Accumulation point and Adherence points of $X \subset R$: $ X = \left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},..., \frac{1}{n},...\right\}$ The Set $X \subset R$: $$ X = \left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},..., \frac{1}{n},...\right\}$$


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*Each value of $X$ set is a point of adherence? Yes.

*Each value of $X$ set is isolated, because is not an accumulation point? Yes

*There is only one accumulation point: $0$?

*$0$ is also an adherence value?
I thinking Iam making a confusion on the last 2 questions: 
The third and fourth question are correct?
I think they are. But I can't formulate a formal answer. 
Any help?
 A: The sequence $\{ 1/n\}_1^\infty $ converges to $0$ so every neighborhood of $0$ contains infinitely many points of the sequence, so $0$ is  a limit point.
$0$ is the only limit point becasue the points of the sequende are isolted points.
A limit point is an adherent point as well, so $0$ is an adherent point for your set. 
A: I assume the metric is Euclidean metric.

for 3rd one

A point $a\in \mathbb R$ is an accumulation point of $X$ if for all $r\gt 0, \{B(a,r)-\{a\}\}\cap X\neq \emptyset$. 
Let $a\in \mathbb R\setminus\{0\}$.
If $a\lt0,$ choose $r=\frac{|a|}{2}$.
If $a\gt1,$ choose $r=\frac{a-1}{2}$.
If $a\in(0,1)\setminus X,a$ lies in between $\frac{1}{n+1},\frac{1}{n}$for some $n\in \mathbb Z^+$ choose $r=\min \{\frac{1}{2}(\frac{1}{n}-a),\frac{1}{2}(a-\frac{1}{n+1})\}$.
If $a\in X,$ $a=\frac{1}{n}$for some $n\in \mathbb Z^+$ choose $r=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+1})$.
Then for all cases, $\{B(a,r)-\{a\}\}\cap X =\emptyset$.

fourth one

Let $r\gt0$. Then by Euclidean property there is $N\in \mathbb Z^+$ s.t $\frac{1}{r}\gt N$.
Then $B(0,\frac{1}{N})\subseteq B(0,r)$. 
So $\frac{1}{N+1} \in  (B(0,\frac{1}{N})\cap X)\subseteq (B(0,r)\cap X )$.
Thus $B(0,r)\cap X \neq \emptyset$ and clerly $B(0,r)\cap \mathbb R\setminus X \neq \emptyset$ Therefore $0$ is an adherence point.
