The Relationship between Reflexive Space, Separable Space and Compactness I'm currently studying Functional Analysis. I am confused about the relationship between reflexive spaces, separable spaces, and compactness of the unit ball (in different space, in different topology).

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*Does the compactness of the closed unit ball in $X^*$ in weak* topology related to $X$ separable? My teacher tolds us that if $X$ is separable, then for any bounded sequence $\{f_n\} \subset X^*$ it contains a converged subsequence. But Banach-Alaogu tells us that the unit ball in $X^*$ is closed in weak* topology.


*How can we prove a space is reflexive? So far as I know, we can prove the uniform convexity property, or instead we can prove the unit ball in $X$ is precompact in weak topology. Are there other ways?


*I'm thinking about the relationship between reflexive and separable. Are there any results about that? How can I understand the connection between them?
 A: *

*The unit ball in the dual space $X^{*}$ is always compact under the weak$^{*}$ topology by the Banach-Alaoglu theorem, however, recall that for a general topological space compactness and sequential compactness are not equivalent. If X is separable one can show (see for example Theorem 3.28 in Haim Brezis book) that the unit ball in the dual space is metrizable and hence also sequentially compact.

*One can sometimes give an explicit characterization of the dual space, as in the case of $L^{p}$, for $1 < p < \infty$. There is also a result that says if $Y$ is a closed subspace of a reflexive space $X$, then $Y$ is reflexive.

*I have no useful comments here.
A: For 3:
If we take $X$ to be a set, we can define $\ell^2(X)$ to be the set of functions $f : X \rightarrow \mathbb{R}$ such that $$
\| f \| = \sum_{x \in X} |f(x)|^2
$$
converges. This space $\ell^2(X)$ is a Hilbert space, and therefore is reflexive. Now, we can define the standard orthonormal basis $(e_x)_{x \in X}$ to have $e_x(x) = 1$ and $e_x(y) = 0$ for $y \in X$ with $y \neq x$. If $x,y \in X$ are different, then $\| e_x - e_y \| = 2$. So $\{e_x\}_{x \in X}$ is a discrete subset of $\ell^2(X)$. 
Now, take $X$ to be your favourite uncountable set. Then $\{e_x\}_{x \in X}$ is a discrete uncountable metric space, and therefore not separable. This proves that $\ell^2(X)$ is not separable, because subspaces of separable metric spaces are separable.
