Let $\ell$ be the set of sequences of real numbers where only a finite number of terms are different from zero $$\ell = \big\{\{x_n\}_{n=1}^\infty :x_i=0\text{ for all but a finite number of }i\text{-s}\big\}.$$ For $x=\{x_n\}$ and $y=\{y_n\}$ in $\ell$, define $$d(x,y)=\sup_{n\in\mathbb{N}}|x_n-y_n|.$$ Let $u_k\in\ell$ be defined by $$u_k=\bigg\{1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{k},0,0,\dots\bigg\}.$$ Show that $\{u_k\}_{k=1}^\infty$ is convergent, or show that $\{u_k\}_{k=1}^\infty$ is not convergent.

My strategy for showing that $\{u_k\}$ is not convergent is to show that it converges to $a=\{1,\frac{1}{2},\frac{1}{3},\dots\}$ which is not contained in $\ell$.

Let $\epsilon>0$ be given. Then we can choose $N=\lceil 1/\epsilon \rceil$, such that if $n\ge N$ then \begin{align} d(u_n,a) &= d\left(\left\{1,\frac{1}{2},\dots,\frac{1}{n},0,0,\dots\right\}, \left\{1,\frac{1}{2},\frac{1}{3},\dots\right\}\right) \\ &= \frac{1}{n+1}<\frac{1}{n}\le \frac{1}{N} \le \epsilon. \end{align}

Denoting the set of all bounded sequences of real numbers by $\ell^*$, i.e. $$\ell^* = \big\{\{x_n\}_{n=1}^\infty:x_i\in\mathbb{R} \text{ for all }i\in\mathbb{N} \big\},$$ it seems clear that $\ell\subset \ell^*$ and that $a\in\ell^*$. Also, since $\{u_k\}\rightarrow a\in\ell^*$ it cannot also converge to some other $b\in\ell^*$. Thus there is no way it can converge in $\ell$.

Does this make sense? Is this a valid strategy for showing that something does not converge in some metric space?

Any help would be greatly appreciated!


2 Answers 2


Yes. The completion $\ell^*$ is a Hausdorff space, so the limit is unique. Since the limit lies in $\ell^*\setminus\ell$, there cannot be any other limit in $\ell$.

Needed to say, what you write as a definition of $\ell^*$ is not the space of bounded sequences, but the space of all sequences. Neither of these two is what you're looking for, because the completion $\ell^*$ contains all sequences $\{x_1,x_2,x_3,\dotsc\}$ such that $x_i\xrightarrow{i\to\infty}0$.

  • $\begingroup$ Are you saying that the completion of $\ell$ is the set of all real sequences ? $\endgroup$
    – mercio
    Mar 2, 2015 at 16:44
  • $\begingroup$ @mercio That seems to be what I'm saying, and it seems to be some bullsh** at the same time. Sorry for that, I'll correct the answer. $\endgroup$
    – yo'
    Mar 2, 2015 at 16:54
  • $\begingroup$ @yo' Sorry for that inapropriated comment but I want to thank you for your help with LaTeX here: math.stackexchange.com/questions/1170253/… $\endgroup$ Mar 2, 2015 at 22:09
  • $\begingroup$ @yo' Unfrontunatly we can't send messages to others and your e-mail isn't shown so that's why I say it here. $\endgroup$ Mar 2, 2015 at 22:10
  • $\begingroup$ @Scientifica The ping from there has reached me! :) $\endgroup$
    – yo'
    Mar 2, 2015 at 22:17

No, you haven't given $\ell^\star$ a topology so it doesn't make sense to say that $\{u_k\} \to a$ in $\ell^*$, and even then you would need to show that $\ell^*$ is separated to that the sequence can't converge to two limits at the same time.

Instead you can do a completely straightforward proof that it doesn't converge :

Let $a \in \ell$. You want to show that $(u_k)$ doesn't converge to $a$, so you need to give an $\epsilon > 0$ such that $d(u_k,a) > \epsilon$ for infinitely many $k$.

Since $a_i$ is nonzero for finitely many integers, there has to be some $i$ such that $a_i = 0$. Then, for any $k \ge i$, $d(u_k,a) \ge |(u_k)_i - a_i| = \frac 1i$.

And so you are done by picking $\varepsilon = \frac 1i$

  • $\begingroup$ Is it not sufficient to equip $\ell^*$ with the metric $d$? I am not really sure what "giving $\ell^*$ a topology" entails, since I am doing a real analysis course in which the word topology has not been mentioned (yet). As for showing that $\{u_k\}$ does not converge to $a$, I agree. However, the problem is to prove that it does not converge to any element in $\ell$. $\endgroup$
    – Morten
    Mar 2, 2015 at 14:29
  • 1
    $\begingroup$ I'll chime in here. By giving $\ell^\star$ a topology he/she probably means that you have to define a distance in $\ell^\star$ in order to speak of convergence in $\ell^\star$. Also, $a$ is an arbitrary element of $\ell^\star$. $\endgroup$
    – Reveillark
    Mar 2, 2015 at 14:38
  • $\begingroup$ @Reveillark That's really a nit-picking. There's the metric $d$ which extends to $\ell^*$ and defines the topology. I really don't like this type of proofs. Yes, you can prove that the sequence doesn't converge "by hand". But it's much more natural to thing in general terms; find a natural way that would work for all similar sequences. $\endgroup$
    – yo'
    Mar 2, 2015 at 14:55
  • $\begingroup$ So I read the question again and you say "bounded sequences" even though the definition at the next line says otherwise. I wrote this under the assumption that $\ell^*$ is the set of all real sequences. However if you really meant $\ell^*$ to be the set of bounded sequences, then your reasoning works but you really should mention that you extend $d$ the obvious way to $\ell^*$ $\endgroup$
    – mercio
    Mar 2, 2015 at 16:42
  • $\begingroup$ @mercio Yes, I left out the part about extending $d$ in the original question for brevity, apologies. Anyway, in order to clear up some confusion on my part I have a silly question to ask: How does $\{\{x_n\}:x_n\in\mathbb{R}\}$ not denote the set of bounded sequences of real numbers? Is a bounded sequence not just any $\{x_n\}$ such that $|x_n|<\infty$ for any $n$? Is this not automatically implied in the above definition, since $\pm\infty\not\in\mathbb{R}$? $\endgroup$
    – Morten
    Mar 4, 2015 at 14:23

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