Regarding an isomorphism between a subspace of $\ell^{\infty}$ and $\ell^1$ Let $c_0$ be the subspace of $\ell^\infty$ consisting of sequences that converge to $0$. Show that $c_0$ is a closed subspace of $\ell^{\infty}$ whose dual space is isomorphic to $\ell^1$. Conclude that $c_0$ is not reflexive
and therefore neither is $\ell^{\infty}$.
The final part seems clear to me, the one regarding reflexivity. For the first part, about closedness, I am not sure which of the methods is appropriate. Would showing that the complement is open work or is there something better? For the isomorphism part I have no idea.
Thank you very much in advance for any help.
 A: As for the closedness. Let $x\in\ell_\infty$ be the limit point of $\{x_n:n\in\mathbb{N}\}\subset c_0$, i.e. $x=\lim\limits_{n\to\infty} x_n$ in $\ell_\infty$. Since convergence in $\ell_\infty$ is the uniform convergence on $\mathbb{N}$ we can "interchange limit signs". More preciesly
$$
\begin{align}
\lim\limits_{k\to\infty}x(k)&=\lim\limits_{k\to\infty}\lim\limits_{n\to\infty} x_n(k)\\
&=\lim\limits_{n\to\infty}\lim\limits_{k\to\infty} x_n(k)\\
&=\lim\limits_{n\to\infty}0\quad (\text{ since } \{x_n:n\in\mathbb{N}\}\subset c_0)\\
&=0
\end{align}
$$
Hence $x\in c_0$. Since $x$ was choosen too be arbitrary limit point of $c_0$, we conclude that $c_0$ is closed.
As for duality. Consider map
$$
I:\ell_1\to (c_0)^*:x\mapsto\left(y\mapsto\sum\limits_{k=1}^\infty x(k)y(k)\right)
$$ 
We need to prove that $I$ is a surjective isometry. Take $x\in\ell_1$, then for all $y\in c_0$ we have
$$
\begin{align}
|I(x)(y)|&=\left|\sum\limits_{k=1}^\infty x(k)y(k)\right|
\leq\sum\limits_{k=1}^\infty |x(k)y(k)|
\leq\sum\limits_{k=1}^\infty \sup\{|x(k)|:k\in\mathbb{N}\}|y(k)|\\
&=\sum\limits_{k=1}^\infty \Vert x\Vert |y(k)|
=\Vert x\Vert\sum\limits_{k=1}^\infty|y(k)|=\Vert x\Vert\Vert y\Vert
\end{align}
$$
Hence $\Vert I(x)\Vert\leq \Vert x\Vert$. Now fix $\varepsilon>0$, then there  exist $K\in\mathbb{N}$ such that $\sum\limits_{k=1}^K|x(k)|>\Vert x\Vert-\varepsilon$. Then consider $y'\in c_0$ defined by
$$
y'(k)=
\begin{cases}
\overline{x(k)}|x(k)|^{-1}&\quad\text{ if }\quad k\leq K\\
0&\quad\text{ otherwise}
\end{cases}
$$
In this case $\Vert y'\Vert=1$ and
$$
\begin{align}
|I(x)(y')|&=\left|\sum\limits_{k=1}^\infty x(k)y'(k)\right|
=\sum\limits_{k=1}^K x(k)\overline{x(k)}|x(k)|^{-1}
=\sum\limits_{k=1}^K |x(k)|
=\sum\limits_{k=1}^K |x(k)|\Vert y'\Vert\\
&\geq (\Vert x\Vert-\varepsilon)\Vert y'\Vert\\
\end{align}
$$
Hence $\Vert I(x)\Vert\geq\Vert x\Vert-\varepsilon$. Since $\varepsilon$ is arbitrary we conclude $\Vert I(x)\Vert\geq\Vert x\Vert$. On the other hand as we proved earlier $\Vert I(x)\Vert\leq \Vert x\Vert$, hence $\Vert I(x)\Vert=\Vert x\Vert$. Since $x$ is arbitrary, then $I$ is an isometry. 
Let's show that $I$ is surjective. For each $n\in\mathbb{N}$ we denote by $e_n$ the vectrofrom $c_0$ defined by
$$
e_n(k)=
\begin{cases}
1&\quad\text{ if }\quad k=n\\
0&\quad\text{ otherwise }
\end{cases}
$$
Now take arbitrary $f\in(c_0)^*$ and consider $x'\in\ell_\infty$ defined by
$x'(k)=f(e_k)$. Fix $K\in\mathbb{N}$ and consider vector $y''\in c_0$ defined by 
$$
y''(k)=
\begin{cases}
\overline{f(e_k)}|f(e_k)|^{-1}&\quad\text{ if }\quad k\leq K\\
0&\quad\text{ otherwise }
\end{cases}
$$ 
Obviously $\Vert y''\Vert=1$. Moreover
$$
\sum\limits_{k=1}^K |x'(k)|=\sum\limits_{k=1}^K f(e_k)y''(k)=f\left(\sum\limits_{k=1}^K y''(k)e_k\right)=f(y'')=|f(y'')|\leq\Vert f\Vert\Vert y''\Vert=\Vert f\Vert
$$
Since $K\in\mathbb{N}$ is arbitrary 
$$
\sum\limits_{k=1}^\infty |x'(k)|\leq\Vert f\Vert
$$
Hence $x'\in\ell_1$. Fix $\varepsilon>0$ and $y\in c_0$. Then there exists $K\in\mathbb{N}$ such that $k>K$ implies $|y(k)|<\varepsilon$. This is equivalent to say that $k>K$ implies $\left\Vert y-\sum\limits_{n=1}^k y(n)e_n\right\Vert<\varepsilon$. Since $\varepsilon$ is arbitrary this is equivlent to $y=\lim\limits_{n=1}^\infty y(n)e_n$. Then we may say
$$
f(y)=f\left(\sum\limits_{n=1}^\infty y(n)e_n\right)=\sum\limits_{n=1}^\infty y(n)f(e_n)=\sum\limits_{n=1}^\infty y(n)x'(n)=I(x')(y)
$$
Since $y\in c_0$ is arbitrary, then $f=I(x')$. Since $f\in (c_0)^*$ is arbitrary $I$ is surjective. 
Since $I$ isometric and surjective this is isometric isomorphism, hence $(c_0)^*\cong\ell_1$.
