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 $a$:

$(\exists \epsilon>0)(\forall N\in \Bbb N)(\exists j\ge N\land |x_{n_j}-a|\ge\epsilon)$


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!

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    $\begingroup$ For some short substantation of your statement it suffice to notice that if a sequence converges, then every subsequence converges to the same limit. Sequence $(\frac{1}{n})_{n=1}^{\infty}$ and hence every subsequence is convergent to $0$ in $[0,1]$, so it cannot be convergent in $(0,1]$. $\endgroup$ – Przemek Jul 16 '17 at 13:48
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    $\begingroup$ @PropositionX Your argument is correct, but Przemek is right, one could simplify the argument. Even in complete metric spaces a bounded sequence need not have a convergent subsequence, for example in the case of infinite-dimensional Banach spaces where for example the unit ball is not compact $\endgroup$ – Peter Melech Jul 16 '17 at 13:53
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    $\begingroup$ @PeterMelech Also, if you take a space with infinitely many points and equip it with the discrete distance $d(x,y)=1$ for all $x\ne y$, that would be a complete metric space. It is a bounded space as well (diameter one) so all sequences are automatically bounded. But it is easy to find sequences that do not have convergent subsequences (for example the sequence terms are pairwise distinct). $\endgroup$ – Jeppe Stig Nielsen Jul 16 '17 at 14:48
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    $\begingroup$ You could also, if you like, take $\{1/n: n\in \mathbb N\}$ to be the whole space....... You can also show that for any$ x>0$ there exists$ r>0$ such that $\{n\in \mathbb N: x_n\in (-r+x,r+x)\}$ is bounded and hence finite. Because a sub-sequence $(x_{n_j})_j $ converges to $ x$ iff for all $r>0$ the set $ \{j\in \mathbb N: x_{n_j}\not \in (-r+x,r+x)\} $ is finite,..... so $ if (x_{n_j})_j$ converges to $x$ then for all $r>0$ the set $\{j\in \mathbb N: x_{n_j}\in (-r+x,r+x) \}$ is infinite. $\endgroup$ – DanielWainfleet Jul 17 '17 at 2:39
  • $\begingroup$ Thanks you all who leave comments. Got it! $\endgroup$ – PropositionX Jul 17 '17 at 3:58

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.


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 $\mathbb{R}$ \ $\mathbb{Q}$


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.


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