Solution: Consider the set \[ A = \{ i + \frac{1}{n} : i = 1,2,3 \text{ and } n \in \mathbb{N} \} . \] If $A^\prime $ denotes the set of all limit points of $A$, then it is clear that $\{ 1,2,3 \} \subset A^\prime $ as every deleted neighbourhood of $1,2$ or $3$ contains a point of $A$. To be more precise, let $\varepsilon >0 $ and $(1-\varepsilon , 1 + \varepsilon )$ be a neighbourhood of $1$. By the Archimedean property, there exists a natural number $n$ such that $\frac{1}{n} < \varepsilon $. Thus, \[ 1 - \varepsilon \lt 1 + \frac{1}{n} \lt 1 + \varepsilon. \] Similarly, for other points. Now we need to show the converse. If $x \notin \{ 1,2,3 \} $. We will find a deleted neighbourhood of $x$ which does not contain any point of $A$. Take $\delta = \min \{ \vert x-1 \vert , \vert x-2 \vert , \vert x-3 \vert \} $.

Consider the set \[ U = \left\{ s \in \mathbb{R} : \vert x-s \vert \lt \frac{\delta}{2} \right\} \] Since the set \[ V = \bigcup_{i=1}^{3} \left( i, i + \frac{\delta }{2} \right) \] contains all but finitely many points of $A$ and it is disjoint from $U$, the set $U$ contains only finitely many points of $A$. Let $a_1, a_2, \dots, a_k$ are the points of $A \cap U$. \[ \gamma = \min \{ \left\vert x - a_i \right\vert , \frac{\delta }{2},~ x\neq a_i \}. \] Then the neighbourhood \[ W = \left\{ y \in \mathbb{R} :\vert y - x \vert \lt \gamma \right\} \] contains no point of $A$ except possibly $x$. Therefore, $x$ is not a limit point of $A$ and hence, $A^\prime = \{ 1,2,3 \} $.