Compactness in weak$^*$-topology on $l_1(\mathbb{N})$ Let $K$ denote the closed unit ball of $l_1(\mathbb{N})$ (considered as a vector space over $\mathbb{C}$). 
Is $\mathrm{co}(\mathrm{Ext}(K))$ (the convex hull of its extreme points) compact in the weak$^∗$-topology on $l_1(\mathbb{N})$? Justify your answer.
(Here $l_1(\mathbb{N})$ is identified as the dual space of $c_0(\mathbb{N})$) 
I know these two facts:  
- $K$ is the norm closure of $\mathrm{co}(\mathrm{Ext}(K))$ 
- Banach-Alaoglu's theorem tells me that $K$ is compact in the weak$^*$-topology on $l_1(\mathbb{N})$
 A: Claim 1: The extreme points of the (closed) unit ball $K=B_{l^1}$ of $l^1(\Bbb N)$ is precisely the set $E=\{ \alpha e_i:i\in\Bbb N, |\alpha|=1 \}$ where 
$e_i=(0,0,\dots,0,1,0,0,\dots)$, consisting of $0$ for all but the $i^{\text{th}}$ coordinate.
Proof: If $e_i = (1-\lambda)x+\lambda y$ for some $x,y\in B_{l^1}$, then $(1-\lambda)x_i+\lambda y_i=1$ and $|x_i|,|y_i|\le 1$. From these we deduce that $x_i=y_i=1$ and hence all other entries of $x,y$ are zeroes. Thus $x_i=y_i=e_i$, similar for $\alpha e_i$.
On the other hand, any point  $x\in B_{l^1}\backslash E$ must have a coordinate such that $|x_i|<1$. We can deduce from this that $x$ is not an extreme point (why?). QED

Having known that, it is easy to see that for any $x\in \text{co}(E)$, $x$ can have only finitely many nonzero terms, so $\text{co}(E) \subsetneq B_{l^1}$.
On the other hand, Krein-Milman theorem states that $\overline{\text{co}} ^*(E)=B_{l^1}$ so $\text{co}(E)$ is not a weak$^*$ closed set.
A weak$^*$ compact set must be weak$^*$ closed because the topology is Hausdorff, hence $\text{co}(E)=\text{co}(\text{Ext}(K))$ is not weak$^*$ compact.

Edit: If you haven't learned the Krein-Milman theorem for a locally convex space yet, then you can use following line of reasoning, given that you knew $\overline{\text{co}}(E)=B_{l^1}$: 
We clearly have
$$
B_{l^1} = \overline{\text{co}}(E) \subset \overline{\text{co}} ^*(E)
$$
since weak$^*$ topology is coarser than the norm topology. Also, the reverse inclusion 
$$
\overline{\text{co}} ^*(E) \subset B_{l^1}
$$ 
comes from the fact that $\text{co}(E) \subset B_{l^1}$ and Banach-Alaoglu ($B_{l^1}$ is weak$^*$ compact and hence weak$^*$ closed). This shows that
$$\overline{\text{co}} ^*(E)=B_{l^1}.$$ 
