# Show that the space $(X,d)$ is not complete and prove its completion

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.

• 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 $1/n^2$.) – Yanko Oct 30 '17 at 21:47
• @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. – Pumpkin Oct 30 '17 at 21:50
• Each $x_n$ is a constant sequence and is therefore not contained in $X$. – amsmath Oct 30 '17 at 21:52
• 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. – Yanko Oct 30 '17 at 21:53
• @amsmath Sorry for the typo I corrected my $x_n$ example now it is not constant? – 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

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