Problem about limit points in the standard topology of $\Bbb R$ I have a problem about topology that is supposed I must prove without the use of any metric in the standard topology of $\Bbb R$ (metrics are not started in the book by now) just using pure topology definitions.
The problem say

Let $S\subset\Bbb R$ and $a\in\Bbb R$. Prove that $a\in \overline S \iff \forall n\in\Bbb N\ \exists x_n\in S: |x_n-a|<\frac1n$

So I take an open interval containing $a$ of the kind $(a-r,a+r):r\in\Bbb R$ and I must show that for any $r$ that I take $r>\frac1n>|x_n-a|\,$ i.e. $\,S\cap(a-r,a+r)\ne\emptyset\,$ what is obvious that can hold for any $r$ that I choose because I can choose any $n$ that prove the inequality.
Now the questions:
1) It is my proof good enough? It lacks something? It is wrong?
2) I cant understand the "$\iff$" very clear, I means that if I change the condition $|x_n-a|<\frac1n$ by $|x_n-a|=\frac1n \,$ I can still write that $\forall n\in\Bbb N\ \exists r\in\Bbb R:\ |x_n-a|=\frac1n<r$ so the above proof still holds but the "iff" statement seems a contradiction because I can write a different "iff" statement that holds the above, so what is wrong here?
Thank you in advance.
 A: For iff proofs you must show both directions. Your forward direction argument seems to be on the right track, i.e., $a\in\overline{S}$ $\Rightarrow$ for every open set $U\subseteq\mathbb{R}$ with $a\in U$, $U\cap S \neq \emptyset$. So in particular, we could take the open sets of the form $(a-\frac{1}{n},a+\frac{1}{n})$ for every $n\in\mathbb{Z}_{>0}$.
For the reverse direction you must start with the assumption that for every $n\in\mathbb{Z}_{>0}$, $(a-\frac{1}{n},a+\frac{1}{n})\cap S\neq \emptyset$. These open sets of the form $(a-\frac{1}{n},a+\frac{1}{n})$ are what's known as a countable basis at $a$ in $\mathbb{R}$. The basis open sets of the standard topology of $\mathbb{R}$ are the open intervals of $\mathbb{R}$. 
To show $a\in\overline{S}$ we need to show that for every open set $U\subseteq\mathbb{R}$ with $a\in U$, $U\cap S\neq\emptyset$. So we need to start with an arbitrary open set $U$ containing $a$ and show it has this property. Since $U$ is open and $a\in U$, there exists a basis open set $(c,d)\subseteq\mathbb{R}$ s.t. $a\in (c,d)\subseteq U$ (this is from the definition of a basis for a topology). So by the Archimedian property of $\mathbb{R}$, there exists an $n\in\mathbb{Z}_{>0}$ with $a\in (a-\frac{1}{n},a+\frac{1}{n})\subseteq (c,d)\subseteq U$. By assumption, $(a-\frac{1}{n},a+\frac{1}{n})\cap S\neq\emptyset$, and thus $U\cap S\neq\emptyset$. This shows $a\in\overline{S}$. 
A: Suppose that $a$ is an element of the closure of set $S$ and pick any open interval, $I\ ,$ that contains point $a$. By the definition of closure, this interval will contain a point ($x_{I}$) of $S$. Since interval $I$ contains both points $a$ and $x$ the distance between them will be less than the length of the interval, $$|x_{I}-a| < \text{Length}(I)\ .$$ Keep in mind that the length of the interval is arbitrary; once you decide upon it, you can always find a corresponding point $x_{I}\ .$
For the converse statement, suppose that no matter what open interval $I$ (containing the point $a$) you choose, it will also contain a point $y_{I}$ of $S\ .$ You want to show that point $a$ is a element of $\bar{S}\ .$ Since you can write $$\bar{S} = S \cup \text{Limit}(S)\ ,$$ where $\text{Limit}(S)$ denotes the set of all limit points of $S$, then you want to show that either $a\in S$ or $a$ is a limit point of $S\ .$ If the chosen point $a$ happens to lie in $S$ then you are done. Otherwise the argument above implies that you can always find a sequence $y_{n}\in S$ converging to $a$; simply pick the open interval $I$ such that $\text{Length}(I) < \frac{1}{n}$ and the corresponding point, $y_{n}\in S\ ,$ will be $(1/n)$-close to point $a$.
