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Definition. A subset of $\mathbb{R}$ is called open if it is a union of open intervals.

Example. $\mathbb{R} \backslash \mathbb{Z}$ is open.

Proof (is from my lecture notes).$\mathbb{R} \backslash \mathbb{Z}=\cup\left\{ \left( n,n+1\right) :n\in \mathbb{Z}\right\}$.

My question is: why $n$ in $\mathbb{Z}$? I think, $n$ cannot be in $\mathbb{Z}$ because we say $\mathbb{R} \backslash \mathbb{Z}$.

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    That's the definition, there is no choice being made, it just is what it is defined to be. – 2017-01-12
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    So you're saying the interval $(5,6)$ is not a subset of $\mathbb{R} \setminus \mathbb{Z}$ because 5 is an integer? – 2017-01-12
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    $(5,6)$ is an element of $\mathbb{R}\ backslash \mathbb{Z}$ because it (by definition) includes everything between $5$ and $6$ except these points by definition. – 2017-01-12

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Seems like you aren't understanding what $n\in \mathbb{Z}$ means. This is simply an element from the integers $\mathbb{Z}$. So by considering $(n,n+1)$ you are always just excluding the integers but getting everything in between. Any open interval is an open set and the union of open sets is open therefore the set in question is open.

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$\mathbb{R} \backslash \mathbb{Z}$ is $\mathbb{R}$ without $\mathbb{Z}$.

This means you have all the numbers, but the integers.

So for any $n \in \mathbb{Z}$, you have all of $]n, n+1[$.