Real Analysis – Constructing a Compact Set with Countable Limit Points

Question

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!

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

Answer 1

What about $$A=\left\{\frac1n+\frac1 m:m,n\in\Bbb N\right\}\cup\{0\}\text{ ? }$$

One can see the $A'$ is $$\left\{\frac 1 n:n\in\Bbb N\right\}\cup \{0\}$$ Thus, let $E=A\cup A'=\bar A$ which is closed, and bounded.