Construct a compact set of real numbers whose limit points form a countable set. I searched and found out that the below is a compact set of real numbers whose limit points form a countable set.
I know the set in real number is compact if and only if it is bounded and closed.
It's obvious it is bounded since $\,d(1/4, q) < 1\,$ for all $\,q \in E.$
However, I'm not sure how this is closed.
Is there any simpler set that satisfies the above condition?
Thank you!
$$E = \left\{\frac 1{2^m}\left(1 - \frac 1n\right) \mid m,n \in \mathbb N\right\}.$$

The limit points are $\{\frac{1}{2^m}\mid m\in \mathbb{N}\}$. These are contained in the set (to get $\frac{1}{2^k}$ (for $k>1$), take $m=k-1$, $n=2$). 
We can tell there are no other limit points, since the closest points to $\frac{1}{2^k}(1-\frac{1}{l})$ (for $l>2$) are $\frac{1}{2^k}(1-\frac{1}{l+1})$ and $\frac{1}{2^k}(1-\frac{1}{l-1})$, so we can isolate them in a neighborhood of radius $\frac{1}{2^k}(1 - \frac{1}{2(l+1)})$.
Edit: As Andre has pointed out, $1/2$ is not in the set, so the problem does not work as stated.
