In class we are given the following definition of a limiting point and some information about it:
Point $a \in \mathbf{R}$ is a limiting point of subset $S \in \mathbf{R}$ if any punctured neighborhood of $a:0< | x-a | <\epsilon$ there is at least 1 point from $S$.
Existence of 1 point implies that there are infinitely many of them. In other words, there exists a sequence $S_n \in S : \lim_{n \to \infty }S_n = a$ In particular, if $S$ is itself a sequence then $a$ is a limiting point of $S$ if there is a convergent subsequence of $S$
So a few things I need confirmation on and clarification are listed below, my knowledge of math isn't very good so could someone please explain it in layman terms.
- What is a punctured neighborhood?
- $0< | x-a | <\epsilon$, $x$ isn't defined in the definition could this be a typo? If so what would be the right inequality or similar part of the definition too this ?
- So in order to find a limiting point, I have to find an infinitely amount of points/ a sequence of points that fit the inequality and are convergent?
- Could someone give me an easy example of finding a limiting point, nothing much comes up when googling "limiting point" that seems to be in the scope of what I know and are too advanced than stuff we see class.
3 Answers
$\begingroup$The first two questions answer each other: The punctured neighborhood of $a$ (with radius $\epsilon$) is defined as the set $$\{x \in \mathbb{R} \mid 0 < |x-a| < \epsilon\}.$$ The term "punctured" comes from the fact that we explicitly exclude $a$; otherwise it would simply be called a neighborhood.
To find a limit point $a$, you should show that for every $\epsilon > 0$, the punctured neighborhood of $a$ contains a point of $S$. Start with a fixed $\epsilon$ and try to construct $s(\epsilon) \in S$; you will most likely get different $s$ for different $\epsilon$.
For easy examples, try these sets: $$\{ \frac{1}{n} \mid n \in \mathbb{N} \}$$ $$\{ n \mod{2} \mid n \in \mathbb{N}\}$$
$\endgroup$ $\begingroup$- A punctured neighborhood of a point $a$ is a set of the type $(a-r,a+r)\setminus\{a\}$, for some $r>0$ (that's the definition used in that text; there are other possibilities).
- There's no typo. It follows from these inequalities that $x\neq a$. That's all.
- No. If you prove that a point $a$ is a limiting point of $S$, then it will follow automatically that every punctured neighborhood $a$ contains infinitely many points of $S$.
- The point $0$ is a limiting point of the set $\left\{\frac1n\,\middle|\,n\in\mathbb{N}\right\}$, because every set $(-\varepsilon,\varepsilon)$ contains numbers of the form $\frac1n$ ($n\in\mathbb N$).
A partial answer
It's useful to see this illustrated. The figure below (I can't see to size it appropriately) is a sequence of points $\color{red}{x_{i}}$ converging to $x$, and the $\color{blue}{\text{blue}}$ circle is the $\varepsilon$-neighbourhood around $x$.
