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Given a set $$S = \left\{\dfrac{1}{a} + \dfrac{1}{b} : \text{ where } a, b \in \mathbf{N}\right\}\;.$$ How could I find the limit points of this set?

My idea is to consider as $a \rightarrow \infty$ and $b \rightarrow \infty$, the sum is equal to $0$. So the first limit point is $0$, but I feel like I'm computing a limit with respect to two variables rather than finding a limit point. I'm a bit confused between limit point and limit.

Limit Point definition

A number $a$ is a limit point of a set $S \subset \mathbf{R} $ if for every $\epsilon > 0$ there is $x \in S$ such that $0 < |a - x| < \epsilon$, that is the set $S \cap (a - \epsilon, a + \epsilon) \setminus \{a\}$ is not empty.

This definition is almost identical to the definition of limit ($\rightarrow \infty$). So in order show $0$ is a limit point, I have to show that for every $\epsilon > 0$, then for there exists a $N = \max(a, b)$, such that $$0 < \bigg|0 - \bigg(\dfrac{1}{N} + \dfrac{1}{N}\bigg)\bigg| < \epsilon$$

I wonder is this approach reasonable? Any idea would be greatly appreciated.

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What you did for $0$ is correct; however, $0$ is not the only limit point; you can take any $\frac 1 a$ and let $b \to \infty$ to prove that $\frac 1 a$ is a limit point as well. –  Lord_Farin Oct 13 '12 at 7:59
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2 Answers

up vote 2 down vote accepted

You have found one of the limit points correctly. As a hint for more: What happens if you fix $a$ and let $b\to\infty$?

Remains to show that these are all limit points, i.e. if $x>0$ and $\frac1x\notin \mathbb N$, the $x$ is not a limit point. For such $x$ you can find $n\in\mathbb N$ with $\frac1{n+1}<x<\frac 1n$. Then find $m$ with $\frac1m<x-\frac1{n+1}$. Now if $\epsilon$ is small enough, namely $<\frac1n-x$ and $<x-\frac1{n+1}-\frac1m$, then a great deal of $S$ are seen immediatly to be off by more than $\epsilon$: All with $a\le n$ and $b\le n$ as well was all with $a\ge m$ or $b\ge m$. This should help you see how to make $\epsilon$ a bit smaller if necessary.

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Thanks a lot for the hint. I guess I need to play with the definition a bit more. –  Chan Oct 13 '12 at 8:04
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HINTS: $\newcommand{\cl}{\operatorname{cl}}$For $n\in\Bbb Z^+$ let $$S_n=\left\{\frac1n+\frac1k:k\in\Bbb Z^+\right\}\;.$$

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

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

  3. Show that $\cl P$ is the set of accumulation points of $S$.

In visualizing $S$, you may find it helpful to show that for each $n>1$, $$S_n\setminus\left(\frac1n,\frac1{n-1}\right)$$ is finite (and you can even calculate exactly how many elements it has). In other words, $S_n$ 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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