Convergence in product spaces

Section 1.8 Convergence in product spaces

The relative virtues of the product and box topologies begin to come to light as we look at issues of convergence and metrics in product spaces. In particular, we see in the context of the spaces \(\R^\omega\) and \(\R^\R\) that different topologies on these product sets give rise to different and valuable notions of function convergence for real-valued functions.

Definition 1.8.1. Pointwise convergence.

Let \(\{X_i\}_{i\in I}\) be a collection of topological spaces, let \(X=\prod_{i\in I}X_i\text{,}\) and let \((f_n)_{n=1}^\infty\) be a sequence of elements of \(X\text{:}\) i.e., using tuple notation, for each \(n\in \Z_+\) we have

\begin{equation*} f_n=(x_{n,i})_{i\in I}\text{,} \end{equation*}

where \(x_{n,i}=f_n(i)\in X_i\text{.}\) The sequence \((f_n)_{n=1}^\infty\) converges pointwise to an element \(f=(x_i)\in X\) if for all \(i\in I\) we have \(x_{n,i}\rightarrow x_i\text{:}\) i.e., \(f_n(i)\rightarrow f(i)\text{.}\)

Remark 1.8.2. Pointwise convergence.

Observe that the definition of pointwise convergence makes no reference to any topology on \(X=\prod_{i\in I}X_i\text{.}\) Below we investigate how it relates to convergence in the box and product topologies.

Example 1.8.3. Sequences in \(\R^\omega\) and \(\R^\R\).

Our understanding of tuples in product spaces as functions pays real dividends when investigating convergence of sequences in these spaces.

For example, if a tuple \((x_i)_{i=1}^\infty\) in \(\R^\omega\) is just a function \(f=\Z_+\rightarrow \R\text{,}\) the graph of which in the real plane is easily visualized: namely, a discrete collection of plotted points \((n, f(n))\) for \(n\in \{1,2,3,\dots\}\text{.}\)

Similarly, a tuple \((x_\alpha)_{\alpha\in \R}\) in \(\R^\R\) is just a function \(f\colon \R\rightarrow\R\text{,}\) which is easily visualized via a graph.

In both cases, sequences of tuples are easily visualized as sequences of functions \((f_n)_{n=1}^\infty\) (via their graphs). Furthermore, for the sequence to converge pointwise pointwise at a given index \(i\) (\(n\in \Z_+\) for \(\R^\omega\text{,}\) and \(r\in \R\) for \(\R^\R\)) is simply for the sequence of values \((f_n(i))_{n=1}^\infty\) to converge.

Theorem 1.8.4. Pointwise convergence and product topology.

Let \(\{(X_i, \mathcal{T}_i)\}_{i\in I}\) be a collection of topologies, let \(X=\prod_{i\in I}X_i\) be their product, and let \((f_n)_{n=1}^\infty\) be a sequence of elements of \(X\text{.}\)

If \(f_n\rightarrow f\) in the box or product topology, then \(f_n\) converges to \(f\) pointwise.
In the product topology we have

\begin{equation*} f_n\to f \iff f_n \text{ converges to } f \text{ pointwise}\text{.} \end{equation*}

Proof.

Suppose \(f_n\rightarrow f\) in either topology. Fix \(i\in I\) and let \(U_i\ni f(i)\) be any open set containing \(f(i)\in X_i\text{.}\) The set \(U=U_i\times \prod_{j\ne i}U_j\) is an open neighborhood of \(f\) in both topologies. By convergence there is an \(N\) such that \(f_n\in U\) for all \(n\geq N\text{.}\) It follows that \(f_n(i)\in U_i\) for all \(n\geq N\text{.}\) This proves that \(f_n(i)\rightarrow f(i)\) for all \(i\in I\text{.}\)
Suppose the sequence \((f_n)\) converges to \(f\) pointwise. To show \(f_n\rightarrow f\) in the product topology it is enough to show that for any base open set of the form \(U_{i_1}\times U_{i_2}\times\cdots \times U_{i_m}\times \prod_{k\ne i_k }X_k\) there is an \(N\) such that if \(n\geq N\text{,}\) then \(f_n\in U\text{.}\) Since \((f_n)\) converges to \(f\) pointwise, for each \(i_j\text{,}\) \(1\leq j\leq m\) there is an \(N_{i_j}\) such that if \(n\geq N_{i_j}\text{,}\) then \(f_n(i_j)\in U_{i_j}\text{.}\) Setting \(N=\max\{ N_{i_j}\}_{j=1}^m\text{,}\) it follows that if \(n\geq N\) then \(f_n(i_j)\in U_{i_j}\) for \(1\leq j\leq m\text{.}\) Since \(f_n(j)\in X_j\) for all \(j\ne i_j\) we conclude that \(f_n\in U\) for all \(n\geq N\text{,}\) as desired.

Example 1.8.5. Convergent sequences in \(\R^\omega\).

Let \(X=\R^\omega\text{,}\) where \(\R\) is equipped with the standard topology, and let \((f_n)_{n\in \Z_+}\) be the sequence defined as

\begin{equation*} f_n=(1/n, 1/n,1/n,\dots) \end{equation*}

for all \(n\geq 1\text{.}\)

Visualize the sequence \((f_n)\) in the real plane.
Decide whether the sequence converges pointwise. If so, what does it converge to?
Decide whether the sequence converges in the product topology, and whether it converges in the box topology.

Solution.

It is relatively easy to see that the sequence \(f_n\) converges pointwise to \(f=(0,0,\dots)\text{.}\) Since \(\R\) with the standard topology is Hausdorff, it follows that this pointwise limit is unique. By Theorem 1.8.4, \((f_n)\) also converges in the product topology to \(f\text{.}\) We claim \((f_n)\) does not converge at all in the box topology. First observe, that the only candidate for the limit of this convergence is \(f=(0,0,\dots,)\text{,}\) using (1) of Theorem 1.8.4. Now consider the open set

\begin{equation*} U=(-1,1)\times (-1/2, 1/2)\times (-1/3,1/3)\times\cdots=\prod_{n=1}^\infty(-1/n, 1/n)\text{.} \end{equation*}

We have \(f=(0,0,\dots, )\in U\text{,}\) and yet not only is it not the case that the sequence \((f_n)\) eventually lies in \(U\text{,}\) in fact we have \(f_n\notin U\) for all \(n\in \Z_+\text{.}\)

Definition 1.8.6. Standard bounded metric.

Let \((X,d)\) be a metric space. The function \(\overline{d}(x,y)=\min\{d(x,y), 1\}\) defines a metrix on \(X\text{,}\) called the standard bounded metric corresponding to \(d\text{.}\) The two metrics \(d, \overline{d}\) induce the same topology on \(X\)

Proof.

See Munkres for a proof that \(\overline{d}\) is a metric, and that \(d\) and \(d'\) induce the same topology.

Topological specimen 12. Uniform topology.

Let \(X=\R^I\) for some set \(I\text{,}\) where \(\R\) is equipped with the standard topology. Let \(\overline{d}(x,y)=\min\{\abs{x-y},1\}\) for all \(x,y\in \R\text{.}\) The function

\begin{align*} d_{\operatorname{uni}}\colon X\times X \amp \rightarrow \R\\ \left((x_i)_{i\in I}, (y_i)_{i\in I}\right) \amp\mapsto \sup\{\overline{d}(x_i,y_i)\colon i\in I\} \end{align*}

is a metric called the uniform metric on \(X\text{.}\) The uniform topology is the topology induced by \(d_{\operatorname{uni}}\text{.}\)

Proof.

See Munkres for a proof that \(d_{\operatorname{uni}}\) is a metric.

Theorem 1.8.7.

Let \(X=\R^I\) for some set \(I\text{,}\) where \(\R\) is equipped with the standard topology, and let \(\mathcal{T}_{\operatorname{prod}}, \mathcal{T}_{\operatorname{uni}}, \mathcal{T}_{\operatorname{box}}\) be the product, uniform, and box topologies on \(\R\text{.}\) We have

\begin{equation*} \mathcal{T}_{\operatorname{prod}}\subseteq \mathcal{T}_{\operatorname{uni}}\subseteq \mathcal{T}_{\operatorname{box}}\text{.} \end{equation*}

When \(I\) is infinite these inclusions are all strict.

Proof.

Take any open set \(U\) in the product topology and any \(x=(x_i)_{i\in I}\) lying in \(U\text{.}\) There is an open neighborhood of \(x\) lying in \(U\) of the form \(B=\prod_{j=1}^n (x_{i_j}-\epsilon_{i_j},x_{i_j}+\epsilon_{i_j})\times \prod_{i\ne i_j}\R\text{,}\) where we may further assume \(\epsilon_{i_j}< 1\) for all \(1\leq j\leq n\text{.}\) Letting \(\epsilon=\min\{\epsilon_{i_j}\}\text{,}\) I claim we have \(B_\epsilon(x)\subseteq B\text{.}\) Indeed, for any \(y\in B_{\epsilon}(x)\text{,}\) we have

\begin{align*} \sup\{\overline{d}(y_i,x_i)\} < \epsilon \amp \implies \min\{\abs{y_{i_j}-x_{i_j}},1\}< \epsilon \text{ for all } 1\leq j\leq n \\ \amp \implies \abs{y_{i_j}-x_{i_j}}< \epsilon \text{ for all } 1\leq j\leq n \amp (\epsilon < 1)\\ \amp\implies \abs{y_{i_j}-x_{i_j}}< \epsilon_{i_j} \text{ for all } 1\leq j\leq n \\ \amp \implies y\in B \end{align*}

This proves that \(\mathcal{T}_{\operatorname{prod}}\subseteq \mathcal{T}_{\operatorname{uni}}\)

For the next inclusion, take any open set \(U\) in the uniform topology. For any \(x\in U\) we can find \(\epsilon > 0\) such that \(B_{\epsilon}(x)\subseteq U\text{.}\) Let \(V=\prod_{i\in I}(x_i-\epsilon/2, x_i+\epsilon/2)\text{,}\) which is open in the box topology. Given any \(y=(y_i)\in V\text{,}\) we have \(\overline{d}(y_i, x_i)\leq \abs{y_i-x_i}< \epsilon/2\text{.}\) It follows that in the uniform metric we have

\begin{equation*} d_{\operatorname{uni}}(y,x)=\sup\{\overline{d}(y_i, x_i)\}\leq \epsilon/2< \epsilon\text{,} \end{equation*}

showing that \(V\subseteq B_\epsilon(x)\subseteq U\text{.}\) This proves \(\mathcal{T}_{\operatorname{uni}}\subseteq \mathcal{T}_{\operatorname{box}}\text{.}\)

The fact that both inclusions are strict when \(I\) is infinite is left as a homework exercise.

Theorem 1.8.8. Product topology on \(\R^\omega\).

Let \(X=\R^\omega\text{,}\) where \(\R\) is equipped with the standard topology, and let \(\overline{d}(x,y)=\min\{\abs{x-y},1\}\) for all \(x,y\in \R\text{.}\) The function

\begin{align*} D\colon X\times X \amp \rightarrow \R\\ \left((x_i)_{i=1}^\infty, (y_i)_{i=1}^\infty\right) \amp\mapsto \sup\left\{\frac{\overline{d}(x_i,y_i)}{i}\colon i\in \Z_+\right\} \end{align*}

is a metric on \(\R^\omega\) that induces the product topology.

Proof.

See Munkres. The proof that the product topology is equal to the topology induced by this metric is quite illustrative, and will be a useful model for some of your homework exercises.

Remark 1.8.9. When is the product topology on \(\R^I\) metrizable?

Theorem 1.8.8 shows that the product topology on \(\R^\omega\) is metrizable. It turns out that when \(I\) is uncountable, this is no longer true. (As we will see later.) In particular, the product topology on \(\R^\R\) is not a metric topology. The product topology on \(\R^\R\) is valuable to us, since it is the natural setting to study pointwise convergence of functions (1.8.4). Consider this an argument for studying topologies beyond metric topologies.

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