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.

  • $\begingroup$ To clarify, is $k$ fixed? $\endgroup$ – Madhav Nakar Nov 27 '19 at 16:33
  • $\begingroup$ Yes , k is an integer $\endgroup$ – 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 $\delta A = \bar{A} /\ Int(A)$?

  • $\begingroup$ No ..i did not learn that. How it can be related? $\endgroup$ – Almaa Nov 27 '19 at 16:25
  • $\begingroup$ 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? $\endgroup$ – Madhav Nakar Nov 27 '19 at 16:29
  • $\begingroup$ 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. $\endgroup$ – Almaa Nov 27 '19 at 16:35
  • $\begingroup$ Is it connected to that if we take k=0 then we'll get that 0 is the only limit point of A? $\endgroup$ – Almaa Nov 27 '19 at 16:47
  • $\begingroup$ Yes, take $k = 0$, then 0 is the only limit point. Similarly take $k = 1$, then 1 is the only limit point. $\endgroup$ – 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 $\operatorname{cl}(A)=A$.


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