Is $S$ is open in $\mathbb{R}^2$? let  $S =  \{(x,y) \in \mathbb{R}^2 :  x\notin \mathbb{Z} , y \notin \mathbb{Z} \}$ . Is  $S$ is  open in $\mathbb{R}^2$ ?
My attempt : yes , i was thinking  about the  theorem that  complements  of closed set is  open
Is its  true  ?
 A: Here is a cute way: Consider the continuous function $f\colon\mathbb{R}^2\to\mathbb{R}$ given by $f(x,y)=\sin(\pi x)\sin(\pi y)$. Then $f(x,y)\neq0$ if and only if $\sin(\pi x)\neq0$ and $\sin(\pi y)\neq0$ which occurs if and only if $(x,y)\in S$. Since $f$ is continuous, the set of points $(x,y)$ where $f(x,y)\neq0$ is open. Thus, $S$ is open.
A: $$S = \bigcup_{i, j\in Z} A(i, j),\ A(i, j) = \{(x, y): i<x<i+1,  j<y<j+1\} $$
It sufficies to prove that $A(i, j)$ are open, but it's trivial since Chebyshev and Euclidean metrics are equivalent, and $A(i, j)$ is a ball in Chebyshev metric with center $(i+1/2, j+1/2)$ and radius $1/2$.
A: The functions $f(x,y)=x, g(x,y)=y$ are both continuous maps from $\mathbb R^2$ to $\mathbb R.$ Observe that $\mathbb Z$ is a closed subset of $\mathbb R,$ which implies $\mathbb R\setminus Z$ is open in $\mathbb R.$ By continuity, both $f^{-1}(\mathbb R\setminus Z),g^{-1}(\mathbb R\setminus Z) $ are open. Hence so is their intersection, which is exactly the set $S.$
A: You can certainly find an open ball disjoint from $\Bbb R^2\setminus S$, centered at any $(x,y)\in S$.  Just let $r=\operatorname {min}(d(x,\Bbb Z), d(y,\Bbb Z))$.  Then $B_{(x,y)}(r)$ is such a ball.
