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.
Best Answer
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.