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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 $\Phi$?

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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 $\Phi$

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  • $\begingroup$ $ 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
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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 $\|x-x_n\| \to 0$.

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