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!
$$E = \left\{\frac 1{2^m}\left(1 - \frac 1n\right) \mid m,n \in \mathbb N\right\}.$$
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.
The idea you used is good. I will do the same thing more slowly, building the set step by step in order to retain control over the geometry.
The backbone of the set is, like yours, the sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$. This sequence has limit $0$, so we throw in the number $0$.
Now near $\frac{1}{n}$ for every $n$, we produce a sequence that has limit $\frac{1}{n}$, and that does not mix in unpleasant ways with other sequences. Note that the distance from $\frac{1}{n}$ to $\frac{1}{n+1}$ is $\ge \frac{1}{2n^2}$.
So for every $n$, throw in the numbers $\frac{1}{n}+\frac{1}{(2n^2)(2^m)}$, where $m\ge 1$.
Another way: Start with the set $A$ of numbers $n+\frac{1}{2^m}$, where $n$ and $m$ range over the positive integers. This set does the job beautifully. Only one minor flaw: the set is not compact. Let $B$ be the set of all reciprocals of numbers in $A$, together with $0$. This does the job.
What about the trivial unitary set as in $\{ 0 \}$?
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.
After 11 years, I can prove the simplest solution works.
Let $$ A = \left\{\frac 1n+\frac 1m:m,n\in\Bbb N\right\}, B = \left\{\frac 1n:n\in\Bbb N\right\} $$ and $$E = A \cup B \cup \{0\}$$
It suffices to prove $E$ is compact, because then it is closed, and since it's countable, so are its limit points.
Let $O$ be an open cover of $E$. Any set in it containing $\frac 1n,n\in\Bbb N$ is open, so it will contain every $\frac 1n < x < \frac 1n+ \varepsilon$ for some $\varepsilon > 0$. Due to the Archimedean Property, there exists a natural $m > \frac 1 \varepsilon$. Thus there are only a finite number of $m$'s where $m < \frac 1 \varepsilon$, or $\frac 1m > \varepsilon$. This means for any $\frac 1n \in E,n\in\Bbb N$, the set of $O$ containing $\frac 1n$ contains every $\frac 1n + \frac 1m$ except for a finite number of $m\in\Bbb N$.
The set in $O$ containing $0$ is also open, so it must contain every $0 < x < \varepsilon$ for some positive $\varepsilon$. Using the same reasoning, this set must contain every element of $B$ except for a finite amount. Let $C$ be every element of $B$ that is not in the set in $O$ containing $0$. $O$ must contain a set surrounding every element of $C$ too, meaning $B$ and $\{0\}$ have a finite subcover $F$ in O.
The sets of $O$ containing each element $n$ of $C$ must also contain every $\frac 1n + \frac 1m$ except for a finite number. This means in total, there are only a finite number of these not in the subcover, so add the set of $O$ containing each of them to $F$.
This transitivity doesn't apply to the set of $O$ containing $0$, since its elements are countable, not finite. We've established that for each $n$ in this set, there must be finite $m$'s where $\frac 1n + \frac 1m$ is not in the set of $O$ containing $\frac 1n$. Otherwise, $F$ obviously contains $\frac 1n + \frac 1m$.
Let $M$ be the max value of these finite leftovers: the maximum $m$ for each element $\frac 1n$ of the set of $O$ containing $0$ where $\frac 1n + \frac 1m$ is not also in the set. $M$ must in turn have a maximum value $s$. Thus any $\frac 1n + \frac 1m$ is covered by $F$ if $m > s$, or conversely $n > s$. There are obviously only a finite number of $m, n$ pairs that do not satisfy this, so adding $\frac 1n + \frac 1m$ to $F$ will produce a finite subcover of $E$.
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
