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

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?

• 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. – Thomas Andrews Jun 25 '13 at 12:38
• 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 $\frac{1}{q}=\frac{1}{2q}+ \frac{1}{2q}$. – Thomas Andrews Jun 25 '13 at 12:40
• Nice observation, @Thomas! Make it an answer? – Cameron Buie Jun 25 '13 at 12:41
• A COMPLETE PROOF IS HERE: math.stackexchange.com/questions/930646/… – Gregory Grant Apr 2 '15 at 14:01

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.
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 $\{0\}$.