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!
$$E = \left\{\frac 1{2^m}\left(1 - \frac 1n\right) \mid m,n \in \mathbb N\right\}.$$

For each positive integer m, put
Aₘ = {2⁻ᵐ} ∪ {2⁻ᵐ+2⁻ⁿ , m,n∈ N⁺}.
Am has only one limit point in 2⁻ᵐ. Put
A = {0} ∪ (∪₁infinity Aₘ)
For any fixed positive integer m, since
2⁻ᵐ⁻¹+3⁻ⁿ<2⁻ᵐ
it is clear that A has no other limit points apart from the limit points
S={0,1/2,1/4,...................}
Apparently, S is countable.
It remains to prove that A is compact. Since A ⊂ [0, 1], A is bounded. Also, S ⊂ A
implies that A is closed
