In my efforts to improve my mathematical rigour I am trying to understand each precise step in the proofs I attempt as well as look at the solution for. In this instance I am trying to understand the process in establishing whether a point is a limit point in $\mathbb{R}$. As the title says I'm trying to understand how finiteness is concluded in the following proof: Find the limit points of the set $\{ \frac{1}{n} +\frac{1}{m} \mid n , m = 1,2,3,\dots \}$
In the first and second solution both authors arrive at the conclusion: $$0<\frac{1}2(x-\frac{\epsilon}2)<\frac1m$$ and $$0<\frac{\epsilon}2<\frac1n$$
How do these establish that there are only a finite number of elements in the respective sets? What am I missing in understanding finiteness?
And to follow this to make sure my reasoning is correct, once we establish that there is only a finite number of elements, then by definition we can find a $\delta$ - neighbourhood around any of the finite points such that it contains no other points of $S$ which then implies the point $x$ is not a limit point.
Both of the answers to your questions identify the set $A=\{0\}\cup\{\frac{1}{n} : n \in \mathbb{N}\}$ as the complete list of limit points of your set. Checking that these are limit points is easy, but it remains to show there are no others.
The explanations choose a point $x$ not in $A$, build a special interval around it, and then conclude the inequalities you mention for any point in this special interval which has the form $\frac{1}{n}+\frac{1}{m}$. The inequalities bound $m$ and $n$ above, which means there is a strictly finite number of points of the form $\frac{1}{n}+\frac{1}{m}$ in this special interval, and therefore no point in it can be a limit point; it it were, there would be an infinite sequence of terms of that form converging to it, which is not possible.
I hope that helps!