Clarification on a proof involving cluster point Definition of cluster point- Let $A \subseteq \mathbb{R}$. A point $c\in\mathbb{R}$ is a cluster point of $A$ if for evert $\delta>0$ there exists at least one point $x\in A$, $x\neq c$ such that $|x-c|<\delta$.
Theorem- A number $c\in\mathbb{R}$ is a cluster point of a subset $A$ of $\mathbb{R}$ iff there exists a sequence $(a_n)$ in A such that lim$(a_n)=c$ and $a_n \neq c$ for all $n\in\mathbb{N}$. I'm stuck on understanding the foward direction.
Proof: $(\rightarrow)$"If $c$ is a cluster point of $A$ then for any $n\in\mathbb{N}$ the $\frac{1}{n}$-neighborhood $V_{\frac{1}{n}}(c)$ contains atleast one point $a_n\in A$ distinct from $c$. Then  $a_n\in A$, $a_n \neq c$, and $|a_n-c|<\frac{1}{n}$ implies  lim$(a_n)=c$." 
What I understand is that if we let $\delta>0$ then it follows by the Archimdean property that there exists a $k\in\mathbb{N}$ such that $\frac{1}{k}<\delta$. It follows if $n\geq k$ then $\frac{1}{n}<\frac{1}{k}<\delta$. Since $c$ is a cluster point there exists a $x_{n}\in A$ where $x_{n}\neq c$ such that $|x_{n}-c|<\frac{1}{n}<\delta$ since $\frac{1}{n}>0$. Thus by the definition of convergence we have $lim (a_n)=c$. Would this be a correct way of understanding the proof to this theorem?
 A: What you are doing is finding another $1/n$ when there's already one provided in the proof.
I will just list a variety of points here that will hopefully illuminate what's going on.
(1) A cluster point $c$ of a subset $A$ is one that, no matter how small an open interval you put around it, will have at least one other point of $A$ nearby inside that interval.
(2) In the proof, a decreasing sequence of neighbourhoods is being used to construct a sequence that converges to $c$. This means that you can choose a point that is not $c$ for each $n$. It doesn't matter whether or not the sequence you're building has a pattern, it's terms are none the less getting closer and closer to $c$.
(3) For each $n$, you can pick one element of $A$ that is not $c$ and call it $a_n$. Given that this $a_n$ is inside $V_{1/n}$ you have $|a_n - c|<1/n$ and this is true for all $n$.
(4) By the convergence of the sequence $\{1/n\}$ to zero, given a $\delta>0$, there is an $N$ so that for all $n>N$, $1/n < \delta$. This is what implies $|a_n - c| < 1/n < \delta$ for all $n>N$.
(5) Thus, given a $\delta > 0$ there exists an $N$ with $n>N$ implies $|a_n - c| < 1/n < \delta$ and this gives convergence of $a_n$ because it's true no matter how small your $\delta$.
A: Maybe you're thinking about it too hard. For any positive number $\delta$, you're promised one point $x =x _\delta$ different from $c$. In particular, for any $n\in\Bbb N$, $n^{-1}$ is a positive number, so you're promised a point $x=x_n$ different from $c$. And since $x_n\in(c-n^{-1},c+n^{-1})$, we have $$|x_n-c|<n^{-1}$$ so that $x_n\to c$. You are right that to show that $x_n\to c$ we use the Archiemedean property of the real numbers. Given $\varepsilon >0$ all we need to do is take $n$ large enough so that $n^{-1}<\varepsilon$.
