compact set of real numbers whose limit points form a countable set

Question

I came up with the following example of a set that I believe is compact and whose limit points form a countable set.

Let $E_k=\cup_{n=2}^\infty \{\frac{1}{n}+k\}\cup \{k\}. $ Note that for each $k$, $E_k$ is a compact set and the set of limit points of $E_k$ is precisely $\{k\}$.

Then $E=\cup_{k=0} ^\infty E_k$ is a compact set (I want to say this is definitely true) whose limit points are the set of nonnegative integers, which is a countable set.

Does this example work?? Thank you!!

Answer 1

Your set $E$ is not compact, because it’s unbounded. However, a modification of it will work. For $k\ge 2$ let

$$E_k=\left\{\frac1k\right\}\cup\left\{\frac1k+\frac1m:m>k(k-1)\right\}\;,$$

and let $E=\{0\}\cup\bigcup_{k\ge 2}E_k$; I’ll leave it to you to show that this works.