The space of convergent sequences is not dual to any space.

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I've been asked to prove that

(a) The space of sequences over some field $\mathbb{F}$ converging to zero $c_0$ equipped with sup norm $\|x\|=\sup_{k\in\mathbb{N}}|x_k|$ is not dual to any vector space.

(b) The space of sequences over some field $\mathbb{F}$ that converge $c$ equipped with sup norm $\|x\|=\sup_{k\in\mathbb{N}}|x_k|$ is not dual to any vector space.

I have succesfully solved part a by the following:

First, notice that the unit ball in $c_0$ has no extreme points. This is a little tedious but straightforward to prove.

Suppose $c_0$ is dual to some space $X$. Then by Alaoglu's theorem, the unit ball of $c_0$ is $w^*$ compact. It follows then by Krein-Milman that the convex hull of the extremal boundary of the unit ball is dense in the unit ball. However the convex hull of the extremal boundary is empty and thus contradiction.

I haven't been able to get the same argument for part b to work. The extreme points of the unit ball of $c$ are $\{x\in c : |x_i|=1 \forall i\}$. To the best I can tell, unlike part a the convex hull of the extremal boundary is dense in the unit ball.

I'd be really grateful for any direction or ideas anyone has. I'm beginning to doubt the truth of the result I'm trying to prove.

For reference, the version of Krein-Milman I have is:

Consider a vector space $X$ equipped with a weak topology induced by a separating space $X^*$ of functionals on $X$. Then for each convec compact subset $C$ of $X$, the convex hull of the extremal boundary $\delta C$ of $C$ is dense in $C$.

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Of course if $\Bbb F$ is just "some field" none of this makes any sense; there's no notion of convergence available to use in the definitions of $c$ and $c_0$. Probably we should instead assume that $\Bbb F$ is $\Bbb R$ or $\Bbb C$.

You're right that the convex hull of the extreme points of the unit ball of $c$ is dense in the unit ball; in fact if $\Bbb F=\Bbb C$ every point of the unit ball is a convex combination of two extreme points.

Note that $$c=c_0\oplus C,$$where $C$ is the space of all constant sequences. So if $c$ were the dual of $X$ then $c_0$ would be the dual of $C^\perp\subset X$.