# irrationals, metrizable, Michael Line

• May 20th 2009, 01:19 AM
xboxlive89128
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.
• May 20th 2009, 11:34 AM
Opalg
Quote:

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.