# Thread: irrationals, metrizable, Michael Line

1. ## irrationals, metrizable, Michael Line

Prove that the irrationals have usual neighborhoods and the rationals are isolated is metrizable.

This is the flip-flop of the Michael Line. I do not see how to prove this. I know we want to use Urysohn Metrization Theorem, but I do not see how to use it now. Thanks in advance.

2. Originally Posted by xboxlive89128
Prove that the irrationals have usual neighborhoods and the rationals are isolated is metrizable.

This is the flip-flop of the Michael Line. I do not see how to prove this. I know we want to use Urysohn Metrization Theorem, but I do not see how to use it now. Thanks in advance.
If I understand the question correctly, you only need to check two things in order to apply the Urysohn theorem to this space:
1) The space is regular (it is easily seen to be Hausdorff);
2) The space is second-countable.

An open sets in the space is the union of a usual open set with a subset of the rationals. So a closed set is a usual closed set less a subset of the rationals. If F is such a set then $F = F_0\cap Q^c$, where $F_0$ is closed in the usual sense and $Q\subseteq\mathbb{Q}$. If $x\notin F$ then either $x\notin F_0$ (in which case you can separate x from F by a usual open set), or $x\in Q$ (in which case {x} is an open set separating x from F). That proves (1).

For (2), you can take a countable base for the usual topology on $\mathbb{R}$ and add to it all the singleton sets $\{\{x\}:x\in\mathbb{Q}\}$ to get a countable base.