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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.

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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

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.

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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 $\mathbb R$.

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

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