Verify whether the set $\mathbb{N} \times \mathbb{Z}$ is open or closed (or both) on $\mathbb{R^2}$ with standard Euclidean metric?

Question

How does this set look like? How do we choose an $\epsilon$ radius?

Answer 1

$\mathbb{N}\times \mathbb{N}$ is closed in $\mathbb{R}\times \mathbb{R}$, since for any $(x,y)\in \mathbb{R}\times \mathbb{R}$, we may find an open ball $G$ in $\mathbb{R}$ such that $G/ \lbrace (x,y) \rbrace \cap \mathbb{N}$ is empty set. For any $(x,y)\in \mathbb{R}\times \mathbb{R}$, define $G=\lbrace (a,b):d((a,b),(x,y))<1\rbrace$. Then clearly $G$ is open and it also contains $(x,y)$ such that $G/ \lbrace (x,y) \rbrace \cap \mathbb{N}$ is empty. Therefore any $(x,y)\in \mathbb{R} \times \mathbb{R}$ is not alimit point of $\mathbb{N}\times \mathbb{N}$ or in other words $\mathbb{N}\times \mathbb{N}$ contains each of its limit points. Hence it is closed. It is not closed since $(1,1)\in \mathbb{N}\times \mathbb{N}$ but any open set containing $(1,1)$ must contains some irrationals so that $\mathbb{N}\times \mathbb{N}$ can not be open.

Proceeding in a similar way may go for $\mathbb{N}\times \mathbb{Z}$.

Answer 2

Hint: if a real number $x$ is not an integer then it lies striclty between two consecutive integers, say $n <x<n+1$. Then $|y-x| <\min \{x-n,n+1-x\}$ implies that $y$ is not an integer. Use this idea to show that the complement of $\mathbb N \times \mathbb Z$ is open. Hence $\mathbb N \times \mathbb Z$ is closed. It is not open because $(1,0)$ belongs to this set and every open ball around this point contains points of the form $(1,\frac 1 k)$ which don't belong to $\mathbb N \times \mathbb Z$ .