A point $a$ in a metric space $M$ is called a value of adherence of sequence $(x_{n})$ in $M$, when $a$ is a limit of the one subsequence of $(x_{n})$. Let $V$ the set of values of adherences of $(x_{n})$ and, for each $k\in\mathbb{N}$, let $F_{k}$ the adherence of set $\{x_{n};n\geq k\}$. Show that $V=\displaystyle\bigcap_{k=1}^{\infty}{F_{k}}$ and conclude that $V$ is a closed subset of M.
How I can prove that X is closed? I need help. Regards
A sketch of proof that the set of sub-sequential limits is a closed set:
Let $\{x_n\}$ be a sequence. Let $E^{*}$ be the set of all sub-sequential limits of $\{x_n\}$. We shall show $E^{*}$ contains its limit points. Let $a \in \overline{E^{*}}$. Then, there exists a sequence of elements of $E^{*}$ converging to $a$.
But there is a sequence of elements converging to each element of $E^{*}$. Ergo, we can find a subsequence that converges to $a$.
Do you need a hint to proceed?
I think that i could manage a way to show that $V$ is a closed subset.
Denote by $A^{C}$ the complementary set of $A$ in $M$. Since you have proven that $V=\bigcap_{k=1}^{\infty}{F_{k}}$, by De Morgan law it follows that: $$V^C=\bigcup_{k=1}^{\infty}{F_{k}}^C$$ Since every $F_k$ is closed, it is true that $F_{k}^C$ is open. Since any arbitrary union of open sets is open, we conclude that $V^C$ is an open set. Therefore, $V$ must be a closed subset of $M$.
