# Limit point of the set $G= \{1/n: n \in \mathbb N\}$?

In $\mathbb{R}$ , what is the limit point of the set $G= \{1/n: n \in \mathbb N\}$?

Definition: A point $p \in X$ is a limit of of E if every neighborhood of $p$ contains a point $q \in E$ $q\neq p$

I do understand $0$ is a limit point but shouldn't it be the case that all the points in [0,1] are limit points by the definition?

• Why do you think that $\frac 23$ is a limit point, for example? Note that it is not in $G$, in case you are confused. – астон вілла олоф мэллбэрг Nov 16 '17 at 11:16
• I gave an answer very similar to your question. See this – Juniven Nov 16 '17 at 11:40

I do understand 0 is a limit point but shouldn't it be the case that all the points in [0,1] are limit points by the definition?

Definitely not.

Consider a point $a\in G$, then there exists $n\in\mathbb N$ such that $a=\frac1n$. Choose $\delta<\frac1n-\frac1{n+1}$ and you get $G\cap (a-\delta,a+\delta)=\{a\}$, so by definition $a$ is no limit point of $G$.

If you consider $a\in (0,1]\setminus G$ then there exists $n\in\mathbb N$ such that $\frac1{n+1}<a<\frac1n$. Choose $\delta<\min\{\frac1n-a,a-\frac1{n+1}\}$ and you get $G\cap (a-\delta,a+\delta)=\emptyset$. By definition $a$ is no limit point of $G$.

• Ok I see but to be clear so only limit point is 0, right? – Pumpkin Nov 16 '17 at 11:23
• Yes, $0$ is a limit point. – Mundron Schmidt Nov 16 '17 at 11:24
• Alternative approach to the concept of limit points is that there should be infinitely many points of the set in the nhd of the limit point, which can only be possible for $0$ here. – Shatabdi Sinha Nov 16 '17 at 11:54

To be a limit point of a set you should find at least one point of the set in any punctured ball around that point.

So now can you find such a point for example $(B(1,\frac 13)- \{1\}) \cap G$?

To prove that no point other than zero can be a limit point use contradiction. Assume $a \not = 0$ is a limit point. Then as $\frac 1n \to 0$ we have that $\exists n_0$ s.t. $n \ge n_0 \implies \left| \frac 1n \right| < \frac{|a|}{2}$. So then you can find only finitely many elements of $G'$ in the ball $B(a,\frac{|a|}{2})$. Let the the minimum distance from those point to $a$ be L. Note that $L$ always exists as we have only finitely many points. Then $B(a,L) - \{a\}$ doesn't contain any element of $G$. Hence $a$ can't be a limit point.

Do notice that the sequel $\frac{1}{n}$ converges to zero when $n$ goes to infinity. Therefore it would only make sense to check if $0$ is a limit point of $G$, which you said you did.

But now, why is $[0,1]$ all limit points of $G$? Take, for instance, the middle point between $1$ and $\frac{1}{2}$, which is $\frac{1+\frac{1}{2}}{2}=\frac{3}{4}$. Is there any neighbourhood of of this point that does not intersect $G$? Take an open interval $\left( \frac{3}{4}-\epsilon,\frac{3}{4}+\epsilon\right)$, for any epsilon between $0$ and $\frac{1}{4}$, and you'll see that this interval does not intersect $G$.

Same argument can be made for all points in $[0,1]$ which is not $0$ and is not in $G$.

• Well, are you sure that every point of $G$ is a limit point? If so the have that $\lim \sup \frac 1n = 1$ and $\lim \inf \frac 1n = 0$, but these can't be different as the limit obviously is $0$. – Stefan4024 Nov 16 '17 at 11:24
• @Marra I understood the interval [0,1] but is everypoint in G is a limit point? According to proof Mundron Schmidt makes it is not. – Pumpkin Nov 16 '17 at 11:24
• Yes, you are both correct. I had the idea of closure of the set $G$ in my head instead of limit points. I'll correct. – Marra Nov 16 '17 at 11:26

$0$ is the limit here. Also note that

A point is a limit point iff it contains infinitely many points of the set in it's neighborhood.

So, that can not be every other point of the set except $0$ here. Another important remark that might help you is, a limit point may not be unique like the limit. There may be several limit points in fact.