Why isn't $\mathbb{Z}$ an open set of the set of real numbers in the standard topology? In class my professor said that the interior of $\mathbb{Z}$ is the empty set. That is because the interior is the union of all open sets contained in $A$. But why is it not an open set?
Can someone explain (without using metric definitions because we have not covered that)?
 A: If you take an open interval centered around an integer $n$, it looks like $(n-\varepsilon, n + \varepsilon)$ for some real number $\varepsilon$. But $(n-\varepsilon, n + \varepsilon)$ will never be contained in $\mathbf{Z}$ for any $\varepsilon$; there will always be a non-integer real number in $(n-\varepsilon, n + \varepsilon)$. So $n$, which was arbitrarily chosen, cannot be in the interior of $\mathbf{Z}$.
A: Standard topology on $\mathbb{R}$ is generated by the open intervals $(a, b)$ for $a, b\in\mathbb{R}$. Therefore
$$
A = \bigcup_{k\in\mathbb{Z}} (k, k+1)
$$
is an open set. However, $A$ is the complement of $\mathbb{Z}$. Therefore, $\mathbb{Z}$ is closed. Finally, the only clopen sets in the standard topology are $\emptyset$ and $\mathbb{R}$ (a hallmark of a connected space). Therefore, $\mathbb{Z}$ is not open.
A: If you know that a base for the standard topology on $\mathbb R$ is the collection of all intervals of the form $(a,b): a<b$ then it's easy to show that $\mathbb Z$ is not open, for if so, then there would be a basis element $(a,b)$ of the aforementioned type such that $(a,b)\subseteq \{0\}$. Using the definition of open set, this would mean that there are $a<c<d<b$ such that $(c,d)\subseteq (a,b)$ and now $c=0=d$ which is impossible since $c<d.$
