# Set of all limit points of the set $S=\{n+\frac{1}{3m^{2}};\,n,m \in \mathbb{N}\}$

I came across the problem that says:

Find the set of all limit points of the set $S=\left\{n+\frac{1}{3m^{2}};\,n,m \in \mathbb N\right\}$. There are four options:

(a) $\mathbb{Q}$.

(b) $\mathbb{Z}$.

(c) $\mathbb{R}$.

(d) $\mathbb{N}$.

Could someone point me in the right direction? Thanks in advance for your help.

-
Can't read this. You'll have to add latex if you want answers. – Rudy the Reindeer Dec 7 '12 at 6:45
And maybe, mayyyybe, what you think and what you tried? – Did Dec 7 '12 at 6:49
case1. put n=fixed and vary m through 1,2,3....Then n becomes a limit point of S. Case 2. Put m=fixed and vary n through 1,2,...Then what will happen? and also i am confused about case 3 where both m and n vary through 1,2,3,4... – learner Dec 7 '12 at 6:54

If you want just the answer, since $x>0, \ \ \forall x \in S$ eliminate a,b,c.
To prove that any element $n \in \mathbb N$ is a limit point of $S$ find a sequence $(x_m)_{m\in \mathbb N}$ in $S$ with $x_m\neq n, \ \forall m \in \mathbb N$ and $x_m\xrightarrow[m\to\infty]{}n$.
To prove that an element $r \not\in \mathbb N$ is not a limit point of $S$ find an $\epsilon>0$ small enough s.t. $(r-\epsilon,r+\epsilon)\cap S$ is finite.