Prove $(x_n)$ has a subsequence convergent to $x$.

Question

I’m trying to do this one but I don’t know how to start.

Let $(x_n)$ be a sequence on a metric space $(X,d)$, and $x$ a limit point of the image set of the sequence. Prove $(x_n)$ has a subsequence convergent to $x$.

My definition of limit point is that a point $x$ is a limit point on a space $S\subset X$ if every neighbourhood of $x$ contains a point of $S$ different of $x$.

I don’t know how to build the convergent subsequence with this assumptions.

Thanks for your help.

Answer 1

For every positive integer $k$ the set $\{n\in\mathbb N\mid d(x_n,x)<\frac1k\}$ is infinite.

(If not then it can be shown that $x$ is not a limit point of $\{x_n\mid n\in\mathbb N\}$)

So for $k=1$ choose some $n_1$ with $d(x_{n_1},x)<\frac11=1$.

Then for $k=2$ choose some $n_2>n_1$ with $d(x_{n_2},x)<\frac12$.

Et cetera.

Answer 2

Let denote : $B_\epsilon = \{y \in X \mid d(y-x) < \epsilon \}$ this is an open ball with center $x$ and radius $\epsilon$.

Since $x$ is a limit point of the $(x_n)$ it means that for all $\epsilon$ we can find an $N_{\epsilon}$ such that : $x_{N_{\epsilon}} \in B_\epsilon \backslash \{x\}$.

Now let's construct a striclty increasing function $\phi : \mathbb{N} \to \mathbb{N}$ such that : $x_{\phi(n)} \to x$.

Let's choose an $\epsilon > 0$, and let $\phi(0) = N_{\epsilon}$. Now suppose that the function $\phi$ has been construct for all $n \leq N$. Then we can take : $0 < \epsilon'= \frac{1}{K} < \min \{d(x_{\phi(n)}-x) \mid n \leq N \}$. And we can let : $\phi(N+1) = N_{\epsilon'}$.