I am reading the solution of user Gregory Grant from this thread(Find the limit points of the set $\{ \frac{1}{n} +\frac{1}{m} \mid n , m = 1,2,3,\dots \}$) and I am having trouble following the argument.
Zero is the limit of $\frac1n+\frac1n$ as $n\rightarrow\infty$. And $\frac1n$ is the limit of $\frac1n+\frac1m$ as $m\rightarrow\infty$. Thus $A=\{0\}\cup\{\frac1n\mid n\in\mathbb N\}$ are all limit points. We will show these are the only limit points.
Everything ok so far.
Let $x\not\in A$. Then there is an $\epsilon>0$ such that the interval $[x-\epsilon,x+\epsilon]$ has no point of the form $\frac1n$ (or 0).
So far I perfectly understand.
Therefore, the sum $\frac1n+\frac1m$, with least one of $\frac1n$ or $\frac1m$ less than $\epsilon/2$, must be at a distance of at least $\epsilon/2$ from $x$.
Here is where I stop following. I am unclear on the wording of the author when he says "least". Does he mean "at least" or "the least"? I guess it's sort of the same thing. Either way, I dont see why we can assume this. What if both $\frac{1}{n}$ and $\frac{1}{m}$ are greater than $\frac{\epsilon}{2}$? What does this even mean since $\epsilon$ is arbitary?
So the only such numbers in the interval $[x-\epsilon/2,x+\epsilon/2]$ must have $\frac1n\geq\epsilon/2$ and $\frac1m\geq\epsilon/2$. So $n\leq2/\epsilon$ and $m\leq2/\epsilon$. Hence there are only a finite number of numbers of the form $\frac1n+\frac1m$ in the interval $[x-\epsilon/2,x+\epsilon/2]$. Thus $x$ is not a limit point.
I can see how this conclusion is desirable, and why it suffices to show this. I just don't have the connection with the previous step so it is worthless to me at this time.
Thank you for your help.
In case anyone is interested I now know where the assumption comes from. It seems really obvious now, but if both $\frac{1}{n}$ and \frac{1}{m} are greater than or equal to $\frac{\epsilon}{2}$, then there are only finitely many combinations of $n$ and $m$ which satisfy this anyway, so we cannot have any limit points of this set in the interval (since we would need an infinite number of points)...