{j\in J_k}A_j\right)\) with respect to the index set \(K\)

  \[\prod_{k\in K}\left(\prod_{j\in J_k}A_j\right)\rightarrow\prod_{j\in J_k}A_j\]

  is denoted by \(\pr_k\),

• The \(j\)-th projection of the product \(\prod_{j\in J_k}A_j\) with respect to the index set \(J_k\)

  \[\prod_{j\in J_k}A_j\rightarrow A_j\]

  is also denoted by \(\pr_j\),

• The \(i\)-th projection of the product \(\prod_{i\in I}A_i\) with respect to the index set \(I\)

  \[\prod_{i\in I}A_i\rightarrow A_i\]

  is also denoted by \(\pr_i\)

While this may cause some confusion when read in text, in diagrams the source and target are both explicit, so there is no ambiguity.

Since \((J_k)_{k\in K}\) is a partition of \(I\), for each \(i\in I\) there exists a unique \(k\in K\) such that \(i\in J_k\). We now define the function \(\pr_{ik}\) as the composition

\[\pr_{ik}:\prod_{k\in K}\left(\prod_{j\in J_k}A_j\right)\overset{\pr_k}{\longrightarrow}\prod_{j\in J_k}A_j\overset{\pr_i}{\longrightarrow}A_i\]

Then by the universal property of the product \(\prod_{i\in I}A_i\), there exists a map \(\phi:\prod_{k\in K}\left(\prod_{j\in J_k}A_j\right)\rightarrow\prod_{i\in I}A_i\) making the following diagram commute:

Similarly, by the universal property of the product \(\prod_{k\in K}\left(\prod_{j\in J_k}A_j\right)\) with respect to the index set \(K\), there exists a map \(\psi:\prod_{i\in I}A_i\rightarrow\prod_{k\in K}\left(\prod_{j\in J_k}A_j\right)\) making the following diagram commute:

Then \(\phi\circ\psi\) and \(\psi\circ\phi\) are both identity functions, and thus they furnish the desired bijection.

For instance, let us verify that \(\phi\circ\psi\) is the identity on \(\prod_{i\in I}A_i\). It suffices to show that for every \(i\in I\) the following diagram commutes:

The universal property of the product asserts that there is a unique function \(\prod_{i\in I}A_i\rightarrow \prod_{i\in I}A_i\) making the above diagram commute. Clearly the identity on \(\prod_{i\in I}A_i\) also makes the diagram commute, so by uniqueness it must equal \(\phi\circ\psi\).

Now from

\[{\pr_i}\circ(\phi\circ\psi)=({\pr_i}\circ\phi)\circ\psi={\pr_{ik}}\circ\psi={\pr_i}\circ({\pr_k}\circ\psi)={\pr_j}\circ{\pr_{J_k}}=\pr_j\]

we obtain the desired conclusion. (The last equality regards \(\pr_j\) as the projection onto \(\{j\}\subseteq I\).) Although this equation appears complicated, it is merely the formulaic transcription of the statement that the following diagram commutes:

Let \((A_i)_{i\in I}\) and \((B_i)_{i\in I}\) be families with the same index set, and let a family of functions \((g_i:A_i\rightarrow B_i)_{i\in I}\) be given. Define \(u_f:I\rightarrow\bigcup_{i\in I}B_i\) by \(i\mapsto g_i(f(i))\); then \(u_f(i)\in B_i\), and therefore \(u_f\in\prod_{i\in I}B_i\).

Definition 4 The function \(f\mapsto u_f\) defined above is called the product of the \((g_i)\) and is denoted by \(\prod_{i\in I}g_i\).

Proposition 5 Let \((A_i)_{i\in I}\), \((B_i)_{i\in I}\), and \((C_i)_{i\in I}\) be three families, and let \((f_i)_{i\in I}\) and \((g_i)_{i\in I}\) be families of functions from \(A_i\) to \(B_i\) and from \(B_i\) to \(C_i\), respectively. Then

\[\prod_{i\in I} (g_i\circ f_i)=\left(\prod_{i\in I} g_i\right)\circ\left(\prod_{i\in I}f_i\right)\]

holds.

Proof

There is nothing to explain beyond the following two commutative diagrams:

and

Since it is obvious that the product of the \(\id_{A_i}\) is \(\id_{\prod A_i}\), the preceding proposition makes it clear that the product of injective functions is injective, and the product of surjective functions is surjective.

Distributivity Between Operations

When two or more operations are defined, whether distributivity holds between them is an important question.

Proposition 6 Let \(((A_{k,i})_{i\in J_k})_{k\in K}\) be a family of families of sets. Suppose further that \(K\neq\emptyset\) and \(J_k\neq\emptyset\) for all \(k\in K\). Then for \(I=\prod_{k\in K} J_k\neq\emptyset\),

\[\bigcup_{k\in K}\left(\bigcap_{i\in J_k}A_{k,i}\right)=\bigcap_{f\in I}\left(\bigcup_{k\in K}A_{k,f(k)}\right),\quad\bigcap_{k\in K}\left(\bigcup_{i\in J}A_{k,i}\right)=\bigcup_{f\in I}\left(\bigcap_{k\in K}A_{k,f(k)}\right)\]

hold.

Proof

First, let \(x\in \bigcup_{k\in K}\left(\bigcap_{i\in J_k}A_{k,i}\right)\). We must show that \(x\in \bigcap_{f\in I}\left(\bigcup_{k\in K}A_{k,f(k)}\right)\), i.e., that \(x\in \bigcup_{k\in K}A_{k,f(k)}\) for every \(f\in I\). Since \(x\in \bigcap_{i\in J_k}A_{k,i}\) for some \(k\in K\), we have \(x\in A_{k,f(k)}\). Hence \(x\in \bigcup_{k\in K}A_{k,f(k)}\) for all \(f\), and the inclusion follows.

To establish the reverse inclusion we argue by contrapositive. Suppose \(x\not\in \bigcup_{k\in K}\left(\bigcap_{i\in J_k}A_{k,i}\right)\). Then for every \(k\in K\) we have \(x\not\in \bigcap_{i\in J_k}A_{k,i}\). Thus there exists some \(i\) such that \(x\not\in A_{k,i}\) for every \(k\). Choosing \(f\in I\) so that \(f(k)\) is such an \(i\), we obtain \(x\not\in\bigcup_{k\in K}A_{k,f(k)}\), so \(x\) does not belong to the right-hand side. The second equality is proved similarly.

Distributivity also holds between product and union, and between product and intersection, as follows; the proof is almost identical to the above, so we omit it.

Proposition 7 Let \(((A_{k,i})_{i\in J_k})_{k\in K}\) be a family of families of sets, and define \(I\) as in the previous proposition. Then

\[\prod_{k\in K}\left(\bigcup_{i\in J_k}A_{k,i}\right)=\bigcup_{f\in I}\left(\prod_{k\in K}A_{k,f(k)}\right),\quad\prod_{k\in K}\left(\bigcap_{i\in J}A_{k,i}\right)=\bigcap_{f\in I}\left(\prod_{k\in K}A_{k,f(k)}\right)\]

hold.

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References

[Bou] N. Bourbaki, Theory of Sets. Elements of mathematics. Springer Berlin-Heidelberg, 2013.

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