Rational points in the Alexandroff line

Question

Let $X$ be the subset of the long line consist of rational points with the topology inherits from the long line.

Is $X$ a metrizable space?

Answer 1

As stated in the comments, one can show that $\omega_{1}$ is not paracompact hence non-metrizable. However, one can also give a direct proof that $\omega_{1}$ is not metrizable without resorting to paracompactness. Here is one way to prooe that $\omega_{1}$ is not metrizable. Suppose to the contrary that $d$ is a metric inducing the topology on $\omega_{1}$. The for each limit ordinal $\alpha$ and $n>0$, there is some $f_{n}(\alpha)<\alpha$ such that $d(\beta,\alpha)<\frac{1}{n}$ whenever $f_{n}(\alpha)\leq\beta\leq\alpha$. If $C$ is the set of all limit ordinals, then $C$ is a club set and $f_{n}$ is regressive. Therefore there is some stationary set $S_{n}$ and ordinal $\alpha_{n}$ such that $f_{n}(\alpha)<\alpha_{n}$ for each $\alpha\in S_{n}$. In particular, since $S_{n}$ in unbounded, whenever $\alpha,\beta>\alpha_{n}$, there is some $\gamma>\alpha,\beta$ with $f_{n}(\gamma)<\alpha_{n}<\alpha,\beta$. Therefore, $d(\alpha,\gamma)<\frac{1}{n}$ and $d(\beta,\gamma)<\frac{1}{n}$, so $d(\alpha,\beta)<\frac{2}{n}$. In particular, if $\alpha=\sup\{\alpha_{n}|n\in\omega\}$, then whenever $\beta,\gamma>\alpha$, we would have $d(\beta,\gamma)<\frac{2}{n}$ for all $n$, so $d(\beta,\gamma)=0$. This is a contradiction. In particular, since the rational points on the long line contains $\omega_{1}$, the set of rational points on the long line is not metrizable either.

Another way to prove $\omega_{1}$ is not metrizable will be to argue that every continuous map $f:\omega_{1}\rightarrow\mathbb{R}$ is bounded (this is easy), then show $\omega_{1}$ is not compact (this is also easy). However, any metric space where every continuous real-valued function is bounded is compact. Therefore $\omega_{1}$ is not metrizable.