Because we use the metric topology of .
Every point of is a limit point of .
And every point of is a limit point of .
Can you answer your questions now?
BTW: The set is pathwise connected. It is actually easy to exhibit the path between two points.
I found a proof:
PlanetMath: $\mathbb{R}^2 \setminus C$ is path connected if $C$ is countable
Thanks a bunch Plato...
The set is the set of all ordered pairs in that have at least one irrational coordinate. The is a point which has both coordinates irrational.
Suppose that is any other point. One of its coordinates is irrational, say it’s q.
Using the three points construct two line segments: .
Both line segments are subsets of because at least on coordinate of each point is irrational.
The first segment connects and the second connects .
Therefore the set is clearly pathwise connected.