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)?
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 $\mathbf{Z}$.
One point of view is that it's closed (complement is open), and if $\mathbb  Z $ were clopen, then $\Bbb R$ wouldn't be connected.