Let $$S=\left\{\frac{1}{n}+\frac{1}{m}:m,n\in{\Bbb N}\right\}$$ and $S'$ be the set of limit points of $S$. All the results I've found on Google or Math.SE only give the following $$ \left\{\frac1n:n\geq 1\right\}\cup\{0\}\subset S'. $$

Here is my question: Is $$ \left\{\frac1n:n\geq 1\right\}\cup\{0\}\supset S' $$ also true?

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    $\begingroup$ Yes; here's a sketch: Convince yourself the only limit points must be in $[0,1]$. Now pick a point in $[0,1]$. If it's of the form $1/n$ or $0$, then we're done. Otherwise, the point must fall in $(\frac{1}{k+1}, \frac{1}{k})$ for some $k \in \mathbb{N}$. Finally, convince yourself we cannot get arbitrarily close to such a point. (Posted as a comment since many details are omitted.) $\endgroup$ – Benjamin Dickman Oct 4 '13 at 1:01
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    $\begingroup$ But $S$ contains $2$? $\endgroup$ – Tunococ Oct 4 '13 at 1:12
  • $\begingroup$ @Tunococ: according to wikipedia it is demanded that any neighbourhood of the limit point contain a point of the set other than the limit point if that happens to lie in the set (so points like 5/6 etc are also excluded). $\endgroup$ – doetoe Oct 4 '13 at 1:18
  • $\begingroup$ @doetoe Oh you're right. That definition certainly makes more sense in the context of this problem. $\endgroup$ – Tunococ Oct 4 '13 at 8:35
  • $\begingroup$ A COMPLETE PROOF IS HERE: math.stackexchange.com/questions/930646/… $\endgroup$ – Gregory Grant Apr 2 '15 at 14:01

I think it is. Since we are dealing with a subset of a metrical space, the limit points are real numbers that are the limit of a sequence of elements in that set, unequal to the limit itself.

Such a sequence $(a_i)$ is of the form $({1\over n_i} + {1\over m_i})$, so we have two sequences of natural numbers $(n_i)$ and $(m_i)$ (not uniquely determined by $(a_i)$ but that doesn't matter).

If one of these can get arbitrarily large, say a subsequence of $(n_i)$ goes to infinity, then a subsequence of $(a_i)$ goes to the limit of the corresponding subsequence of the sequence $({1\over m_i})$, which exists, because otherwise the original sequence wouldn't have a limit either. This gives the limit points of the form ${1\over m}$.

If neither grows arbitrarily large, there is only a finite number of possibilities for $a_i = {1\over n_i} + {1\over m_i}$, so the limit must be an element of the sequence, and by definition is not a limit point.


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