Exercise 19.6

Proof. Suppose $\{x_i\}\to x$, and fix some index $\gamma$. Then, for any neighborhood $U_{\gamma }\ni {\pi }_{\gamma }(x)$, letting $U=\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ where $U_{\alpha }=X_{\alpha }$ for all $\alpha \ne \gamma$, there exists $N\in \mathbb{N}$ such that $x_i\in U$ for all $i\ge N$, and so ${\pi }_{\gamma }(x_i)\in {\pi }_{\gamma }(U)=U_{\gamma }$ for all $i\ge N$, i.e., $\{{\pi }_{\gamma }(x_i)\}\to {\pi }_{\gamma }(x)$. Note that this direction does not depend on the topology being the product or box topology.

In the other direction, suppose $\{{\pi }_{\alpha }(x_i)\}\to {\pi }_{\alpha }(x)$ for all $\alpha$. We take an arbitrary neighborhood $V$ of $x\in \prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$; it then contains a basis element of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ containing $x$, which is a product of open sets $\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$. In the case of the product topology, there then exist only finite $U_{\alpha }⊊X_{\alpha }$, and for these open sets there exist $N_{\alpha }\in \mathbb{N}$ such that ${\pi }_{\alpha }(x_i)\in U_{\alpha }$ for all $i\ge N_{\alpha }$ for each $\alpha$. $N_{\alpha }=1$ works for all other $\alpha$. Thus, we can take $N=\max <!--\; FUNCTION\; APPLICATION\; -->(N_{\alpha })$; then, $x_i\in \prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }\subseteq V$ for all $i\ge N$.

We construct a counterexample for this direction in the case of the box topology. Let ${ℝ}^{\mathbb{N}}$ be the box topology on the product of copies of $ℝ$ indexed by $\mathbb{N}$, and let

Then, for each $\alpha \in \mathbb{N}$, $\{{\pi }_{\alpha }(x_i)\}\to (0,0,0,\dots )=:x$, but this sequence does not converge in the box topology, for the open set

in the box topology contains $x=(0,0,0,\dots )$, but does not contain any $x_i$. □

Proof. $\text{(⇒)}$ First suppose that the sequence $x_1,x_2,\dots$ converges to $x$ and consider any $\beta$. Also suppose that $U$ is any neighborhood of ${\pi }_{\beta }(x)$. Define

so that $B=\prod <!--\; FUNCTION\; APPLICATION\; -->B_{\alpha }$ is a basis element of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ since each $B_{\alpha }$ is open. Note that $B$ is a basis element of both the box and product topologies since possibly $B_{\alpha }\ne X_{\alpha }$ for only one $\alpha$ (i.e. for $\alpha =\beta$). We also clearly have that $x\in B$ so that $B$ is a neighborhood of $x$ in $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$. Since the sequence $x_1,x_2,\dots$ converges to $x$, we have that there is an $N\in {\mathbb{Z}}_{\text{+}}$ where $x_n\in B$ for all $n\ge N$. So consider any such $n\ge N$ so that $x_n\in B=\prod <!--\; FUNCTION\; APPLICATION\; -->B_{\alpha }$. Hence ${\pi }_{\alpha }(x_n)\in B_{\alpha }$ for all $\alpha$, and in particular ${\pi }_{\beta }(x_n)\in B_{\beta }=U$. This suffices to show that the sequence ${\pi }_{\beta }(x_1),{\pi }_{\beta }(x_2),\dots$ converges to ${\pi }_{\beta }(x)$ as desired since $U$ was an arbitrary neighborhood.

$\text{(⇐)}$ Now suppose that the sequence ${\pi }_{\alpha }(x_1),{\pi }_{\alpha }(x_2),\dots$ converges to ${\pi }_{\alpha }(x)$ for every $\alpha$. Let $U$ be any neighborhood of $x$ in $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$. Then there is a basis element $B=\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$ of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ where $x\in B$ and $B\subset U$. Since $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ is the product topology, each $U_{\alpha }$ is open but only a finite number of them are different from $X_{\alpha }$. Suppose then that $J$ is the index set of $\alpha$ and that $I\subset J$ is the finite subset where $U_{\alpha }=X_{\alpha }$ for all $\alpha \notin I$.

Then for any $\beta \in I$ we have that ${\pi }_{\beta }(x)\in U_{\beta }$ since $x\in B=\prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }$, hence $U_{\beta }$ is a neighborhood of ${\pi }_{\beta }(x)$. Then, since ${\pi }_{\beta }(x_1),{\pi }_{\beta }(x_2),\dots$ converges to ${\pi }_{\beta }(x)$, there is an $N_{\beta }\in {\mathbb{Z}}_{\text{+}}$ where ${\pi }_{\beta }(x_n)\in U_{\beta }$ for all $n\ge N_{\beta }$. So let $N=\underset{\alpha \in I}{\max <!--\; FUNCTION\; APPLICATION\; -->}{N}_{\alpha }$, noting that this exists since $I$ is finite. Consider any $n\ge N$ and any $\alpha \in J$. If $\alpha \in I$ then we have that $n\ge N\ge N_{\alpha }$ so that ${\pi }_{\alpha }(x_n)\in U_{\alpha }$. If $\alpha \notin J$ then of course we have that ${\pi }_a(x_n)\in X_{\alpha }=U_{\alpha }$. Hence either way we have that ${\pi }_{\alpha }(x_n)\in U_{\alpha }$ so that $x_n\in \prod <!--\; FUNCTION\; APPLICATION\; -->U_{\alpha }=B$ and hence also $x_n\in U$ since $B\subset U$. Since $n\ge N$ was arbitrary and $U$ was an arbitrary neighborhood of $x$, this shows that $x_1,x_2,\dots$ converges to $x$ as desired. □

Proof. Define

for $i,j\in {\mathbb{Z}}_{\text{+}}$. Now define a sequence $x_1,x_2,\dots$ in ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in {\mathbb{Z}}_{\text{+}}}ℝ={ℝ}^{\omega }$ by ${\pi }_i(x_j)=x_{\mathit{ij}}$. With the box topology on ${ℝ}^{\omega }$ we claim that each coordinate sequence ${\pi }_i(x_1),{\pi }_i(x_2),\dots$ converges to $0$ but that the sequence $x_1,x_2,\dots$ does not converge to the point $0=(0,0,\dots )$.

First, it is easy to see that each coordinate sequence ${\pi }_i(x_1),{\pi }_i(x_2),\dots$ converges to $0$ since, for fixed $i$, there is always an $N\in {\mathbb{Z}}_{\text{+}}$ large enough such that $j>i$ and ${\pi }_i(x_j)=x_{\mathit{ij}}=1∕(j-i)$ is small enough to be within any fixed neighborhood of $0$ for all $j\ge N$. To show that the sequence $x_1,x_2,\dots$ does not converge to $0$ though, consider the neighborhood $U=\prod <!--\; FUNCTION\; APPLICATION\; -->U_k$ of $0$ where every $U_k=(-1,1)$. We note that clearly $U$ is open in the box topology since each $U_k$ is a basis element of $ℝ$ and therefore open. For any $N\in {\mathbb{Z}}_{\text{+}}$ we then have that ${\pi }_N(x_N)=x_{\mathit{NN}}=1$ so that clearly ${\pi }_N(x_N)\notin (-1,1)=U_N$ and hence $x_N\notin \prod <!--\; FUNCTION\; APPLICATION\; -->U_k=U$. This suffices to show that the sequence does not converge, but it does not even come close to converging since there are actually no points in the sequence that are even in this quite large neighborhood of $0$! □