In $(\mathbb{R}, \tau_{st})$, we can write $\mathbb{R} \setminus \mathbb{Z} = \bigcup_{n \in \mathbb{Z}} (n,n+1)$, and hence $\mathbb{R} \setminus \mathbb{Z}$ is an open set. Thus the complement, $\mathbb{Z}$, is closed.
In $(\mathbb{R}^2, \tau_{st})$ I am not sure how to write an expression for $\mathbb{R}^2 \setminus \mathbb{Z}^2$ in terms of open balls. So the first part of my question is whether we can write this in a 'neat' expression like above?
I can try to write $\mathbb{R}^2 \setminus \mathbb{Z}^2 = \mathbb{R}^2 \setminus \bigcup_{n,m \in \mathbb{Z}} \big\{(n,m)\big\}$. So knowing that $\big\{(n,n)\big\}$ is a closed set, we have the complement of the union of closed sets. But this is an infinite union, so I can't actually conclude that $\mathbb{Z}^2 = \bigcup_{n,m \in \mathbb{Z}} \big\{(n,m)\big\}$ is actually closed in this manner.
I know that we can say $\mathbb{Z}^2$ is closed by arguing its set of limit points is empty, but is there any other way I can show this? In particular, is there a way to show (analogous to the $\mathbb{R}^1$ case above) that $\mathbb{R}^2 \setminus \mathbb{Z}^2$ is closed, without relying on an argument by limit points? I.e. Can I show the set as an explicit construction of union of open balls?
Best Answer
$\mathbb{R}^2 \setminus \mathbb{Z}^2$ is the union of the open balls centered on points both having irrational coordinates and radius small enough to not intersect $\mathbb{Z}^2$.
That makes a lot of balls... but it works!