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

Question

This is exercise $2.13$ in Rudin.

Can't we simply define such set as $[a, b]$, with all members being rational? It is bounded, and closed (proof is straightforward), and the limit points are all members of the set since $\mathbb{Q}$ is dense on an interval; and since rational numbers are countable then all the limit points are. Am I missing something? Thanks in advance.

Answer 1

if you want limit points should be finite then {1/n| n $\in \mathbb{N} $}$\cup$ {0} in case of limit points countably infinite:- for each n$\in \mathbb{N}$ define $M_n$ = { 1/n +1/r | r$\in \mathbb{N}$ and 1/r < 1/(n-1) - 1/(n)} $\cup$ {1/n}, define $ M = \bigcup_{n>1} M_n$. claim: the required set is $M \cup {0}$... this set is compact since closed and bounded and set of limits points are {1/n| n $\in \mathbb{N} $}$\cup$ {0} which is countable

Answer 2

You have stated why this doesn't work. The limit points comprise the entire interval $[a,b]$, which is not a countable set.