Let $S=\{-m + {1\over n}\}$ where $m ,n \in \mathbb N$ , and prove that $\{-1,1\}$ are the limit points.
My approach : I understand what a limit point is - that every neighbourhood of the point contain some point of Set $S$.
if $n\to\infty$ and $m=1$ then surely $-1$ is the limit point as every neighborhood of $-1$ should contain some point of $S$.
I am not able to prove the same for $1$ , as I can find a neighborhood $[0.5,2]$ which has no element of $S$.
Is my logic correct? also how do I develop a mathematically rigorous proof of same?
The set $L$ of limit points of $S$ is $-\mathbb N = \{-m : m \in \mathbb{N}\}$.
Indeed, $x_n = -m + \frac{1}{n}$ converges to $-m$.
Indeed, let $a \in L$. Note that $S \subseteq \bigcup_{m \in \mathbb N} J_m$, where $J_m=[-m,-m+1)$ are disjoint intervals. Therefore, there is exactly one interval $J_m$ containing $a$. But in $J_m$ the elements of $S$ form a monotone sequence converging to $-m$. Therefore, $a=-m \in -\mathbb N$.