Convergence in product topology 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? 
 A: Here is my usual abstract-nonsense answer.


*

*The product of topological spaces satisfies (or rather, is defined by) the universal property: A continuous map into the product is nothing else than a family of continuous maps into the factors.

*A sequence in a space $X$ is a (continuous) map $\mathbb{N} \to X$, and it converges iff this map has an extension to a continuous map $\mathbb{N}^+ \to X$, where $\mathbb{N}^+ = \mathbb{N} \cup \{\infty\}$ is the one-point compactification, where the basic-open neighborhoods of $\infty$ are cofinite. The value at $\infty$ is the limit of the sequence.

*From 1. we get $\hom(\mathbb{N}^+,\prod_\alpha X_\alpha) = \prod_\alpha \hom(\mathbb{N}^+,X_\alpha)$, and from 2. we see that this means that a sequence in $\prod_\alpha X_\alpha$ is just an $\alpha$-indexed family of sequences in the $X_\alpha$, and that a limit of this sequence is the same as the $\alpha$-indexed family of some limits of the individual sequences.

A: 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.
