Box topology on $\prod_{n=1}^\infty\mathbb{R}$ 
Let $X$ denote $\prod_{n=1}^\infty\mathbb{R}$, the Cartesian product of countably infinitely many copies of $\mathbb R$ (which is just the set of all infinite sequences of real numbers), endowed with the box topology. Now, let $X^+\subset X$ be the subset consisting of the sequences of strictly positive real numbers, and let $z$ denote the zero sequence, that is, the one whose terms are $z_i = 0$ for all $i$. Show that $z$ is in the closure of $X^+$, but there is no sequence of elements of $X^+$ converging to $z$.

I guess I did the first part. The closure of $X^+$ is $$\bigcap_{\substack{\text{closed }S\,\subseteq X,\\ X^+\subseteq S}}S.$$ But these subsets are of the form $\prod[-E,+\infty)$ for every $E\geq 0$, right?  When $E = 0$, we have the required. Is it right? And what about the second part?
 A: Your argument for the first part is basically ok . Not all closed sets containing $X^+$ are  as you describe; but given any closed set containing $X^+$, there is a closed set of your form contained in it. 
For the first part, it may be easier to show that any open set containing  $z$ must contain an element of $X^+$.  Note that such a set must contain an open set of the form $\prod_{i=1}^\infty O_i$ where each $O_i$ is an open set in $\Bbb R$ containing $0$.
For the second part, suppose that $(x_i)$ is a sequence in $X^+$. In the $i^{\rm th}$  copy of $\Bbb R$ in the product, let $O_i$ be an open set centered at zero such that $x_i(i)\notin O_i$. What can you say about the open set $O=\prod_{i=1}^\infty O_i$?
A: Recall that: From [Wikipedia][1]


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One of the most important properties of first-countable spaces is that
  given a subset $A$, a point $x$ lies in the closure of $A$ if and only
  if there exists a sequence $\{x_n\}$ in $A$ which converges to $x$.

A nice consequence of the result proven above is that $X$ (which is $\mathbb{R}^\omega$ with the box topology) is not first countable and thus not metrizable (all metric spaces are first countable). 
