Subsequence of Sequence in Metric Space with Limit

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $\sequence {x_n}$ be a sequence in $M$.

Let $x$ be a limit point of $S = \set {x_n: n \in \N}$, the set of members of $\sequence {x_n}$.

Then $\sequence {x_n}$ has a subsequence which converges to $x$.

Proof
By Finite Subset of Metric Space has no Limit Points‎, $S$ is infinite (or it has no limit points).

We may assume that $x_n$ never equals $x$.

Otherwise we delete all instances of $x$ from $\sequence {x_n}$ and create a new sequence $x_m$ which is a subsequence of $x_n$ which does never equal $x$.

Then $S \setminus \set x$ is still infinite, and it still has $x$ as a limit point.

So, since $x$ is a limit point of $S$, there is an integer $\map n 1$, say, such that $x_{\map n 1} \in \map {B_1} x$, where $\map {B_1} x$ is the open $1$-ball of $x$.

Suppose we choose the integers $\map n 1 < \map n 2 < \cdots < \map n k$ so that $x_{\map n i} \in \map {B_{1/i} } x$ for $i = 1, 2, \ldots, k$.

Now we put $\epsilon = \min \set {\dfrac 1 {k+1}, \map d {x_1, x}, \map d {x_2, x}, \map d {x_{\map n k}, x} }$.

Since $x_n \ne x$ for any $n$, it follows that $\epsilon > 0$.

Since $x$ is a limit point of $S$, there exists an integer $\map n {k+1}$ such that $x_{\map n {k+1} } \in \map {B_\epsilon} x$.

Now $\epsilon$ has been chosen so as to force $\map n {k+1} > \map n k$, since $\map d {x_i, x} \ge \epsilon$ for all $i \le \map n k$.

Also, note that $x_{\map n {k+1} } \in \map {B_{\frac 1 {k+1}} } x$.

This completes the inductive step in constructing a subsequence.

This converges to $x$, because $\forall k_0 \in \N_{>0}$ and $\forall k \ge k_0$ we have $\map d {x_{\map n k}, x} < \dfrac 1 k \ge \dfrac 1 {k_0}$.

Hence the result.