Find the set of accumulation points of $\{1+\frac{1}{n} : n\in N\} $ with proof Find the set of  accumulation points of $\{1+\frac{1}{n} : n\in N\} $
Hello, I need help understanding what is an accumalation point of a set. 
From the set  $\{1+\frac{1}{n} : n\in N\} $, I can see that $1$ is an accumulation point but I cannot see why $1+\frac{1}{n}$ with $n$ really really large is not accumlation point . For example $1+\frac{1}{1000000} = 1.000001 $ is not an accumulation point.

Let $E\subseteq R$ and $x\in R $. We say that $x$ is an accumulation point of $E$ 
  if $\forall \epsilon>0, (x-\epsilon,x+\epsilon)\cap E\backslash x \ne \emptyset $



*

*If $x-\epsilon<1$,


By the Archimedean property there exists an $n$  such that  $n>\frac{1}{\epsilon}  $ .
then $\epsilon>\frac{1}{n}$.
$1+\frac{1}{n} \in (1-\epsilon,1+\epsilon)$
how can I proof from here that $1+\frac{1}{n}$ is not an accumulation point.
Thank you
 A: I'll prove that $1+1/m$ is not an accumulation point of the set
$$
S=\left\{1+\frac{1}{n}\;\middle|\;n\in\mathbb{N},n>0\right\}
$$
All I need to find is $\varepsilon>0$ such that
$$
\left(1+\frac{1}{m}-\varepsilon,1+\frac{1}{m}+\varepsilon\right)
\cap S=\left\{1+\frac{1}{m}\right\}
$$
that is, the only solution of
$$
1+\frac{1}{m}-\varepsilon<1+\frac{1}{n}<1+\frac{1}{m}+\varepsilon
$$
is $n=m$.
The two inequalities are the same as
$$
\frac{1}{m}-\varepsilon<\frac{1}{n}<\frac{1}{m}+\varepsilon
$$
It is not restrictive to assume $\varepsilon<1/m$, so the inequalities become
$$
\frac{m}{1+m\varepsilon}<n<\frac{m}{1-m\varepsilon}
$$
and it suffices to show we can take $\varepsilon$ so that
$$
\frac{m}{1-m\varepsilon}-\frac{m}{1+m\varepsilon}<1
$$
because in this case the interval cannot contain more than one integer (and it always contains $m$).
This becomes
$$
m+m^2\varepsilon-m+m^2\varepsilon<1-m^2\varepsilon^2
$$
that is
$$
m^2\varepsilon^2+2m^2\varepsilon-1<0
$$
which is certainly satisfied for
$$
0<\varepsilon<\frac{-m^2+\sqrt{m^4+m^2}}{m^2}=\sqrt{1+\frac{1}{m^2}}-1
$$
An explicit value is $\varepsilon=\frac{1}{2m^2}$.
A: A point $p$ is an accumulation point of a set $S$ in a metric space if every neighborhood of $p$ contains a point of $S$ not equal to $p$. Let $S=\left\{1+\frac1n:n\in\mathbb N\right\}$. For any $\varepsilon>0$, we may choose $N>\frac1\varepsilon$, so that $1+\frac1N\in (1-\varepsilon,1+\varepsilon)$, and hence $1$ is an accumulation point of $S$. 
If $p\in\mathbb R$, $p\ne1$, there are three cases to consider. If $p>2$, set $\delta=|2-p|$. Then $(p-\delta,p+\delta)\cap S=\varnothing$. If $p<1$, set $\delta=|1-p|$, and similarly $(p-\delta,p+\delta)\cap S=\varnothing$. If $1<p<2$, let $m=\min\left\{n\in\mathbb N:1+\frac1n\leqslant p\right\}$. Then for any $q\in S$, $q\ne p$, we have $$|q-p|\geqslant \left|\frac1N-\frac1{N-1}\right| =\left|\frac1{N(N-1)}\right|>\frac1{N^2}.$$
Choose $\delta<\frac1{N^2}$, then $(p-\delta,p+\delta)\cap (S\setminus\{p\})=\varnothing$, so again $p$ is not an accumulation point of $S$.
