Is $\mathbb{Q}$ and $\mathbb{R}-\mathbb{Z}$ is open in $\mathbb{R}$ with proof?

Question

Is $\mathbb{Q}$ and $\mathbb{R}-\mathbb{Z}$ is open in $\mathbb{R}$ with proof?

Definition "$X$ is open if for every point $x$ in $X$ there exist $r>0$ such that $B(x,r)$ is in $X$".

Answer 1

If $r>0$ and $x\in\mathbb{Q}$, then $B(x,r)\cap(\mathbb{R}\setminus\mathbb{Q})\ne\emptyset$. So $\mathbb{Q}$ is not open. It's not closed, either, because for $x\in\mathbb{R}\setminus\mathbb{Q}$ and $r>0$, $B(x,r)\cap\mathbb{Q}\ne\emptyset$.

The key fact is that any open interval in $\mathbb{R}$ contains both rational and irrational numbers, which can be expressed by saying that $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are both dense in $\mathbb{R}$, so neither is closed, hence neither is open.

On the other hand $$ \mathbb{R}\setminus\mathbb{Z}=\bigcup_{m\in\mathbb{Z}}(m,m+1) $$ is a union of open intervals, so $\mathbb{R}\setminus\mathbb{Z}$ is open.

Answer 2

Hint

Let $x \in \mathbb{R} \setminus \mathbb{Z}$. Since $x$ is real, there is an integer $n$ such that $n \le x < n+1$. Since $x \not \in \mathbb{Z}$, $x \ne n$. Thus $n < x < n+1$. Now to find an open ball around $x$ contained in the set, choose $$ \epsilon = \min \left\{ x-n, n+1 - x \right\} $$ and consider the open ball $B(x, \epsilon)$.

Answer 3

If $x\in\mathbb{Q}$, $\forall \epsilon>0$ there exists an irrational number $r\in B(x,\epsilon)$ $\Rightarrow B(x,\epsilon)\cap(\mathbb{R}-\mathbb{Q})\neq\varnothing$ $\Rightarrow B(x,\epsilon)\nsubseteq\mathbb{Q} $ . The proof of this is shown in almost every basic analysis courses and it is usually accepted in the rest of courses.

Let $x\in \mathbb{R}-\mathbb{Z}$. Then there exist two consecutives integer numbers $m$ and $m+1$ such that $m<x<m+1$. Let $a=min${$|x-m|,|x-(m+1)|$}. Then $B(x,a)\cap\mathbb{Z}=\varnothing \Rightarrow B(x,a)\subset\mathbb{R}-\mathbb{Z} $

According to this, $\mathbb{Q}$ is not open and $\mathbb{R}-\mathbb{Z}$ is open.