# Banach-Alaoglu Theorem/Proof 2

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## Theorem

Let $X$ be a separable normed vector space.

Then the closed unit ball in its normed dual $X^*$ is sequentially compact with respect to the weak-$\ast$ topology.

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## 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 $\closedint {-1} 1$.

- $\map \FF B$

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 3

$R \sqbrk {B^*}$ is a closed subset of $\map \FF B$.

$\Box$

### Lemma 4

$R$ is a homeomorphism from $B^*$ with the weak* topology to its image:

- $R \sqbrk {B^*}$

seen as a subset of $\map \FF B$ with the product topology.

$\Box$

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.

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$\blacksquare$

## Also known as

The **Banach-Alaoglu Theorem** is also known just as **Alaoglu's Theorem**.

## Source of Name

This entry was named for Stefan Banach and Leonidas Alaoglu.