If $L$ is a limit point of the sequence $(x^{(n)})^∞_{n=m}$, then $L$ is an adherent point of the set $\{x^{(n)}:n\ge m\}$. Let $(x^{(n)})^∞_{n=m}$ be a sequence of points in a metric space $(X, d)$, and let $L ∈ X$. Show that if $L$ is a limit point of the sequence $(x^{(n)})^∞_{n=m}$, then $L$ is an adherent point of the set $\{x^{(n)}:n\ge m\}$. Is the converse true? 

Assume that $L$ is a limit point of  $(x^{(n)})^∞_{n=m}$ 

Then $\forall N\ge m$ and $\forall \epsilon&gt;0$, $\exists n\ge N $ such that $d((x^{(n)}, L) \le \epsilon$

Closure definition of an adherent point: We say that $x_0$ is an adherent point of E if for every radius $r&gt;0$, the ball $B(x_0,r)$ has a non-empty intersection with $E$. 

It is also equivalent to say that $\exists$ a sequence $(x^{(n)})^∞_{n=m}$ in E which converges to $x_0$ with respect to the metric $d$. 

In these defintions, are $x_0$ and L equivalent? Also could I get some help in formally writing this down?
Use the closure definition. Choose $\epsilon &lt;r$. Then the point $x^{(n)}$ in the definition of limit point belongs to E and $B(L,r)$.
It seems to me the converse is not true. For a counterexample,  we can construct a sequence like {0, 1, -1, 1, -1, 1, -1, ... } in the metric space {R, d:|x-y|}, then 0 is the adhere point of the sequence set. However, 0 is not the limit point of this sequence because |1 - 0| is bigger than an arbitrary $\epsilon$ which is smaller than 1. Could you please help check whether there is anything wrong in this statement?