To be clear, I mean $$ {\mathbb R}^\infty = \bigcup_{n} {\mathbb R}^n $$ with the colimit topology, where ${\mathbb R}^n\to {\mathbb R}^{n+1}$ is the inclusion with the last coordinate 0.
It seems that this space is not metrizable, as I read from here
Direct limit $\mathbb{R}^\infty$ is not paracompact? (false)
that it is not locally metrizable. It could be a simple reason, but I don't see it.
Your help is appreciated.
A metric space $(X,d)$ is always "first countable". Namely, every point $x\in X$ admits a countable set of open sets (specifically $B_{\frac{1}{n}}(x) : =\{y\in X : d(x,y)<\frac{1}{n}\}$) with the property that any $x\in U\subseteq \mathrm{X}$ constains $B_{\frac{1}{n}}(x)$ for some $n$.
We will show that our space is not first countable. Let $0\in \mathbb{R}^\infty$ be the zero element and let $U_n\subseteq \mathbb{R}^\infty$ be countably many open sets containing $0$. We construct an open set $U\subseteq \mathbb{R}^\infty$ around $0$ that does not contain any of the $U_n$'s.
By definition, $U_n\cap \mathbb{R}^n$ is an open subset of $\mathbb{R}^n$ for all $n$. Therefore, it contains an open subset of the form $V_n:=(-a_{n,1},a_{n,1})\times...\times (-a_{n,n},a_{n,n})$ where $(-a_{n,i},a_{n,i})$ are intervals around zero for all $1\leq i \leq n$.
Let $U = \{ x\in \mathbb{R}^\infty : -a'_{n,n}<x_n<a'_{n,n} \text{ for all } n\}$ where $0<a'_{n,n}<a_{n,n}$.
We first observe that $U$ is open. Indeed, $U\cap \mathbb{R}^m = (-a'_{1,1},a'_{1,1})\times...\times (-a'_{m,m},a'_{m,m})$ is open.
Now we prove that $U$ contains no $U_n$. Indeed, let $n\in\mathbb{N}$ be arbitrary. By construction $U_n\cap \mathbb{R}^n$ contains $V_n$. Since $(a'_{n,n},a_{n,n})\subseteq (-a_{n,n},a_{n,n})$, there exists some $x\in U_n$ so that $x_n\in (a'_{n,n},a_{n,n})$, but then $x\not\in U$. As required.
