Let $x_1,x_2, \ldots$ be a sequence of points of the product space $\prod X_\alpha$. Show that the sequence converges to the point $x$ if and only if the sequence $\pi_\alpha (x_1), \pi_\alpha (x_2)\ldots$ converges to $\pi_\alpha (x)$ for each $\alpha$. Is this fact true for box topology instead of product topology?
But what does it mean to converge in product topology? Is it that a common neighborhood exists for all points in the sequence?
Best Answer
Let $U\subseteq X_\alpha$ be an open neighbourhood of $\pi_\alpha(x)$. By definition of product topology, the set $V:=\prod U_i\subseteq \prod X_i$ with $U_\alpha=U$ and $U_i=X_i$ for $i\ne \alpha$ is an open neighbourhood of $x$. Hence almost all $x_n$ are in $V$, hence almost all $\pi_\alpha(x_n)$ are in $U$. This just says that $\pi_\alpha(x_n)\to \pi_\alpha(x)$.
For the other direction suppose that $\pi_\alpha(x_n)\to \pi_\alpha(x)$ for all $\alpha$. Let $U$ be an open neighbourgood of $x$. Wlog. we may assume that $U$ is the product of open sets $U_\alpha$ of $X_\alpha$ where $U_\alpha=X_\alpha$ for almost all $\alpha$ (as these special sets form a basis of the product topology). Then for each $\alpha$, we have that almost all $\pi_\alpha(x_n)$ are in $U_\alpha$, that is there are $n_\alpha\in\mathbb N$ such that $\pi_\alpha(x_n)\in U_\alpha$ for all $n>n_\alpha$. Since $U_\alpha=X_\alpha$ for almost all $\alpha$, we can choose $n_\alpha=0$ for almost all $\alpha$. Therefore we can consider $N=\max\{n_\alpha\mid \alpha\}=\max\{n_\alpha\mid U_\alpha=X_\alpha\}\in\mathbb N$ because we are taking the maximum only over finitely many natural numbers. Then for $n>N$ we have that $\pi_\alpha(x_n)\in U_\alpha$ and hence $x_n\in U$. This shows that $x_n\to x$.
With box topology, the above argument fails at the bolded step. More concretely, note under the box topology the product of infinitely many discrete two-point spaces is discrete and find an explicite counterexample from that.