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 $\forall \epsilon>0$, $\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 $r>0$, 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 $\epsilon <r$. Then the point $x^{(n)}$ in the definition of limit point belongs to E and $B(L,r)$.