Alaoglu's Theorem

Theorem
The closed unit ball of the dual of a normed space is compact with respect to the weak* topology.

Proof
Let $X$ be a normed vector space.

Denote by $B$ the closed unit ball in $X$.

Let $X^*$ be the dual of $X$.

Denote by $B^*$ the closed unit ball in $X^*$.

Let $\map {\mathcal F} B = \closedint {-1} 1^B$ be the topological space of functions from $B$ to $\closedint {-1} 1$.

By Tychonoff's Theorem, $\map {\mathcal F} B$ is compact with respect to the product topology.

We define the restriction map:
 * $R: B^*\to \map {\mathcal F} B$

by $\map R \psi = \psi \restriction_B$.

Lemma 1
$\map R {B^*}$ is a closed subset of $\map {\mathcal F} B$.

Lemma 2
$R$ is a homeomorphism from $B^*$ with the weak* topology to its image $\map R {B^*}$ seen as a subset of $\map {\mathcal F} B$ with the product topology.

Proof
Thus by lemma 2, $B^*$ in the weak* topology is homeomorphic with $\map R {B^*}$.

This is a closed subset of $\map {\mathcal F} B$ (by lemma 1) and thus compact.