Prove by counterexample that a bounded sequence in a metric space need not have a convergent subsequence I am trying to prove that a bounded sequence in a metric space need not have a convergent subsequence.
My counterexample is: Consider the metric space:  ($(0,1]$,standard norm on $\Bbb R$ restrict to $(0,1]$) 
and the sequence $1,\frac{1}{2},\frac{1}{3},...$ which is bounded. As $0\notin (0,1]$,the sequence is not convergent.
I thought about proving that if the sequence has a convergent subsequence, it must converge to $0$, then as $0\notin (0,1]$, we have a contradiction. 
Here is my attempt:
Suppose, in order to get a contradiction, that a subsequence of the sequence $x_n=\frac{1}{n}$, call it $(x_{n_j})$, is a convergent subsequence. Suppose it converges to $a\in (0,1]$, then from the definition of convergence:
$(\exists a\in (0,1])(\forall \epsilon>0)(\exists N\in \Bbb N)(j\ge N\implies |x_{n_j}-a|<\epsilon)$
To obtain the contradiction, we prove that for this $a$:
$(\exists \epsilon>0)(\forall N\in \Bbb N)(\exists j\ge N\land |x_{n_j}-a|\ge\epsilon)$
Proof:
We know that $\frac{1}{n}$ can be less then any $a>0$ if we take $n$ large, so we have a term $x_{n_l}$ of $(x_{n_j})$ such that $x_{n_l}<a$.
Set $\epsilon=a-x_{n_l}$, then for all $N\in \Bbb N$, take $j\ge l$, then we have $|a-x_{n_j}|>\epsilon$ since the sequence $\frac{1}{n}$ is decreasing.
Could some please check if the argument above is correct? Thanks in advance! 
 A: Take as a  metric space the sequence space $(l^1(\mathbb{R}),d)$ where 
$l^1(\mathbb{R})=\{x_n| \sum_{n=1}^{\infty}|x=(x_1,x_2....)| < \infty ,x_n$ is a sequence of real numbers $\}$ and $d(x,y)=\sum_{n=1}^{\infty}|x_n-y_n|$
Then take the subset $c_{00}=\{x|x$ has a finite number of non zero terms 
$\}$. So you have the subspace $(c_{00},d)$ of $(l^1(\mathbb{R}),d)$.
Thus a sequence  in $c_{00}$ has the form $x=(a_1,a_2....a_m,0,0......)$
Now take the sequence $x_n=(1, \frac{1}{4},\frac{1}{9}...\frac{1}{n^2},0,0...) \in c_{00}$ which is a Cauchy sequence and bounded but does not converge in $c_{00}$
If it had a convergent subsequence in $c_{00}$ then it would converge in $c_{00}$ which is a contradiction.
If you want a simpler example take the subspace  $(\mathbb{R}$ \ $\mathbb{Q},d)$ of $(\mathbb{R},d)$
where $d$ is the usual metric and the sequence $a_n=\frac{\sqrt{2}}{n} \in  \mathbb{R}$ \ $\mathbb{Q}$.
$a_n$ is a bounded cauchy sequence in $\mathbb{R}$ \ $\mathbb{Q}$ but it does not have a convergent subsequence in $\mathbb{R}$ \ $\mathbb{Q}$
A: That's correct. If a metric space is not complete, than a nonconvergent Cauchy sequence is bounded and has no convergent subsequence.
The explanation is simpler if you know about completion: in the completion $\hat{X}$ of the space $X$, the given Cauchy sequence will converge to a point $p\in\hat{X}\setminus X$, so any subsequence will converge to $p$.
In your case the completion of $(0,1]$ is (isometric to) $[0,1]$ and the sequence converges to $0$.
An example in a different direction is the sequence $a_n=n$ in $\mathbb{R}$ with the metric
$$
d(x,y)=\min\{|x-y|,1\}
$$
Here every sequence is bounded, but the given sequence does not converge.
A: I see only one minor issue in the proof. In the very last paragraph, you say "then for all $N\in\Bbb N,$ take $j\ge l,$" which will be enough to make sure that $|a-x_{n_j}|\geq\epsilon,$ as you say. However, remember that we also want $j\ge N.$ To fix this, say "then for any $N\in\Bbb N,$ take $j=\max(l,N),$" instead. Do you see why we need to change "all" to "any"? Does this choice of $j$ make sense?
One could also prove (more briefly) that a convergent sequence has all of its subsequences converging to the same limit. Hence, since your sequence converges outside the space, so do all of its subsequences.
