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? 
 A: 
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 $G$.
A: 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.
A: $0$ 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.
A: 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 $G$.
