Separable version of Banach Alaoglu I have a question regarding the following statement:
Let $X$ be a separable Banach space, then every bounded sequence in $ X^\ast $ has a weak$^\ast$-convergent subsequence.
I am sure that it suffices $X$ to be a normed vector space. However in most of my literature they state that it should be Banach. How comes so? Or am I wrong?
 A: You're right, $X$ doesn't have to be complete.  
By the Banach-Alaoglu theorem, the closed unit ball of $X^*$ is weak$^*$-compact.  Since $X$ is separable, the closed unit ball of $X^*$ is metrizable in the weak$^*$-topology.  If the sequence is bounded, we can reduce to the case it is contained in the closed unit ball of $X^*$, and the rest follows.
A: The proof of this theorem relies on two facts:
1) The unit ball $B_{X^*}$ is weak* compact for every normed space $X$.
2) The unit ball $B_{X^*}$ is weak* metrizable for every separable normed space $X$.
Indeed, if $(x_n^*)$ is a bounded sequence in $X^*$, then there exists $M>0$ such that the sequence $(x_n^*/M)$ is in $B_{X^*}$. By compactness and metrizability of $B_{X^*}$ with respect to the weak* topology, we have sequential compactness. Thus there is a subsequence $(x_{n_k}^*/M)$ such that $x_{n_k}^*/M \to x^*$ in weak* for some $x^*\in B_{X^*}$. Therefore $x_{n_k}^*\to Mx^*$ in weak*.
This goes to show that $X$ only needs to be a separable normed space. Some authors are only concerned with Banach spaces, so they fail to state results in their most general form.
