Explain the theorem regarding convergent sequences (read details) . Theorem is:

Let {$p_n$} be a sequence in a metric space X.
{$p_n$} converges to p $\in$ X if and only if every neighbourhood of p contains all but finitely many of the terms of {$p_n$}.

I understood that every neighbourhood of p contains finite terms of {$p_n$} and also other way.
Please correct me if I am wrong.

But if we take {1/n : n$\in$ N}, where X is R, then there could be some neighbourhood of x=0(which is point of convergence of sequence) and it has infinitely many points.

I am not getting where I am making mistakes, Please find and correct me.

Explain proof of this theorem or give a hint or any intuitive way to think about this theorem.
NO : all but means : only a finite number of $p_i$ are "outside" the neighbourhood of $p$.

This means that we can find an $N$ such that :


  for all $n &gt; N$, all $p_n$ are "inside" the neighbourhood.
N  is all of definite  equations  expandation  N>n이 성립하므로 {p}는 n의 부분적인  원소집합으로  치환가능하기때문에 N은 {p}

Bot translation of Korean text:


  This is because the establishment of the {p} is a set of n elements in a partial substitution is possible because N is {p}