Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to find the interior, accumulation points, closure, and boundary of the set

$$ A = \left\{ \frac1n + \frac1k \in \mathbb{R} \mid n,k \in \mathbb{N} \right\} $$ and use the information to determine whether the set is bounded, closed, or compact.


So far, I have that the interior is empty, but not sure how to prove it. My thoughts are to fix $n$ and then the accumulation points would be $\left\{ \frac 1n \mid n \in \mathbb{N} \right\}$. But I'm not sure if that is correct. Then, I believe that the boundary is $[0,2]$. Can someone confirm that?

Any help would be appreciated.

share|improve this question
    
For the boundary, consider $\frac{7}{4}$, which is not in $A$ or in $A$'s boundary, so there must be a point in the boundary which is less than it but greater than or equal to $\frac12+\frac12$ –  Henry Oct 5 '12 at 20:22
add comment

2 Answers

HINTS: $\newcommand{\cl}{\operatorname{cl}}$For $n\in\Bbb Z^+$ let $$A_n=\left\{\frac1n+\frac1k:k\in\Bbb Z^+\right\}\;.$$

  1. Clearly $A_n\subseteq A$, so every accumulation point of $A_n$ is an accumulation point of $A$; what is the unique accumulation point of $A_n$?

  2. For $n\in\Bbb Z^+$ let $p_n$ be the unique accumulation point of $A_n$, and let $B=\{p_n:n\in\Bbb Z^+\}$. Every accumulation point of $B$ is an accumulation point of $A$; why? What is the unique accumulation point of $B$?

  3. Show that $\cl B$ is the set of accumulation points of $A$.

In visualizing $A$, you may find it helpful to show that for each $n>1$, $$A_n\setminus\left(\frac1n,\frac1{n-1}\right)$$ is finite (and you can even calculate exactly how many elements it has). In other words, $A_n$ is almost a subset of the interval $\left(\frac1n,\frac1{n-1}\right)$. This makes it a lot easier to see where the accumulation points are.

share|improve this answer
add comment

Tip for the interior: a point $x \in A$ is an inner point, if and only if, there exists a neigbourhood of $x$ that is completly in $A$. $A$ is not only a subset of $\mathbb R$ but also of $\mathbb Q$. Now remember that $\mathbb Q$ is dense in $\mathbb R$.

share|improve this answer
1  
Perhaps more to the point: The complement of $\mathbb{Q}$ is dense in $\mathbb{R}$. –  Harald Hanche-Olsen Oct 5 '12 at 19:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.