Limit points and the product topology Let $(x_n)_{n \in \mathbb{N}^+}$ be a sequence of points in $\mathbb{R}^{\mathbb{N}}$ such that $x_n = (1, \frac{1}{n}, \frac{1}{n^2}, \cdots)$. Show that $x := (1,0,0,0,\cdots)$ is a limit for $(x_n)$ in the product topology, but not the box topology.

I first want to show it is a limit point in the product topology. 
Using the definition of a limit point, $x := (1,0,0,0,...)$ is a limit point in $\mathbb{R}^{\mathbb{N}}$ (with the product topology) iff every open set $x\in U$ in the product topology contains some point of $(x_n)$. I don't really know how to show  with open $U$, $U$ some other point of $(x_n)$ which is not $x$. Advice on how to do that? 
 A: Hints: For the first part you can notice that it is enough to show for every basic open set containing $x$ contains a point of the sequence. But what does a basic open set in product topology look like?
For the second one, try to construct a basic open set around the point $x$ such that no point of the sequence lies in it. What is it you can do in box topology which you could not in product topology? 
Hope it helps.
A: It is equivalent to show that for every basic open set $U$ of $\mathbb R^\mathbb N$, $U$ contains some $x_n$. Indeed, we have that any open set $U$ is a union $\bigcup U_i$ of basic open sets. Thus, if $U$ contains $x_n$, there is some $i$ such that $U_i$ contains $x_n$.
Now remember that a basis of open sets for the product topology is the family of sets of the form $U_1\times U_2\times \dots$ where $U_i$ is an open interval in $\mathbb R$ and all but finitely many $U_i$ are $\mathbb R$ itself. For example, $\{(x_0,x_1,\ldots) \in\mathbb R^\mathbb N \mid x_2\in (-2,2), x_5\in (0,1)\}$ is open in $\mathbb R^\mathbb N$, since it is $\mathbb R\times \mathbb R \times (-2,2)\times \mathbb R\times \mathbb R\times (0,1)\times\mathbb R\times\dots$.
Back to the question. It remains to show that given a basic open set which contains $(1,0,0,\dots)$, it contains some $x_n$. Let $U=\prod U_i$ be basic open. If $U_i$ is not $\mathbb R$, let $n_i$ be so that $\forall n\geq n_i, \frac1{n^i}\in U_i$. Let $N=\max\{n_i\}$, the maximum being defined since there are only finitely many $i$ so that $U_i\neq\mathbb R$. Now note that for all $n\geq N$, we have that $x_n\in U$. 
Actually, note that this proves a more general result: let $x_n^i$ be the $i$th coordinate of the sequence $(x_n)_{n\in\mathbb N}$. Then we have that $x_n\rightarrow x$ in the product topology if, and only if, for all $i\in\mathbb N$ we have $x_n^i\rightarrow x^i$. Since we indeed have that $\frac1{n_i}\rightarrow 0$ for $i\geq 1$ and $\frac{1}{n^i}\rightarrow 1$ when $i=0$, this proves the first statement.
For the second part, I give you a hint: the fact that we could define the max above is what fails in the box topology. Consider the open neighbourhood of $(1,0,0,\ldots)$ that is $\prod_{i\geq 0} (-\frac1{i!n^i}, \frac1{i!n^i})$.
