What can you say about the limit of $( x_n)$ as $n$ approaches infinity? I have a metric space $M$ and a convergent sequence $(x_n)$ in $M$. I also have $a$ in $M$ such that:
1) the set $\{a, x_1, x_2, \dots, x_n,\dotsc\}$ is closed.
2) the set $\{x_1, x_2, \dots, x_n, \dots\}$ is neither open nor closed. 
What would this say about $\lim_{n\to\infty} x_n$? 
This is puzzling me.
Would this limit exist? Any hints would be great! 
 A: Let's forget, for the moment, the information that the sequence is convergent.
The closure of a set is the union of the set with the set of its limit points. Since $X=\{x_n:n\in\mathbb{N}\}$ is not closed, but $X\cup\{a\}$ is, we conclude that $a$ is a limit point of $X$. Thus there is a sequence in $X$ that converges to $a$ and this can be taken as a subsequence of $(x_n)$ in the following way:
Set $n_1=1$; for $k>1$, define 
$$
n_k=\min\{n:d(x_n,a)<1/k, n>n_{k-1}\}
$$
Note that $X\cap B_{1/k}(a)$ is infinite for every $k\in\mathbb{N}$, because $a$ is a limit point, so the set we take the minimum of is not empty.
Then it's trivial to verify that the subsequence $(x_{n_k})$ converges to $a$.
Can we say the whole sequence converges to $a$? No. Define
$$
x_n=\begin{cases}
1/n & \text{if $n$ is even,}\\
n   & \text{if $n$ is odd}
\end{cases}
$$
Then $(x_n)$ satisfies your hypothesis with $a=0$, but it doesn't converge to $0$.
However, if we add the hypothesis that the whole sequence is convergent, then it converges to the same limit as any of its subsequences. So surely
$$
\lim_{n\to\infty} x_n=a
$$
because a subsequence converges to $a$.
A: I would say that a subsequence of $\{x_i\}$ has limit $a$.
A: Since the second set is not closed it does not contain all of its limit points. Therefore, there exists a sequence (say $y_n)$ in the set 2 that converges to a limit (say $b\in M$ and $b\neq x_j$). But this sequence is also in the first set. As the first set is closed, $b\in$ set $1$ but $b\neq x_j$, so b must be a! That is there exists a sequence of $x_j 's$ such that converges to $a$.
