I'm trying to understand the proof where $E \subset X$ and if $p\in Cl(E)$, then $\exists$ a sequence $(p_n) \in E$ s.t $p_n \rightarrow p$. The proof is as follows:
Assume $p \in Cl(E)$. Then $\forall n \in \mathbb{N}$, $N_{1/N} (p) \cap E \neq \emptyset$. So we may choose $p_n \in N_{1/N} (p) \cap E$.
Let $\epsilon > 0$, and let $N = 1/\epsilon$. Then $\forall n \geq N$, $d(p_n, p) < 1/n \leq 1/N = \epsilon$.
Where $N_{1/N} (p)$ is used to the symbolize the neighborhood around p with radius $1/N$.
I'm really confused behind the choice of 1/N. Why did we select that for our radius of the neighborhood? I have a hunch it might be something to do with the Archimedean property, but then doesn't the Archimedean property state that:
$\exists n \in \mathbb{N}$ such that $n \cdot d(p_n, p) > 1 \rightarrow d(p_n, p) > 1/n$?
Help would be much appreciated!
A point $p$ is in the closure of $E$ if every open ball around $p$ contains a point of $E$. Of course it may be that this point is $p$ itself (then we can construct the sequence just by setting $p_n := p \ \forall n$, but this is of course not guaranteed.
But we have this property that for every open ball around $p$ we can find a point that is in $E$. This means intuitively, that no matter how small a distance we choose, there is always a point in $E$ that is no further away from $p$.
So how to construct the sequence that converges against $p$? $\displaystyle \frac{1}{n}$ has the property that the larger $n$, the smaller $\displaystyle \frac{1}{n}$. So we set $p_n := $ a point in $E$ that is also in the open ball around $p$ with radius $1/n$. You can imagine this as a ball around $p$ that gets smaller and smaller the higher our $n$ goes.
Our intuition says that the $p_n$ must finally converge against $p$. To state this formally correctly, given $\epsilon > 0$, find an $N \in \mathbb{N}$ with $d(p_n, p) < \epsilon \ \forall n \geq N$. Supposing that the "$\epsilon$ is very small, choose a very big $N$" (so that the ball around $p$ becomes very small as well). Thus the choice that $\displaystyle N \gt \frac{1}{\epsilon}$ (for very small $\epsilon$, $\displaystyle \frac{1}{\epsilon}$ becomes very large), and the desired property follows.