Show that $(C,d)$ is a completion of $(\Phi,d)$ 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 $\Phi$?
 A: 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 $\Phi$
A: 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 $\|x-x_n\| \to 0$.
