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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.

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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 $1$.

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  • $\begingroup$ Very clear explanation, this got me over the last hump! $\endgroup$
    – Chris
    Commented Nov 16, 2015 at 13:52
  • $\begingroup$ Actually, I don't understand it! I get lost during the third paragraph. I see each of $m,n \geq q + 1$. But they could be very very large. Why then must the smaller be less than $2(q+1)$ while these increase without bound? $\endgroup$
    – Chris
    Commented Nov 16, 2015 at 17:37
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    $\begingroup$ Suppose that the smaller of $m$ and $n$ is $\ge 2(q+1)$. Then the sum $\frac{1}{m}+\frac{1}{n}$ is $\le \frac{1}{2(q+1)}+\frac{1}{2(q+1)}$, that is, the sum is $\le \frac{1}{q+1}$. But the distance of $\frac{1}{q+1}$ from $b$ is, by definition of $\epsilon$, greater than or equal to $\epsilon$. Much more informally, if both $m$ and $n$ are very very large, then the sum of their reciprocals is smaller than $\frac{1}{q+1}$, and therefore nowhere near $b$. The $\frac{1}{2(q+1)}$ makes explicit what very very large means. $\endgroup$ Commented Nov 16, 2015 at 17:43
  • $\begingroup$ But I thought we wanted to find exactly that, a sum $\leq \frac{1}{1+q}$? $\endgroup$
    – Chris
    Commented Nov 16, 2015 at 17:47
  • $\begingroup$ No, recall that $b$ is between $\frac{1}{q+1}$ and $\frac{1}{q}$, say between $\frac{1}{43}$ and $\frac{1}{42}$. We want to show there are only finitely many sums of reciprocals that are "close" to $b$. We can't use numbers both $\ge \frac{1}{86}$, for then the sum of their reciprocals is $\e \frac{1}{43}$, which is not near enough to $b$. So one of $m$ and $n$ must be between $\frac{1}{86}$ and $\frac{1}{43}$. For any choice of these, there are only finitely many $\frac{1}{n}$ that have a chance of getting us close enough to $b$. $\endgroup$ Commented Nov 16, 2015 at 17:54

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