# 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.

• To clarify, is $k$ fixed? – Madhav Nakar Nov 27 '19 at 16:33
• Yes , k is an integer – Almaa Nov 27 '19 at 16:36

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 $$\delta A = \bar{A} /\ Int(A)$$?

• No ..i did not learn that. How it can be related? – Almaa Nov 27 '19 at 16:25
• Then I am assuming you learned that the definition of the boundary is that it is a set of points where every neighborhood of those points contains a point in $A$ and a point not in $A$. Am i correct? – Madhav Nakar Nov 27 '19 at 16:29
• we defined a to be a limit point of A (in Topological space) if for every neighborhood U of a : $A\cap U \{a} \neq \emptyset$.. what you said seemsto be correct cuz the intersection would not be empty. – Almaa Nov 27 '19 at 16:35
• Is it connected to that if we take k=0 then we'll get that 0 is the only limit point of A? – Almaa Nov 27 '19 at 16:47
• Yes, take $k = 0$, then 0 is the only limit point. Similarly take $k = 1$, then 1 is the only limit point. – Madhav Nakar Nov 27 '19 at 17:02

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 $$\operatorname{cl}(A)=A$$.