Closure of a sequence I'm working on a problem for a class and I'm a bit confused on what exactly the question is asking, the question is as follows,
Suppose  $(x_n)$ is a sequence in $\Bbb R$. Prove that $\bigl\{a \in \Bbb R : \text{there  is  a subsequence  }(x_{n_{k}}) \text{ with } (x_{n_{k}}) \to  a \bigr\}= \bigcap^{\infty}_{n=1}  \overline{\{ x_n,x_{n+1},x_{n+2},...\}}$
I know that this involves dealing with the closure of a set which we defined as,
Suppose $A\subseteq R$, we define the closure of $A$ denoted by $\overline{A}$ by $\overline{A}=\{x \in R: \exists \ (a_n)$ in $A$ such that $(a_n) \to x \}$.
Thanks for the help in advance as I'm confused on exactly what to show.
Edit note: Had to adjust as R is the set of the reals.
 A: Here is a sketch of a solution. I leave some details (why?) for the OP.

*

*If $a\in \bigcap^{\infty}_{n=1}  \overline{\{ x_n,x_{n+1},x_{n+2},...\}}$, there is $n_1\geq1$ such that $|x_{n_1}-a|<\frac12$ (why?). Then, by induction, suppose we have chosen $x_{n_1},\ldots,x_{n_k}$ with $n_1\leq\ldots\leq n_k$ and $n_k\geq k$ such that
$$ |x_{n_j}-a|<\frac{1}{2^j},\qquad j=1,\ldots,k$$
Then, for $k+1$, one can choose $n_{k+1}\geq\max(k+1,n_k)$  (why?) such that
$$|x_{n_{k+1}}-a|<\frac{1}{2^{k+1}}$$
We have constructed a subsequence $x_{n_k}\xrightarrow{k\rightarrow\infty}a$.


*On the other direction, suppose $x_{n_k}\xrightarrow{k\rightarrow\infty}a$. Then, for any $\varepsilon>0$, there is $K$ such that $k\geq K$ implies $|x_{n_k}-a|<\varepsilon$. As $n_k\geq k$
$$B(a;\varepsilon)\cap\{x_\ell:\ell\geq k\}\neq\emptyset,\qquad \forall k\tag{why?}$$
As this holds for any $\varepsilon>0$, this means that $a\in\overline{\{x_\ell:\ell\geq k\}}$ for all $k\in\mathbb{N}$
A: I think that you need to prove that the set of the $a$'s, defined as the limits of subsequences of $x_n$, is equal to the infinite intersections of the closures of the sets defined by the subsequences $\{ x_n, x_{n+1}, x_{n+2} \}$
