# Proof on not metrizable

How to prove that space $\mathbb{R}_w$, the countably infinite product of $\mathbb{R}$ in the box topology, is not metrizable? I have tried finding a solution to this problem, but failed. Kindly help me find this answer.

At no point of this product in the box topology, the space is first countable: e.g. for $p=(p_1, p_2,p_3,\ldots)$: suppose $\{U_n: n \in \mathbb{N}\}$ is a local base at $p$. Every $U_n$ contains an open box around $p$ and as all open sets in $\mathbb{R}$ are unions of open intervals, for each $m$:

$$\exists r^{(m)}_1,r^{(m)}_2, \ldots,r^{(m)}_i,\ldots: p \in \prod_i (p_i - r^{(m)}_i, p_i + r^{(m)}_i) \subseteq U_m$$.

Then define $O = \prod_i (p_i - \frac{1}{2} r^{(i)}_i, p_i + \frac{1}{2}r^{(i)}_i)$ which is box-open, and contains $p$. But no $U_n$ can be a subset of $O$ ($U_n$ fails at the $n$-th coordinate), and so the $U_n$ cannot form a local base at $p$.

All metric spaces do have local countable bases everywhere: $\{B(p, \frac{1}{n}):n \in \mathbb{N}\}$ will do.

So this box product is not metrisable.

Hint: Do it by contradiction method. You must have learnt the sequence lemma: It says " If A a subset of a topological space X and there is a sequence in A converging to a point x then $x\in cl (A)$. The converse holds if X is metrizable". So assuming $\mathbb {R} ^{\omega}$ is metrizable try to find a contradiction to this theorem(i.e., prove the converse of the above Lemma isn't true.)
Try it with the set $$A=\{(x_1,x_2,\cdots)| x_i>0 for\, all \, i\in \mathbb {N}\}$$.
$\mathbf {0}$ is a limit point of this set but there is no sequence of points in A that converges to $\mathbf {0}$ in box topology. Hope it helps?