Construct a compact set of real numbers whose limit points form a countably infinite set. I have seen examples of sets that have these properties, like:
$$A=\left\{\frac1n+\frac1 m:m,n\in\Bbb N\right\}\cup\{0\}$$
And it is clear that 0 and all 1/n are limit points. However, how does one show that there are no other limit points?
I am completely stuck here. I have found many examples of sets that have these properties, but always run into trouble showing there are no other limit points. For example, I also tried:
$$A=\{0\}\cup\left\{\frac1n:n\in\Bbb N\right\}\cup\left\{\frac{n}{kn+1}:k,n\in\Bbb N\right\}$$
And showed that 0 and all 1/n are limit points, but I am lacking in how I can show that the $n/(kn+1)$ terms are not. To me, this means showing that there is some deleted ball around each of them which contains no element of A.
Thank you so much for your help!
 A: Consider the first set $A$ in your question. Pick a sequence of points of $A$ that converges to some point in $\mathbb R$. Either the sequence has infinite intersection with some interval $[1/n, 1/(n+1))$, or it does not. 
If the first option holds, check that the sequence converges to $1/n$, because a subsequence of it does (namely, the subsequence whose values are in the interval). 
The other option is that each of these intervals contains only finitely many points of the sequence. Check then that the sequence converges to $0$.

By the way, much more complicated countable compact subsets of $\mathbb R$ are possible. The typical examples are modeled after the countable successor ordinals. See also this answer.
A: I'll tackle your first question.  Let $x$ be a limit point of $A$.  Then there exist sequences $(n_i)$ and $(m_i)$ with $1/n_i + 1/m_i \to x$ as $i \to \infty$ and $1/n_i + 1/m_i \neq x$ for all $i$.  $(n_i)$ has a subsequence along which $n_i \to \infty$ or $n_i$ is constant.  In the former case, along a subsequence, $1/n_i \to 0$, so either $m_i \to \infty$ and $x=0$ or $m_i = m \neq 0$ and $x = 1/m$.  In the latter case, assume $n_i = n$ for all $i$ so $x = 1/n + \lim_{i\to\infty}1/m_i$.  Either $\lim_{i\to\infty}1/m_i = 0$ or $\lim_{i\to\infty}1/m_i = 1/m$ for some $m \in \mathbb{N}^+$.  The latter case is impossible, for then $1/n_i + 1/m_i = x$ for some $i$.
I might have been a little sloppy with sequences vs. subsequences and subscripts but I think the argument is correct.
A: This doesn't answer your questions about the set $A$, but rather gives you an alternative way to construct a set with the desired properties. For each $m\in\Bbb n,$ let $$C_m=\left\{\frac1m+\frac1{nm(m+1)}:n\in\Bbb N\right\},$$ and let $$C=\{0\}\cup\bigcup_{m\in\Bbb N}C_m$$ You should be able to show that the sets $C_m$ lie in pairwise-disjoint intervals of the form $$\left(\frac1m,\frac2m-\frac1{m+1}\right],$$ and that each $C_m$ has only $\frac1m$ as a limit point. From these facts, you can show that $0$ is the only other limit point of the set $C$.
A: Take a sequence $(a_n)$ in $A$ that converges. For every $\epsilon$ there are only finitely many elements $a\in A$ with $|a|>\epsilon$, so either the sequence becomes eventually constant, or $a_n \to 0$.
