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I am stuck on the following problem:

The number of limit points of the set $\left\{\frac1p+\frac1q:p,q \in \Bbb N\right\}$ is which of the following:

  1. $1$

  2. $2$

  3. Infinitely many

  4. Finitely many

If I take $p$ to be fixed (say=$k$) and let $q \to \infty$, then the limit point is given by $\frac{1}{k}$. Since $k$ is an arbitrary natural number, the number of limit points is infinite. The same case can be continued after taking $q$ to be fixed (say=$k_1$). I think option 3 is the right choice. Am I on the right track? Can someone give further explanation?

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  • $\begingroup$ When fixing $p$, why give it a name other than $p$? Also, fixing $q$ is the same as fixing $p$ - it doesn't get you any new limit points. $\endgroup$ – Thomas Andrews Jun 25 '13 at 12:38
  • $\begingroup$ But your answer is correct. There are infinitely many limit points. There is only $1$ limit point that is not in the set, however - note that $1/q$ is in your set since $\frac{1}{q}=\frac{1}{2q}+ \frac{1}{2q}$. $\endgroup$ – Thomas Andrews Jun 25 '13 at 12:40
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    $\begingroup$ Nice observation, @Thomas! Make it an answer? $\endgroup$ – Cameron Buie Jun 25 '13 at 12:41
  • $\begingroup$ A COMPLETE PROOF IS HERE: math.stackexchange.com/questions/930646/… $\endgroup$ – Gregory Grant Apr 2 '15 at 14:01
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Your answer is correct. You have shown that there are infinitely many limit points, which is enough information to answer the multiple choice question.

You can prove that the set of limit points of this set is $\{0\}\cup\{1/k:k\in\mathbb N\}$. Since $\frac{1}{k} = \frac{1}{2k} + \frac{1}{2k}$, that means that $0$ is the only limit point that is not in your original set.

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I think the answer is $\{0\}$ because limit points means nbd of open interval contain at least one points other than point. Let $k=1,2,3,...$ then $\frac{1}{k} = 1,1/2,1/3...$ the nbd of $\{0\}$contained in an open interval other than $\{0\}$. So, the limit point of $\frac{1}{k}$ is $\{0\}$.

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    $\begingroup$ Hello, and welcome to Math StackExchange. You might want to clarify your answer a little bit: what are you defining "nbd" to be? $\endgroup$ – daOnlyBG Jan 17 '15 at 19:56

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