Separable Spaces and Alaoglu's Theorem Review

# Separable Spaces and Alaoglu's Theorem Review

We will now review some of the recent material regarding separable spaces and Alaoglu's theorem.

- On the
**Helly's Theorem**page we said that a topological space $X$ is said to be**Separable**if $X$ contains a countable and dense subset.

- We then proved Helly's theorem, which states that if $X$ is a separable normed linear space then every bounded sequence of continuous linear functionals in $X^*$ has a subsequence that weak* converges in $X^*$.

- On the
**If a Normed Linear Space X* is Separable then X is Separable**page we proved that if $X$ is a normed linear space and $X^*$ is separable then $X$ is also separable.

- On the
**A Reflexive Linear Space X is Separable IFF X* is Separable**page we then proved that if $X$ is a reflexive normed linear space then $X^*$ is separable if and only if $X$ is separable.

- Then, on the
**Closed Subspaces of Reflexive Spaces are Reflexive**page we proved that if $X$ is a reflexive normed linear space and $Y \subseteq X$ is a closed subspace then $Y$ is also reflexive.

- On the
**Separable Criterion for the Compactness and Sequential Compactness of the Closed Unit Ball of X* in the Weak* Topology**page we proved that if $X$ is a separable normed linear space then the closed unit ball in $X^*$ is compact and sequentially compact with respect to the weak* topology.

- We then proved
**Alaoglu's Theorem**which states that if $X$ is a normed linear space then the closed unit ball in $X^*$ is compact with respect to the weak* topology.

- On the
**Every Normed Linear Space is Isometrically Isomorphic to C(K) where K is a Compact Hausdorff Space**page we proved an interesting result. We proved that if $X$ is a normed linear space then $X$ is isometrically isomorphic to the space of continuous functions $C(K)$ for some compact Hausdorff space $K$.

- On the
**Every Bounded Sequence in a Reflexive Space Has a Weak Convergent Subsequence**page we proved that if $X$ is a reflexive normed linear space then every bounded sequence in $X$ has a subsequence that is weak convergent.