Give an example of a set that has $\{0,1,1/2,1/3,...\}$ as its accumulation points. Give an example of a set that has $\{0,1,1/2,1/3,...\}$ as its accumulation points. My guess is that I can take, sequences that converge to each of the limits so in particular, I could take $x_{n1}=(1/n),x_{n2}=1-1/n,x_{n3}=1/2-1/n$ and in general $x_{nk}=1/(k-1)-1/n.$ The set $\{\{x_{nk}\}|k=1,2,3...\}$ would give me the desired set of accumulation points. Does this work?
 A: Let us consider a bijective map $\varphi:\mathbb{N}^+\to \mathbb{N}^+\times\mathbb{N}^+$, $\varphi(n)=\left(\varphi_1(n),\varphi_2(n)\right)$ and define the sequence $\{a_n\}_{n\geq 1}$ through
$$ a_n = [0;\varphi_1(n),\varphi_2(n)] = \frac{1}{\varphi_1(n)+\frac{1}{\varphi_2(n)}}.$$
It is pretty clear the continued fractions of the form $[0;1,\text{whatever}]$ accumulate towards $[0;1]=1$, the continued fractions of the form $[0;2,\text{whatever}]$ accumulate towards $[0;2]=\frac{1}{2}$ and so on. Zero is also an accumulation point for the continued fractions of the form $[0;m,m]$.
The structure of the ordinary continued fractions allows to state that $0,1,\frac{1}{2},\frac{1}{3},\ldots$ are the only accumulation points of our sequence. Your construction does not immediately allow to state the same.

This should be well-known, if it is not, in your case, you may prove it as an interesting sub-exercise.
We may take $\varphi$ as
$$ \varphi(n)=\left(\left\lfloor\frac{\sqrt{8n-7}+1}{2}\right\rfloor,n-\frac{1}{2}\left\lfloor\frac{\sqrt{8n-7}+1}{2}\right\rfloor\left(\left\lfloor\frac{\sqrt{8n-7}+1}{2}\right\rfloor-1\right)\right). $$
A: Let  $$\{\frac{1}{n}|n \in \Bbb{N}\}\cup \{0\}=A$$
Take the set $$A+A=\{x_1+x_2|x_1,x_2 \in A\}$$
A: Another interesting natural example: 
Let $A_m=\left\{  \frac{1}{m} -\frac{1}{n}\cdot(\frac{1}{m}-\frac{1}{m+1}) \right\}_{n\ge2}$
Observe that $\frac{1}{n}\cdot[\frac{1}{m}-\frac{1}{m+1}]$ is always less than $\frac{1}{2} \cdot[\frac{1}{m}-\frac{1}{m+1}]$ which is the half the distance between $\frac{1}{m}$ and $\frac{1}{m+1}$.
Now $A= \cup_{m=1}^{\infty} A_m$ is the desired set. 
