M compact. $(x_n)$ converges if and only if $(x_n)$ has only one closure point. Show, with one example, that the compacity is necessary.
My attempt:
1) Of course if $(x_n)$ converges, $(x_n)$ has only one closure point, because every subsequence of $(x_n)$ converges to the same point as $(x_n)$.
2) Given a sequence $(x_n)$ such that $(x_n)$ has only one closure point, suppose that that $(x_n)$ doesn't converges. Then, we have two options:
There are $(x_{n_j})$ and $(x_{n_i})$ subsequences of $(x_n)$ such that $(x_{n_j})$ converges to $a$, $(x_{n_i})$ converges to $b$, $a \neq b$. Absurd, because then $a$ and $b$ are closure points.
There is a subsequence $(x_{n_j})$ such that $\lim_{j \rightarrow \infty} x_{n_j} = \infty$. Absurd, because since M is compact, M is limitated.
3) Example: $(x_n) = (1, 1, 2, 1, 3, 1, 4, 1, 5, 1,...)$. Then 1 is the only closure point, but $(x_n)$ does not converge.
I am not sure that $2)$ is correct and I also think there is an easier way to do it. Can someone help me? Thanks!
For 2, let $M$ be compact and assume $(x_n)$ does not converge. We shall see that $(x_n)$ has at least two closure points. As $M$ is compact, it has at least one closure point $x\in M$. As $x_n\not\to x$, there exists an open neighbourhood $U$ of $x$ such that not almost all $x_n$ are in $U$. So infinitely many $x_n$ (i.e., a subsequence) are in $M\setminus U$. As this is compact again, the subsequence has a closure point $y\in M\setminus U$. Then $x,y$ are distinct closure points of $(x_n)$.