Show that $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ with the box topology is Hausdorff but not metrizable. 
Show that $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ with the box topology is Hausdorff but not metrizable. 

$\Bbb{R}$ must be Hausdorff. For $x_1, x_2 \in \Bbb{R}$ (where $x_1 \not= x_2$), if $d$ denotes the distance between these two points, then we can choose an open ball $U_1$ with radius $d/2$ around $x_1$ and an open ball $U_2$ again of radius $d/2$ around $x_2$. Obviously, these two open balls will not intersect at any point. 
So let $(x_1, x_2, ...), (x'_1, x'_2, ...) \in \Bbb{R^n}$. Then For each $x_i, x'_i \in \Bbb{R}$ (where $i \in \Bbb{N}$), there exist open balls $U_i$ and $U'_i$  in $\Bbb{R}$ around $x_i$ and $x'_i$, respectively, such that the open balls do not intersect. But then $(U_1 \times U_2 \times ...) \cap (U'_1 \times U'_2 \times ...) = \emptyset$. 
We can show that $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ is not metrizable by proving that it is not first countable. 
The first countability condition tells us that $\forall x \in X$ $\exists \{ U_n \}_{n\ \in \Bbb{N}} \subseteq  P(X)$ such that 
(i) $\forall n \in \Bbb{N}$, $U_n$ is a neighborhood of $x$.  
(ii) For every neighborhood $V$ of $x$ $\exists n \in \Bbb{N}$ such that $U_n \subseteq V$. 
So let's pick the point $(0,0,0,...) \in \displaystyle\prod_{\Bbb{N}} \Bbb{R}$. We know that, for all $n \in \Bbb{N}$ (in each $\Bbb{R}$) $(-1/n, 1/n)$ is a neighborhood of $0$.
So $(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})...$ is a neighborhood of $(0,0,0,0,...)$. 
But in order to have $U_n \in \{U_n\}_{n \in \Bbb{N}}$ such that $U_n \subseteq (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})...$ for all $n$, we know that the sequence $\{(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})...\}_{n \in \Bbb{N}} = \{U_n\}_{n \in \Bbb{N}}$. 
But a sequences approaches infinity, and as $n \rightarrow \infty$ $(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})... = \{0\} \times \{0\} \times ...$, which is a closed neighborhood of $(0,0,0,...)$, meaning that the 1st condition of first countability fails.
Is my answer correct? If not, can anybody give me a hint?
Thanks. 
 A: Your solution to Hausdorffness is not quite correct.  Note what happens if your points are $\mathbf{x} = \langle  1 , 2, 3, \ldots \rangle$ and $\mathbf{x^\prime} = \langle 3 , 3 , 3 , \ldots \rangle$.  (How do you choose the neighbourhoods $U_3$ and $U^\prime_3$?)
Recall that the usual product topology on $\mathbb{R}^\mathbb{N}$ is Hausdorff.  Perhaps we can use this fact (or the proof of this fact.)

Your solution to the non-first-countability of the space is also somewhat off.  You appear to be showing that one particular countable family of neighbourhoods of $\mathbf{0} = \langle 0 , 0 , \ldots \rangle$ is not a neighbourhood base.  But you instead have to show that every countable family of neighbourhoods of $\mathbf{0}$ is not a neighbourhood base.
What you want to do is start with any old countably family $\{ U_n : n \in \mathbb{N} \}$ of open neighbourhoods of $\mathbf{0}$, and produce an open neighbourhood $V$ of $\mathbf{0}$ for which $U_n \not\subseteq V$ holds for all $n$.  As a particularly strong hint, produce $V$ so that for each $n \in \mathbb{N}$ the projection of $V$ onto the $n$th coordinate is a proper subset of the projection of $U_n$ onto the $n$th coordinate.
A: For not metrizable:let $A$ be the set of all sequences with elements in $\mathbb{R}$ with the property that the elements of all these sequences are positive .now the zero sequence(the sequence with zero elements) is a closure point for $A$ because every open set in $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ has a form $U_1 \times U_2 \times ...$ s,t $U_i$ is open in $\mathbb{R}$ and if $U_1 \times U_2 \times ...$ is an open set contains the zero sequence,then its intersection with $A$ is nonempty.
We assume that there exist a sequence ${a_n}$ tends to the zero sequence and $a_n$={$x^n_m$}$_{m\in\mathbb{N}}$.
Now $(-x^1_1,x^1_1)\times (-x^2_2,x^2_2)\times (-x^3_3,x^3_3)...$ Is an Open set contains the zero sequence but It contains no member of the sequence ${a_n}$.
from $\color{red}{Munkres's}\space topology$.
