Constructing a set with exactly three limit points 
Construct a set of real numbers having exactly three limit points.

Progress: 
I know the set $\{ 1/n   + k\}$ has one limit point $k$, but I am unable to justify why.
Please be a bit elaborate in your explanation because I need to understand this question if I face such a question in my exam.
 A: Take a convergent non constant sequence $(a_n)_{n\in \mathbb{N}}$. Then 
the set $\{ a_n : n \in \mathbb{N}\}$ does have exactly one limit point.
Now consider 
\[\bigcup_{i=1}^k \{a_n + i: n \in \mathbb{N}\}\]
this does have exactly $k$ limit points.
A: Hint: The set $\{\frac{1}{n}\mid n\in\mathbb{N}^+\}$ has a single limit point. Can you see how to construct a set which has three limits points from this?
A: Consider the set $A = \{\frac 1 n : n \in \Bbb{Z}^+\} = \{1, 1/2, 1/3, 1/4, ...\}$. This has exactly one limit point, namely $0$.
Next consider the set $\{1 + \frac{1}{n} : n \in \Bbb{Z}^+\}$. This has exactly one limit point, namely $1$. Now consider the union of these two sets: Do you see why it has exactly two limit points? Do you see how to extend this to having $3$, or $n$, limit points?

To show that $A$ actually has $0$ as a limit point, just note that if given $\epsilon > 0$, there exists an $n$ for which $\frac{1}{n} < \epsilon$. So every neighborhood of $0$ intersects $A$ at a point.
Now suppose that $\alpha$ is a limit point of $A$. If $\alpha < 0$, set $\epsilon = \frac{-\alpha}{2}$, and note that $(\alpha - \epsilon, \alpha + \epsilon) \cap A = \emptyset$, a contradiction. Likewise, $\alpha > 1$ leads to a contradiction. If $\alpha = 1$, set $\epsilon = \frac{1}{3}$.
Finally, if $0 < \alpha < 1$, let $n$ be the least integer satisfying $\frac{1}{n} < \alpha$ and choose $\epsilon = \frac{1}{2} (\alpha - \frac 1 n)$.
A: You can prove no other element in the set is a limit point because you can let 1/n be any of the elements and show that there is an r that gives you a neighborhood that doesn't have any of the other the elements in it. Because to be a limit point you have to have another element in the set in the neighborhood. for example, if you have 1/5 then you can make a neighborhood where r is really small then 1/4, and 1/6 wont be in the set and there is no other numbers in the neighborhood. So you would do 1/n-1 and 1/n and 1/n+1 and show that for a very small r like half the distance between 1/n-1 and 1/n then there wont be any elements. It must be for EVERY neighborhood. 
Correct me if I'm wrong.
A: Take the sequence $x_{n+1} = x_n + 1 \mod 3$, with $x_1 =0$. Then take $a_n = x_n + \frac{1}{n}$.
