Constructing a set with exactly three limit points

Solution 1:

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 $\epsilon = \frac{1}{2} (\alpha - \frac 1 n)$.

Solution 2:

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.