An exercise in my Analysis book was to find limit points $X=\{ \frac{1}{n} +\frac{1}{m} \mid n , m \neq 0 , n,m \in\mathbb{Z} \}$. I came up with the answer $\{\frac{1}{n} \mid n \in \mathbb{Z} \setminus {0} \} \cup \{0\}$, and I wish to verify this proof.
(Note: In my book limit points of $X$ are points $x$ that satisfy for all $\epsilon>0$, deleted $\epsilon$ neighbourhoods of $x$ and $X$ have non-empty intersections)
It's pretty clear that $\{\frac{1}{n} \mid n \in \mathbb{Z} \setminus {0} \} \cup \{0\}$ are limit points. I will prove that these are the only ones. Please check if this is correct!
First Step :
Let $X_n=\{ \frac{1}{n} +\frac{1}{m}\ \mid |m|\ge |n|, m \in \mathbb{Z}\}$ for fixed, non-zero $n$. Let us prove that the only limit point of $X_n$ is $\frac{1}{n}$.
$\frac{1}{n}$ is a limit point because of the Archimedean property. If $x$ is a limit point that is not $\frac{1}{n}$, let $\dfrac{1}{x-\frac{1}{n}}$ have $m$ and $m'$ as the closest integers that are not itself. Let $\epsilon_{0}$ be smaller than the minimum of the difference between $x-\frac{1}{n}$,$\frac{1}{m}$ or $x-\frac{1}{n}$, $\frac{1}{m'}$. Then $B_{\epsilon_{0}}(x) \setminus \{x\}$ has an empty intersection with $X_n$.
Second Step:
Let $x>0$ be a positive limit point. Let $Y_n$ denote set of natural $m$'s for which $X_m$ and $B_{\frac{1}{n}}(x) \setminus \{x\}$ have non-empty intersections. $Y_n$ is not empty for any $n$ since $x$ is positive and is a limit point. By the well-ordering property, the sequence $y_n=\min Y_n$ is well defined. Note that $Y_{n+1} \subset Y_n$ so we have $y_n \le y_{n+1}$. If $y_n$ is bounded, it thus converges to a natural number $y$ as all $y_n$ are natural numbers. We may say that $x$ is a limit point of $X_y$, and by Step 1 we have that $x=\frac{1}{y}$. If $y_n$ is not convergent and unbounded, $y_n \to \infty$. But for $k > \frac{4}{x}$, note that $X_{k}$ lies between $[0, \frac{2}{k})$, while for $\epsilon<\frac{x}{2}$, $\epsilon$ neighbourhoods lie in $(\frac{x}{2}, \infty)$. Thus a contradiction, as $Y_k$ is empty if $k > \frac{4}{x}$. We repeat a similar logic for $x<0$. Thus, every limit points of $X$ are of the form $\{\frac{1}{n} \mid n \in \mathbb{Z} \setminus {0} \} \cup \{0\}$
(This seems similar to here, but $n$, $m$ are non-zero integers, not naturals, and I want to verify my own proof.)