Answer to Question #308343 in Real Analysis for Pankaj

Question

Give an example for each of the following:

A set having no limit point.

Answer 1

Definition. A limit point (accumulation point) of a set $S$ in a metric space $X$ is a point $x\in X$ any puncted neighborhood of which contains an element of $S$. The set of all limit points of $S$ is denoted by $S'$.

In other words, $S'$ is the set of all points $x\in X$ at which the set $S$ is locally infinite. Really, if there exists a neighborhood $U$ of $x$ such that $U\cap S$ is finite then, by putting $r=\min\{dist(x,y): y\in U\cap S,y\ne x\}$, we have that a puncted open ball $\dot{B}_r(x)$ contains no elements of $S$. This means that $x$ is not a limit point of $S$. Conversely, if any neighborhood $U$ of $x$ contains an infinite subset of $S$ then a puncted neighborhood $\dot{U}$ of $x$ also contains an infinite subset of $S$, hence, $x$ is a limit point of $S$.

Examples of locally finite sets (i.e. sets having no limit point):

1) any finite set;

2) the set of integers $\mathbb{Z}\subset\mathbb{R}$ (any open interval of length $l$ contains $[l]$ or less integers);

3) the set $S=\{\frac{1}{n}:n\in\mathbb{N}\}$ in the metric space $X=(0,+\infty)$ with the standard metric. Indeed, if $x=\frac{1}{n}\in S$ then the interval $U=(\frac{1}{n+1}, \frac{1}{n-1})$ is a neighborhood of $x$ such that the puncted $U$ contains no elements of $S$. Therefore, $x\notin S'$. If $x\notin S$ and $x>0$ then put $n=[1/x]$.We have $n<1/x<n+1$, thus, $\frac{1}{n+1}<x<\frac{1}{n}$. The interval $U=(\frac{1}{n+1}, \frac{1}{n})$ is an open neighborhood of $x$ containing no elements of $S$. Thus, $x\notin S'$ and therefore, $S'=\emptyset$.