If $p$ is not a limit point of $E$, then it has a neighborhood with at most 1 point of $E$ — why at most 1 point?

Question

I have a doubt about the following proof from PMA Rudin:

Suppose $E \cap K$, and let $q \in K$ be a point which is NOT a limit point of $E$. Then $q$ has a neighborhood $N$ s.t. it has at most one point of $E$ in it (in case $q \in E$).

In other words, $(N - \{q\}) \cap E = \phi$

How can this be true?

I can’t help but think of the scenario where q is in E, then q is on the “border” of the set E. In that case no matter what neighborhood we construct, although it may not be completely inside E, it will intersect E non-trivially.

I can’t help but think of this in terms of 2D euclidean space where a neighborhood is a perfect circle with radius d and we are trying to shrink d as much as possible.

BTW this is coming from the proof for Theorem 2.37.

Answer 1

By definition $q$ is a limit point of $E$ if every neighbourhood $U$ of $q$ contains $e\in E,e\neq q$.

Take the negation of the definition $q$ is not a limit point if there exists a neighbourhood $U$ which does not contain an element $e\in E$, $e\neq q$. Thus if $U$ contains an element of $E$ it is necessarily $q$.

https://en.wikipedia.org/wiki/Limit_point#Definition

Answer 2

Consider the subset $E=\left\{\frac1n:n\in\Bbb Z^+\right\}$ of the compact set $[0,1]$. Every point $\frac1n\in E$ has a nbhd $\left(\frac1n-\frac1{2n(n+1)},\frac1n+\frac1{2n(n+1)}\right)$ whose intersection with $E$ is the singleton $\left\{\frac1n\right\}$: it contains no other point of $E$. In fact, the only point of $[0,1]$ that does not have such a nbhd is $0$, the one limit point of $E$.

In the absence of compactness every point of $E$ can be isolated, meaning that it has a nbhd that contains no other point of $E$: as an example take $E=\Bbb Z$ in the space $\Bbb R$.

This is just a matter of definition: a limit point of $E$ is by definition a point $p$ such that every nbhd of $p$ contains a point of $E\setminus\{p\}$, so $p$ is by definition a non-limit point of $E$ if it has nbhd that does not contain any point of $E\setminus\{p\}$. This might be because $p\notin\operatorname{cl}E$, so that $p$ has a nbhd that contains no points of $E$, or it might be because $p$ is an isolated point of $E$.