Finding Int(A) , cl(A) and A' of a set My set is A = $\{\frac{1}{n} + k : n\in\mathbb{N} , k\in\mathbb{Z}\}$ ,  X= $\mathbb{R}$ with the usual topology.
My attempt:
A point $a\in int(A)$ if A is a neighborhood of $a$ that means that there is an open set $V$ with then $a \in V \subset A$. I took $V$ to be an open ball in $\mathbb{R}$ , so for every point in $A$ , every open interval that contains a also  contains other points that are not in $A$ , that means $Int(A)= \emptyset$.
To find $A'$ (boundary of A) I took an open ball in R and got that $B(a,r)\cap A \setminus\{a\} \neq \emptyset$ , when $a \in A$ , the intersection is not empty cuz the ball contains $A$ and other points too. So that means $A' = A$.
To find $cl(A)$ for every neighborhood $U$ of $a\in A$ , $U\cap A \neq \emptyset$ (an open neighborhood in R is an open interval)  so we can say that $A$ is closed.
I'm not sure of all i solved. Would be grateful if you give me tips or correct me if i used wrong ideas.
 A: I will assume that this is for a fixed $k \in \mathbb{Z}$
It is correct that $Int(A) = \emptyset$ however you have to be careful with the reasoning. $a \in Int(A)$ if there exists an open neighborhood of $a$ that is a subset of $A$. So, you have to show for all open neighborhoods of $a$, they are not a subset of $A$. So, pick any open neighborhood, and as you pointed out, it contains points not in $A$, hence it is not a subset of $A$
The closure of $A$ would be $A \cup \{k\}$. This is because $a $ because $k$ is the only accumulation point of the set $A$
Are you familiar that the boundary of a set is equal to the closure complement the interior ie $\delta A = \bar{A} /\ Int(A)$?
A: Your set is a disjoint set of sequences, the $k$-th converging to $k$ (as $k+\frac1n \to k$ as $n \to \infty$). As in the case of the sequence $\frac1n$ (which is also in there), all points of the sequence are isolated (so not limit points of the set) and the limit ($0$) is a limit point.
You're right $A$ contains no interval so $\operatorname{int}(A) = \emptyset$ indeed. 
$A'=\Bbb Z$, the limits of the sequences and as $A' \subseteq A$, $A$ is closed and $\operatorname{cl}(A)=A$.
