Closure, boundary and interior 
Describe the interior, closure and boundary of the following sets in the real line:

*

*the set of all integers


*the set of all rationals


*the set of all irrationals


*$(0,1)$


*$[0,1]$

Could someone help me through this problem?
 A: Let's do the integers together and then you can try the other ones on your own:
(i) $\operatorname{int}{\mathbb Z}$:
By definition the interior of a set $S$ are all the points $x$ in it such that we can find an $\varepsilon > 0$ such that $B(x,\varepsilon) \subset S$.
Let $\varepsilon > 0$ and $n \in \mathbb Z$. Our $\varepsilon$ is small but we know that if we pick a $k \in \mathbb N$ large enough then we can achieve $\frac{1}{k} &lt \varepsilon$. But then $n + \frac{1}{k} \in B(n, \varepsilon)$ and $n + \frac{1}{k} \notin \mathbb Z$, so $B(n,\varepsilon) \not\subset \mathbb Z$. So the interior of $\mathbb Z$ must be empty.
(ii) $\overline{\mathbb Z} \subset \mathbb R$:
We know that $\overline{\mathbb Z} = \partial \mathbb Z \cup \operatorname{int}{\mathbb Z}$ so let's find out what the boundary $\partial \mathbb Z$ is.
(iii) $\partial \mathbb Z$:
By definition, $\partial \mathbb Z$ are all points $n$ such that every $B(n, \varepsilon)$ has non-empty intersection with both, $\mathbb Z$ and $\mathbb R \setminus \mathbb Z$. Let $n$ be any point in $\mathbb Z$ and $B(n, \varepsilon)$ an epsilon ball around it. Then the intersection $\mathbb Z \cap B(n, \varepsilon)$ contains $n$ hence is non-empty. Also, in (i) we saw that every such epsilon ball has non-empty intersection with $\mathbb R \setminus \mathbb Z$. So $\partial \mathbb Z = \mathbb Z$.
