Subsequence in compact metrical space > Let $X$ be a compact metrical space and $(x_n)\subset X$ a sequence. Show that $x_n\to x$ iff for every convergent subsequence $x_{n_k}\to x$. "$\Leftarrow$" is true since $(x_n)\subset (x_n)$ is a subsequence. "$\Rightarrow$" Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$ with limit $y$. Now I probably need to use the compactness of $X$ to show that $x=y$. However, after looking up the definition of compactness I am slightly confused by how many different definitions there are. Which definition do I use here and how can I obtain $x=y$ with it?

Suppose that you don't have $\lim_{n\to\infty}x_n=x$. Then there is a $\varepsilon>0$ such that, for every $p\in\mathbb N$, there is a natural $n\geqslant p$ such that $d(x_n,x)\geqslant\varepsilon$. So, you can defined a sequence $(n_k)_{k\in\mathbb N}$ as follows:

* $n_1$ is such that $d(x_{n_1},x)\geqslant\varepsilon$;
* $n_{k+1}>n_k$ and $d(x_{k+1},x)\geqslant\varepsilon$.



Then the sequence $(x_{n_k})_{k\in\mathbb N}$ has a convergente subsequence (since the space is compact), whse limit cannot be $x$. This subsequence is a subsequence of the original sequence, but its limit is not $x$.