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Math 344-1,2: Kursobjekt

Aaron Greicius

Section 1.7 Arbitrary products

We now endeavor to give a topological structure to an arbitrary product \(\prod_{i\in I}X_i\) of topological spaces. (See 0.1.15 and 0.1.16 for the definitions of arbitrary tuples and Cartesian products of sets.) As we will see, we can extend the finite product topology to this more general setting in two very different ways.

Topological specimen 11. Product space (arbitrary).

Let \(\{X_i\}_{i\in I}\) be a collection of topological spaces. We define two distinct topologies on the the product \(X=\prod_{i\in I} X_i\text{.}\)
  1. Box topology.
    The box topology on \(X\) is the topology \(\mathcal{T}_{\operatorname{box}}\) generated by the basis
    \begin{equation*} \mathcal{B}_{\operatorname{box}}=\left\{\prod_{i\in I}U_i\colon U_i \text{ open in } X_i\right\}\text{.} \end{equation*}
  2. Product topology.
    The product topology on \(X\) is the topology \(\mathcal{T}_{\operatorname{prod}}\) generated by the basis
    \begin{equation*} \mathcal{B}_{\operatorname{prod}}=\left\{\prod_{i\in I}U_i\colon U_i \text{ open in } X_i, U_i=X_i \text{ for all but finitely many } i\in I\right\}\text{.} \end{equation*}
We have \(\mathcal{T}_{\operatorname{prod}}\subseteq \mathcal{T}_{\operatorname{box}}\text{;}\) when the index set \(I\) is finite, we have \(\mathcal{T}_{\operatorname{prod}}= \mathcal{T}_{\operatorname{box}}\text{.}\)

Proof.

To see that both collections are indeed bases, observe that for any \(x=(x_i)_{i\in I}\) we have \(x=(x_i)_{i\in I}\in \prod_{i\in I}X_i\text{,}\) which is an element of \(\mathcal{B}\) in both cases; and that in both cases if we have \(U=\prod_{i\in I}U_i\text{,}\) \(V=\prod_{i\in I}V_i\in \mathcal{B}\text{,}\) then
\begin{equation*} \left(\prod_{i\in I}U_i \right)\times \left(\prod_{i\in I}V_i \right) =\prod_{i\in I}U_i\cap V_i\text{,} \end{equation*}
is also an element of \(\mathcal{B}\text{.}\)

Remark 1.7.1. Product coarser than box.

It is clear from the definition that we always have \(\mathcal{B}_{\operatorname{prod}}\subseteq \mathcal{B}_{\operatorname{box}}\text{,}\) and hence the product topology is coarser than (contained in) the box topology.

Notation 1.7.2. \(A^\omega\).

Following Munkres, given a set \(A\) and the special index set \(I=\Z_+=\{1,2,3,\dots\}\text{,}\) we write \(A^\omega\) for \(A^{\Z_+}\) as a shorthand.

Example 1.7.3. Infinite binary sequences.

Let
\begin{equation*} X=\{0,1\}^{\omega}=\prod_{i=1}^\infty \{0,1\}=\{(a_i)_{i=1}^n\colon a_i\in \{0,1\}\}\text{.} \end{equation*}
  1. Compare the box and product topologies on \(X\text{,}\) where \(\{0,1\}\) is given the discrete topology.
  2. Compare the box and product topologies on \(X\text{,}\) where \(\{0,1\}\) is given the trivial topology.
Solution.
As observed in Topological specimen 11, the product topology is always contained in the box topology. Let’s see whether the inclusion is strict in either case.
  1. Assume \(\{0,1\}\) has the discrete topology. Since \(\{1\}\) is open in \(\{0,1\}\text{,}\) the set
    \begin{equation*} U=\{1\}\times \{1\}\times \cdots =\{(1,1,\dots )\} \end{equation*}
    is open in the box topology. To see that it is not open in the product topology, take \(x=(1,1,1,\dots, )\) and let \(V\in \mathcal{B}_{\operatorname{prod}}\) contains \(x\text{.}\) By definition, we have
    \begin{equation*} V=\prod_{i=1}^n V_i=V_1\times V_2\times \cdots \times V_N\times \{0,1\}\times \{0,1\}\times \cdots \end{equation*}
    for some \(N\text{,}\) where \(V_{i}\) is an open subset of \(\{0,1\}\text{.}\) Since \(x\in V\text{,}\) we have \(\{1\}\subseteq V_i\ne\emptyset\text{.}\) Since \(V_i=\{0,1\}\) for all \(i\geq N\text{,}\) the element \((1,1,\dots, 1,0,0,\dots)\in V\text{,}\) from which it follows that \(V\not\subseteq U\text{.}\) We conclude that \(U\) is not a union of basis elements of the product topology, and hence is not open in the product topology.
    It is easy to see that \(\{0,1\}^\Z\) is discrete in the box topology. The argument above shows that \(\{0,1\}^\Z\) is not discrete in the product topology, since the singleton \(\{(1,1,\dots, 1)\}\) is not open.
  2. Now equip \(\{0,1\}\) with the trivial topology. Since the only open subsets of \(\{0,1\}\) are \(\emptyset\) and \(\{0,1\}\) it follows that a basis element \(B=\prod_{i=1}^\infty U_i\) in either topology is either empty (if \(U_i=\emptyset\) for any \(i\in I\)) or equal to \(X\) (if \(U_i=\{0,1\}\) for all \(i\in I\)). Since the bases of the two topologies are equal, the topologies are equal. Furthermore, since the only basis elements are \(\emptyset\) and \(X\text{,}\) the topology on \(X\) is just the trivial one: i.e., we have \(\mathcal{T}_{\operatorname{prod}}=\mathcal{T}_{\operatorname{box}}=\mathcal{T}_{\operatorname{trivial}}\text{.}\)

Proof.

  1. We prove each basis axiom separately (for both the box and product basis).
    Axiom 1. Given \(x=(x_i)_{i\in I}\in X\text{,}\) we can find basis elements \(B_i\in \mathcal{B}_{i}\) containing \(x_i\text{.}\) We have
    \begin{align*} x \amp\in B=\prod_{i\in I}B_i\in \mathcal{B}_{\operatorname{box}} \\ x \amp\in B=\prod_{i\in I}X_i \in \mathcal{B}_{\operatorname{prod}} \text{.} \end{align*}
    Axiom 2. This axiom follows readily from the set theoretic fact that
    \begin{equation*} \prod_{i\in I}U_i \cap \prod_{i\in I} V_i=\prod_{i\in I}U_i\cap V_i\text{.} \end{equation*}
    I’ll leave the details to you.
  2. It helps to formulate this result in plain English: we wish to show that the subspace topology of the (box) product topology is equal to the (box) product topology of the subspace topologies.
    Fix a choice of topology \(\mathcal{T}\in \{\mathcal{T}_{\operatorname{box}}, \mathcal{T}_{\operatorname{prod}}\}\text{,}\) let \(\mathcal{B}\in \{\mathcal{B}_{\operatorname{box}}, \mathcal{B}_{\operatorname{prod}}\}\) be the corresponding basis as defined in Topological specimen 11, and let \(\mathcal{T}_A\) denote the corresponding subspace topology on \(A\text{.}\) Recall (1.6.1) that \(\mathcal{B}_A=\{B\cap A\colon B\in \mathcal{B}\}\) is a basis for the subspace topology \(\mathcal{T}_A\text{.}\) Thus
    \begin{align} \amp \tag{1.7.1} \end{align}
    \begin{align} \mathcal{B}_A \amp = \{B\cap A\colon B=\prod_{i\in I}U_i\in \mathcal{B}\} \tag{1.7.2}\\ \amp = \left\{ \prod_{i\in I}U_i\cap A_i\colon \prod_{i\in I}U_i\in \mathcal{B}\right\}\text{,}\tag{1.7.3} \end{align}
    where the last equality follows from the set theory identity
    \begin{equation*} \prod_{i\in I}U_i\cap \prod_{i\in I}A_i=\prod_{i\in I}U_i\cap A_i\text{.} \end{equation*}
    Now observe that when \(\mathcal{B}=\mathcal{B}_{\operatorname{box}}\) the set in (1.7.3) is precisely the basis for the box product topology on the subspaces \((A_i, \mathcal{T}_{i, A_i})\) as described in Topological specimen 11; and similarly, when \(\mathcal{B}=\mathcal{B}_{\operatorname{prod}}\) the set in (1.7.3) is the basis for the product topology on the subspaces \((A_i, \mathcal{T}_{i, A_i})\text{.}\)
    Going back to our plan English formulation, we have shown that the subspace topology of the (box) product topology and the (box) product topology of the subspace topologies share a common basis ((1.7.3)). It follows that the topologies themselves are equal.
  3. Assume \(X_i\) is Hausdorff for all \(i\in I\text{.}\) If \(x=(x_i)_{i\in I}\) and \(y=(y_i)_{i\in I}\) are distinct, then we have \(x_{i_0}\ne y_{i_0}\) for some \(i_0\in I\text{.}\) Since \(X_{i_0}\) is Hausdorff, there are disjoint open neighborhoods \(U_{i_0}, V_{i_0}\) of \(x_{i_0}, y_{i_0}\) in \(X_{i_0}\text{.}\) The sets \(U=U_{i_0}\times \prod_{i\ne i_0}X_i, V=V_{i_0}\times \prod_{i\ne i_0}X_i\text{,}\) are disjoint open basis elements for both topologies containing \(x, y\text{,}\) respectively. Thus \(X\) is Hausdorff. The proof for the \(T_1\) statement is similar.
  4. We prove the equality by showing
    \begin{equation*} x\notin \prod_{i\in I}\overline{A_i} \iff x\notin \overline{\prod_{i\in I}A_i}\text{.} \end{equation*}
    \((\implies)\text{.}\) If \(x=(x_i)_{i\in I}\notin \prod_{i\in I}\overline{A_i}\text{,}\) there is an index \(i_0\in I\) such that \(x_{i_0}\notin \overline{A_{i_0}}\text{,}\) and thus an open set \(U_{i_0}\ni x_{i_0}\) such that \(U_{i_0}\cap A_{i_0}=\emptyset\text{.}\) The set \(U_{i_0}\times \prod_{i\ne i_0}X_i\) is open in both topologies (box and product), contains \(x=(x_i)_{i\in I}\text{,}\) and is disjoint with \(\prod_{i\in I}A_i\text{.}\) We conclude that \(x\notin \overline{\prod_{i\in I}A_i}\) in both topologies.
    \((\impliedby)\text{.}\) If \(x=(x_i)_{i\in I}\notin \overline{\prod_{i\in I}A_i}\) there is an open basis element \(B=\prod_{i\in I}U_i\) in the topology (box or product) containing \(x\) that is disjoint with \(\prod_{i\in I}A_i\text{.}\) It follows that we must have \(U_{i_0}\cap A_{i_0}=\emptyset\) for some \(i_0\in I\) and open \(U_{i_0}\subseteq X_{i_0}\text{,}\) and thus that \(x_{i_0}\notin \overline{A_{i_0}}\text{.}\) We conclude that \(x\notin \prod_{i\in I}\overline{A_i}\text{.}\)
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