Let $E$ be a Riesz space with order unit $u$. With $$\left\|f\right\|_u: = \inf\{\lambda \in [0,\infty): -\lambda u\leq f\leq \lambda u\} $$ $E$ becomes a normed space. The following sets are subsets of the norm-dual of $(E,\left\|\cdot\right\|_u)$ endowed with the weak$^*$ topology. Let
$$K=\{\phi:E\to \mathbb{R}: \phi \text{ is linear and positive}\}$$ $$\Sigma=\{\phi\in K: \phi(u)=1\}$$ $$\Lambda =\text{extreme points of $\Sigma$} $$
One can show that $\Sigma$ is weak$^*$ compact with Alaoglu's theorem. I am trying to understand why (i) $\Lambda$ is weak$^*$-closed, and (2) $\Sigma$ is the convex hull of $\Lambda$.
By Krein-Millman's theorem we know that $\Sigma = \overline{\text{conv($\Lambda)$}}$. So, then we must find out why $\text{conv($\Lambda$)}$ is closed. I know that the convex hull of a compact set need not even be closed, so it does not follow from the compactness of $\Lambda$. Some valuable insight is very much appreciated.