Trying to understand a proof for the limit points of $\cup_{n=0}^{\infty}K_n, K_n = \{(1/n) + (1/m)\mid m=n,n+1,\dots\}$ While reviewing my answer to the exercise of 13 of chapter 2 of Baby Rudin I happened to stumble across a detailed solution manual by Kit-Wing Yu, A Complete Solution Guide to Principles of Mathematical Analysis. In this guide Yu's answer is illustrative, yet I am stuck at understanding why a neighborhood of a point $p \in [0, 1]\setminus K_0, K_0 = \{0, 1, 1/2,\dots\}$ can only contain finitely many elements of the set $K$. Below a paraphrasing of the solution:

We define $K_0 = \{0, 1, 1/2,\dots\}, K_n = \left\{\frac{1}{n} + \frac{1}{m}\mid m=n,n+1,\dots\right\}, n = 1,2,\dots$, and $K = \cup_{n=1}^{\infty}K_n$. By definition $K_0$ and $K_n$ have limit point 0 and $\frac{1}{n}$ respectively. Therefore we have $\{0, 1, 1/2,\dots\}\subseteq K'$. We claim that $K' = K_0$. pause

Yu next shows that necessarily $K' \subseteq [0, 1]$. This is clear to me and it is not used in rest of the proof, so I am going to skip it.

continue Next, suppose that $p \in [0, 1]\setminus K_0$. By this assumption there exits a positive integer $k$ such that $\frac{1}{k+1} < p < \frac{1}{k}$. Since $\frac{1}{n} + \frac{1}{n}\geq \frac{1}{n} + \frac{1}{m}$ and $\frac{1}{n} + \frac{1}{n}\geq \frac{1}{m} + \frac{1}{m}$ for all $m\geq n$, $\frac{2}{k+m}$ is the maximum of the set $K_{k+m}\cup K_{k+m+1}\cup\dots$. Define $\delta = \frac{1}{2}\min\{ p - \frac{1}{k+1}, \frac{1}{k}-p\}$, so $(p - \delta, p + \delta) \subset (\frac{1}{k+1}, \frac{1}{k}).$ If $\delta = \frac{1}{2}(\frac{1}{k} - p)$, then we have $p - \delta = \frac{3p}{2} - \frac{1}{2k} > \frac{1}{2}(\frac{3}{k+1} - \frac{1}{k}) = \frac{2k - 1}{2(k+1)^2} > \frac{2}{k+m}$ for all $m > \frac{4(k+1)^2}{2k - 1}$. In this case, the interval $(p - \delta, p + \delta)$ contains only finitely many elements of the set $K_1 \cup K_2\cup \dots \cup K_{m-1}$. [paraphrasing ends]

I am stuck at understanding what relation does the maximum element of $K_{k + m}\cup K_{k + m + 1} \cup \dots$ have with the elements of the set $K_1\cup K_2 \cup \dots \cup K_{m-1}$, i.e. why are there only finite number of elements of $K$ in the interval $(p - \delta, p + \delta)$. Namely I think the idea is that the lower bound $p - \delta$ is less than the maximum element $\frac{2}{k + m}$ when $m$ is in a certain range of values, so that necessarily some elements of some $K_m$s reside in the interval $(p - \delta, p + \delta)$.  But I can't say for sure.
 A: Sometimes it is too easy to get caught up in minute details and miss the big picture.
All you need for this is a neighborhood $U$ of $p$ whose closure misses every $\frac 1n$. Since $\frac 1{k+1} < p < \frac 1k$, such a neighborhood must exist. $\left(\frac 1{k+1}, \frac 1k\right)$ works for every $\frac 1n$ except $n = k$ or $n = k+1$, and by shrinking it a bit, those two values are missed as well.
Now the maximum shows that for high enough $n, K_n$ does not intersect $\left(\frac 1{k+1}, \frac 1k\right)$ at all, much less the smaller neighborhood of $p$. So there are only a finite number of $K_n$ that intersect with $U$.
Also, in each $K_n$ there is only one limit point, $\frac 1n$. And if you choose any neighborhood of $\frac 1n$, only a finite number of points in $K_n$ lie outside that neighborhood. This is because there will be $\epsilon > 0$ such that $\left(\frac 1n - \epsilon, \frac 1n + \epsilon\right)$ is within the neighborhood, and for all $m > \frac 1\epsilon, \frac1n + \frac 1m$ is within $\epsilon$ of $\frac 1n$.
But $U$ was chosen so that $\Bbb R \setminus \overline U$ is a neighborhood of every $\frac1n, n \ge 1$. Hence each $K_n$ only has a finite number of points in  $U$, and only a finite number of the $K_n$ intersect $U$ at all. And the union of a finite collection of finite sets is finite. Since finite sets do not have limit points, $K' \cap U$ is empty, and $p \notin K'$.
