Let $ \Phi= \{ a_n \in R : a_n=0$ after some $ n \} $ be equipped with the sup metric $d $ defined by $d( a_n, b_n) = \sup_n \lvert a_n-b_n \rvert $.

We are supposed to prove that $ C := \{x_n \in R : \lim x_n =0 \} $ with the same metric $d$ is a completion of $(\Phi, d)$. [Note that you do not need to prove $(C,d) $ is complete.]

So far, I have done the following: Firstly, $(C,d )$ is complete (already given in the question). Secondly , I have to define an isometry $i : \Phi \rightarrow C $ which I define $ i(a_n)= a_n$ as identitity since the metric in the two space is identical. The only argument left is that I have to show that $ \bar{\Phi}$= C $, that is ,closure of the incomplete metric space is equal to our candidate completion metric space.

For this, I tended to use sequential characterisation of closure which is that if $x_n \in C=\bar{\Phi}, $ then $ \exists $ a sequence $ a^{(n)}$ s.t $ \lim_n a^{(n)} = x_n $. This is where I am stuck. How do we show that every sequence in $C$ is a limit of some sequence of sequences of $\Phi$?


NOTE :$(C,d)$ is complete.In order to show that $C$ is completion of $\Phi$ we need to show that $\Phi$ is isometric to a dense subset of $C$.

$\Phi=\{x_n:x_n=0 \text{after finite n}\}$ is dense in $C=\{x_n:\lim x_n=0\}$ .

Let $a_n\in C\implies \lim a_n=0\implies \exists m\in \Bbb N\text{such that }|a_n|<\epsilon \forall n\ge m$.

Consider $p_1=(a_1,0,0,\ldots 0),p_2=(a_1,a_2,0,0,\ldots ,0),\ldots ,p_n=(a_1,a_2,\ldots ,a_n,0,0,\ldots 0)$ and so on .

Note that each $(p_n)_n\in \Phi$.

Also for large $m\in \Bbb N$,$d((p_m)_m,a_n)=\sup _n|p_m-a_n|\le\dfrac{1}{m+1}\to 0\text{as} m\to \infty$

Hence $(p_n)_n\to a_n$. So $\Phi $ is dense in $C$.

Hence $\Phi$ is isometric to $\Phi$ which is a dense subset of $C$. Hence $C$ is a completion of $\Phi$

  • $\begingroup$ $ sup_n \lvert p_m-a_n \rvert = 1/ (m+1) $. this point is not clear to me? can you elaborate on it? $\endgroup$ – Quantes Nov 20 '16 at 7:44
  • $\begingroup$ You can verify it if you take $\frac{1}{n}$ as an example .I am in a hurry now;If you cannot do it wait a while $\endgroup$ – Learnmore Nov 20 '16 at 7:47
  • $\begingroup$ I understand that $ sup$ of that term goes to zero, but i did not understand why it will be exactly that term{ $1/(m+1) $ }. .but this is enough to prove it is a completion thank you very much. $\endgroup$ – Quantes Nov 20 '16 at 7:59
  • $\begingroup$ Hey I am back now.I have made a small edit .Also Notice that since $a_n\to 0$ so $|a_n|\le \frac{1}{n}\forall n\in \Bbb N$@Quantes $\endgroup$ – Learnmore Nov 20 '16 at 8:36
  • $\begingroup$ In case you are satisfied with the answer do accept it by clicking the checkmark $\endgroup$ – Learnmore Nov 20 '16 at 9:11

All you need to do is to show that if $x \in c_0$, then there are $x_n \in \Phi \subset c_0$ such that $x_n \to x$.

Let $x_n =\sum_{k=1}^n x(k) e_k$, note that $x_n \in \Phi$ and $\|x-x_n\| = \sup_{k > n} |x(k)|$. Since $x\in c_0$ we see that $x(k) \to 0$ and so $\|x-x_n\| \to 0$.


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.