Constructing a compact set with countably infinite many limit points I have an exercise to construct a compact set with countably infinite many limit points.
I am trying to use the set:
$$A = \{0\} \cup \{\frac{1}{n}: n=1,2,3,\ldots \}\cup \{\frac{1}{n}+ \frac{1}{m}:n=1,2,3,\ldots ; m =n+1,n+2,\ldots\}$$
The point $0$ and $\frac{1}{n}$ are clearly limit points for all $n$. I am having trouble showing that these are in fact the only limit points. I believe they are, and if I can show these are the only limit points it follows that $A$ is bounded and contains all its limit points therefore it is compact.
Any help is appreciated!
 A: Could you not simply put the indiscrete topology on a countably infinite set?
It would be trivially compact and every point is a limit point.
A: Do you want to construct a compact subset $K$ of $\mathbb{R}$ that
contains countably infinite many limit points? 
Sketch of construction: For each $n\in\mathbb{N}$, let $\delta_{n}=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+1})$
and let 
$$
A_{n}=\{\frac{1}{n+1}+\frac{\delta_{n}}{k}\mid k\in\mathbb{N}\}\cup\{\frac{1}{n+1}\}.
$$
Finally, define $K=\{0\}\cup\bigcup_{n=1}^{\infty}A_{n}$.
It can be verified that $K$ is compact and the set of limit points
of $K$ is precisely $\{0\}\cup\{\frac{1}{n+1}\mid n\in\mathbb{N}\}$. 
A: (I). Exercise: $\{0\}\cup \{1/m: M\in \Bbb N\}$ is a  closed set.
(II).For $n\in \Bbb N$ let $S(n)=\{1/n+x: x=0 \lor n\leq 1/x\in \Bbb N \}.$   Let $B= \{0\}\cup (\;\cup_{n\in \Bbb N}S(n)\;).$ 
Obviously $\{0\}\cup \{1/n:n\in \Bbb N\}\subset B\subset A$. 
We also have $A\subset B.$ For if  $a\in A$ and $a\ne \{0\}\cup \{1/n:n\in \Bbb N\} $ then $a=1/U+1/V$ with $U,V\in \Bbb N .$ Then $a\in S(\min (U,V)).$.... Therefore $B=A$. 
(III). Let  $(x_j)_{j\in \Bbb N}$ be a  sequence in $A$  converging to  $L.$  For $n\in \Bbb N$ let $T(n)=\{j\in \Bbb N: x_j\in S(n)\}.$
(III-1). If  $T(n)$ is finite for every $n$ then for any $r>0$ take $n_r\in \Bbb N$ such that $1/n_r<r/2.$ Then  $\{j\in \Bbb N:x_j\geq r\}$ is a subset of the finite set $\cup \{T(n):n<r_n\} $  . Therefore $L=0.$ 
(III-2). If $T(n_0)$ is infinite let $T(n_0)=\{f(n):n\in \Bbb N\}$ where $f:\Bbb N \to \Bbb N$ is strictly increasing. Let $x_{f(n)}=1/n_0+y_n$ where $y_n=0$ or $n_0\leq 1/y_n\in \Bbb N.$ Then $L-1/n_0=\lim_{n\to \infty}y_n.$ By the Exercise, $L-1/n_0\in \{0\}\cup \{1/m:m\in \Bbb N\}.$ Therefore $L\in A.$
