Is $\mathbb{R}\setminus \mathbb{Z}$ open in $\mathbb{R}$?

Question

Let $U = \mathbb{R}\setminus \mathbb{Z}$. Is $U$ open in $\mathbb{R}$?

My attempt: $U = \mathbb{R}\setminus \mathbb{Z}$ is open in $\mathbb{R}$: Let $x\in U$. Choose $r:=\frac{1}{2}(\text{min}\{x-\lfloor x \rfloor, (\lfloor x \rfloor+1) -x\})$, so $r>0$. Therefore for each $x\in U$ there exist $r>0$ such that $(x-r, x+r)\subset U$. Hence $\mathbb{R}\setminus \mathbb{Z}$ is open in $\mathbb{R}$.

Is my proof correct?

@Exodd $r>0$ because $\lfloor x \rfloor \leq x \leq \lfloor x \rfloor +1$. Is it correct? — user1234, Jun 29 '22 at 18:48
No, $r>0$ because $\lfloor x \rfloor < x < \lfloor x \rfloor +1$. — TonyK, Jun 30 '22 at 14:52
@TonyK Ok sorry my fault, $x$ can not take the integer values. So, $x \neq \lfloor x \rfloor$. — user1234, Jun 30 '22 at 14:55

score 4 · Accepted Answer · answered Jun 29 '22 at 18:04

4

Another approach: $\mathbb{R}\setminus\mathbb{Z}=\cup_{n=-\infty}^{\infty}(n,n+1)=\cdots \cup(-2,-1) \cup (-1,0)\cup (0,1)\cup (1,2)\cup (2,3)\cdots$, is the countable union of open sets, thus is open

answered Jun 29 '22 at 18:04

Guillermo García Sáez

1

Is $\mathbb{R}\setminus \mathbb{Z}$ open in $\mathbb{R}$?

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