Let $X = \left\{(x_n):\sum\limits_{n=1}^\infty n|x_n|<\infty\right\}$, $d(x,y) = \sup|x_n - y_n|$. Show that the space $(X,d)$ is not complete. Prove that the space $c_0$ is its completion.

For the solution, I know that a metric space $(X,d)$ is complete if every Cauchy sequence $(x_n)$ in $X$ converges to a point in $X$ so for the first part I need to find a Cauchy sequence that does not converges to a point in $X$.

For $(x_n) = (1/n^3)_{n=1}^\infty$ and $(y_n) = (1/n^4)_{n=1}^\infty,$ $d(x,y)$ not converges. Thus $X$ is not complete. Is my counter example right?

For the second part, proving $c_o$ is its completion I am having trouble about what kind of approach I should follow.

  • 1
    $\begingroup$ you miss understand that sequence in $X$ is in fact a sequence of sequences (confusing right?). So you want to find a sequence of sequences that does not convergence to a sequence in $X$ (try to find a sequence (of sequences) that converge to the sequence $1/n^2$.) $\endgroup$ – Yanko Oct 30 '17 at 21:47
  • $\begingroup$ @yanko I think it is about my typo. $x_n$ = $(1/n^3,1/n^3,1/n^3,..)$. Yes that is what I want to find. $\endgroup$ – Pumpkin Oct 30 '17 at 21:50
  • $\begingroup$ Each $x_n$ is a constant sequence and is therefore not contained in $X$. $\endgroup$ – amsmath Oct 30 '17 at 21:52
  • $\begingroup$ Right I get it now, so you should carefully read amsmath's comment. Also I would consider sequences that turns out to be 0 at some point. $\endgroup$ – Yanko Oct 30 '17 at 21:53
  • $\begingroup$ @amsmath Sorry for the typo I corrected my $x_n$ example now it is not constant? $\endgroup$ – Pumpkin Oct 30 '17 at 21:54

For the first Q: Note that for if $x_n=0$ for all but finitely many $n$ then $(x_n)_n\in X.$

Let $x[j]=(x_{j,n})_n$ where $x_{j,n}=1/n$ for $n\leq j$ and $x_{j,n}=0$ for $n>j.$

Then $(x[j])_j$ is a Cauchy sequence in $X$. If this sequence converged to $y=(y_n)_n\in X$, then it would be necessary that $y_n=1/n$ for every $n.$ But $(1/n)_n\not\in X.$


This is an old question, but I noticed that its second part, showing that $c_0$ is the completion, was never addressed.

$c_0$ (which denotes the space of sequences converging to $0$, something I wasn't aware of) is equipped with the $\sup$ norm. We show $X$ is a dense subspace of $c_0$. If $x\in c_0$ and $\epsilon >0$, let $K$ be so large that $\vert x(k)\vert <\epsilon/2$ for $k\geq K$ (the notation $x(k)$ is used in place of $x_k$ for more clarity later, when, for example, $x_m(k)$ will mean the $k$-th element of the $m$-th sequence). The sequence $y$ with term

$$ y(k)=\begin{cases} x(k), & k<K \newline 0, & k\geq K \end{cases} $$

clearly satisfies $d(x,y)\leq\epsilon/2<\epsilon$ and is in $X$.

It is well-known that $c_0$ is complete. How to show that a Cauchy sequence of sequences converges is relevant here; below is my own approach to proving this, which is similar to the one in the linked post, except that the steps are in a different order. (I also availed myself of the notation in that post.)

Take a Cauchy sequence $(x_m)_{m\in\mathbb{N}}$ (each $x_m$ being an element of $c_0$) and define a new sequence $y$ by

$$ y(k) = \lim_{m\rightarrow\infty}x_m(k) $$

That the above limit actually exists is guaranteed by the fact that for fixed $k$, the number sequence $(x_m(k))_{m\in\mathbb{N}}$ is Cauchy, as $\vert x_m(k)-x_n(k)\vert\leq d(x_m,x_n)$.

Now, we show at once both that $y$ is bounded and that $(x_m)_{m\in\mathbb{N}}$ converges to $y$ in norm in the space of bounded sequences $\mathcal{l}^\infty$, of which $c_0$ is a subspace. Pick $\epsilon >0$ and $N$ so large that $d(x_m,x_n)< \epsilon/2$ for all $m,n\geq N$. Then, for $n\geq N$:

$$ \vert y(k)-x_n(k) \vert = \lim_{m\rightarrow\infty} \vert x_m(k)-x_n(k)\vert \leq \epsilon/2 $$

This is true for all $k$, so $y$ is bounded (remembering that $x_n$ is, of course, itself bounded) and $d(y,x_n)<\epsilon$.

All that's left to prove is $y\in c_0$. Let $\epsilon >0$ and pick $n$ so that $d(y,x_n)<\epsilon/2$. Then, let $K$ be such that if $k\geq K$, then $x_n(k)<\epsilon/2$. For $k\geq K$, calculate:

$$ \vert y(k)\vert\leq \vert y(k)-x_n(k) \vert + \vert x_n(k) \vert\leq d(y,x_n) + \vert x_n(k)\vert<\epsilon $$

Hence, $\lim_{k\rightarrow\infty}y(k)=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.