$c_0$, the space of sequences converging $0$ is complete with dual $\ell^1(\mathbb{N})$ Let $c_0$ be the space of all complex sequences $(a_n)$ such that $$\lim_{n \to \infty} |a_n| =0$$  with norm $\|(a_n)\|_{c_0} = \sup_{n} |a_n|$.
Is it fair to say that:  
Let $\{(a_n)\}_{n \in \mathbb{N}}$ be a cauchy sequence in $c_0$. Let $\epsilon >0$. Then there exists $N$ such that for $n,m \geq N$, 
$$\|(a_n) - (a_m)\|_{c_0} = \sup_{k} |a_{n,k}-a_{m,k}| < \frac{\epsilon}{2}.$$
Let $n_0 \geq N$. Notice also that since $(a_{n_0}) \in c_0$, there exists $K$ such that, for $k \geq K$, $|a_{n_0,k}| < \frac{\epsilon}{2}$. Therefore, define $(a) \in c_0$ as the sequence $a_k = a_{n_0,k}$ for $k < K$ and $a_k = 0$ for $k \geq K$. Then for $n_0$ as before and  $m \geq N$, $\|(a_m)-(a)\|_{c_0} \leq \|(a_m)-(a_{n_0}) \|_{c_0} +  \|(a_m)-(a_{n_0}) \|_{c_0} < \epsilon$. Thus, $\{(a_n)\}$ converges so $c_0$ is complete.  ?  
I was unsure about this proof since it seems that the cauchy sequence could converge to different limits, but I suppose in the uniform norm of $c_0$, the limits would only differ by $\epsilon$ so the limit is unique.  
Finally I need to show that $\ell^1(\mathbb{N})$ is the dual space of $c_0$. Is it correct to say that, given $(x_n) \in \ell^1$, I want the action of $$(x_n) : c_0 \to \mathbb{C}$$ to be defined as $$(x_n): (a_n) \mapsto \sum_{n \in \mathbb{N}} x_n\overline{a_n}?$$ This clearly converges by comparison, and I'm left to show that all such linear functionals arise this way. It seems prudent to say that a linear functional on $(a_n) \in c_0$ can be determined by it's action on finite truncations of $(a_n)$, and therefore can be identified with some element in $$\operatorname{span}\{e_i\}_{i \in \mathbb{N}}$$ for $e_i$ the orthonormal hilbert basis for $\ell^1(\mathbb{N})$. A very slight hint on how to proceed with this would be appreciated, but please please only a slight hint. I want to proceed on my own.
 A: You are right to be skeptical; your limit $a$ does not work. Hint: Knowing that
$$ \sup_k |a_{n,k} - a_{m,k}| → 0$$
means that for each $k$, $a_{n,k}∈ℝ $ is a ...
Proof that the map $\phi: \ell^1 → c_0^*$ given by $T(x)(a) = \sum a_i \overline{x_i}$ is an isometric isomorphism. 

 We use $ℝ$ but for $\Bbb C$ one just has to find a proper replacement for 'sgn'. Note that  $$ |\phi(a)(x)| = \left|\sum a_i x_i\right| \leq ‖a‖_{c_0} ‖ x‖_{\ell^1}$$ so that $‖\phi(a)‖_{c_0^*}\leq ‖a‖_{c_0}$; also, define $x^n∈ c_0$ with $‖x^n‖_{c_0} = 1$ by $$ x_i^n = (\text{sgn } a_i) \Bbb 1_{i\le n}$$ Then $|\phi(a)(x^n)| = \sum_{i=0}^n |a_i|$, so $‖\phi(a)‖_{c_0^*} \geq \sum_{i=0}^n |a_i|$ for every $n$. Send $n→∞$ to get that $\phi$ is an isometry. Isometries are injective; so now one has to prove its surjective. Pick some $T∈ c_0^*$ and consider the sequence $t=(T(e_i))_{i=1}^\infty$;  were this sequence in $\ell^1$, then by linearity and continuity, $$\phi(t)(a)=\sum T(e_i)a_i = T(\sum a_i e_i) = T(a)$$ i.e. $\phi(t) = T$. Now note that 
$$\sum_{i=0}^n |t_i| = \sum_{i=0}^n T(e_i) \text{sgn }(t_i) = T(\sum_{i=0}^n [\text{sgn }t_i] e_i) \leq ‖T‖_{c_0^*} ‖([\text{sgn }t_i] e_i)\Bbb 1_{i\leq n}‖_{c_0} = ‖T‖_{c_0^*}  $$ 

 which finishes the proof.

