# show $X=[0,1]^\omega$ has no metric which defines the box topology

Let $\omega=|\mathbb{N}|$, and $X=[0,1]^\omega$.

How can I show $X$ has no metric which defines the box topology?

thoughts: I think of looking at $S=(0,1]^\omega$ and its closure (which is $X$), then if assume the existence of such metric - maybe I can try and show there's no sequence in S converging to $(0,0,0,0....)\in X$....so there's no open ball centered in $0\in X$...

any help? :)

The point $0=(0,0,0,\ldots,0,\ldots)$ (or in fact any point in $[0,1]^\omega$) does not have a countable local base, so the space cannot be metrisable (because then $B(x,\frac{1}{n})$ would have been a countable local base at $x$). Suppose that $U_n$ is such a local base at $0$, then for each $n$ pick a sequence $(a^{(n)}_k) \in [0,1]^\omega$ such that $0 \in \prod_k [0, a^{(n)}_k) \subseteq U_n$, this can be done as such sets form a base for the box topology,and $[0,e), e>0$ is a basic open set for $0$ in $[0,1]$.
Then define $O =\prod_k [0, \frac{a^{(n)}_n}{2})$. This is a neighbourhood of $0$ but no $U_n \subset O$, as witnessed by the $n$-th component of $O$ and $U_n$.