Banach-Alaoglu Theorem/Proof 2

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:
 * $\map \FF B = \closedint {-1} 1^B$

be the topological space of functions from $B$ to $\closedint {-1} 1$.

By Tychonoff's Theorem:
 * $\map \FF B$

is compact with respect to the product topology.

We define the restriction map:
 * $R: B^* \to \map \FF B$

by:
 * $\forall \psi \in B^*: \map R \psi = \psi \restriction_B$

Lemma 4
Thus by Lemma 4, $B^*$ in the weak* topology is homeomorphic with $R \sqbrk {B^*}$.

This is a closed set of $\map \FF B$ (by Lemma 3) and thus compact.

By the Eberlein-Šmulian Theorem, this is sequentially compact.