Find the derived set of $\{\frac{1}{n} + \frac{1}{m}: m,n \in \mathbb{N}\}$ and prove it is such. It is easy to see the derived set is $A' = \{\frac{1}{n}: n \in \mathbb{N}\}\bigcup\{0\}$. To prove these are the only elements of the derived set we need to show that the shape of the derived set can only be $\frac{1}{n}$ or $0$. We can see the derived set is bounded above by $1$ and below by $0$. So we look for points between $0$ and $\frac{1}{N}$ where $N$ is the largest number and $\frac{1}{n}$ and $\frac{1}{n+1}$. I know I need to show that only a finite number of points exist between these points but I am having trouble doing so.
 A: It is I think clear that are no negative limit points. Let $b$ be a positive real other than $0$ or a fraction $\frac{1}{n}$. We will show that $b$ cannot be a limit point of $A$. Note that either (i) $b\gt 1$, or (ii) there exists a positive integer $q$ such that $\frac{1}{q+1}\lt b\lt \frac{1}{q}$.  
We deal with Case (ii) because it feels marginally harder. Let $\epsilon$ be the smaller of $\frac{1}{q}-b$ and $b-\frac{1}{q+1}$. We claim there are only finitely many numbers of the form $\frac{1}{m}+\frac{1}{n}$ at distance less than $\epsilon/2$ from $b$.
To get within $\epsilon/2$ of $b$, we must use two integers $m$ and $n$ each $\ge q+1$. Let $m$ be the smaller of the two integers. Then $m$ must be $\lt 2(q+1)$. So there are only finitely many possibilities for $m$.
For any such $m$, there are only finitely many $n$ such that $\frac{1}{m}+\frac{1}{n}\gt \frac{1}{q+1}+\epsilon/2$, so only finitely many within $\epsilon/2$ of $b$.
This completes the argument for Case (ii). For Case (i) we do the same thing, except instead of $q+1$ we use $1$.
