# Is completeness necessary? X separable iff weak* topology on closed unit ball of dual is metrizable.

First some references:

• [M] Megginson - An Introduction to Banach Space Theory
• [D] Denkowski, Migórski, Papageorgiou, Socrates - An Introduction to Nonlinear Analysis
• [B] Brezis - Functional Analysis, Sobolev Spaces and Partial Differential Equations
• [DS] Dunford and Schwartz - Linear Operators Part I: General Theory

The following theorem appears in [M, p.231], [D, p.305], [B, p.74], [DS, p.426]. I'm sure it appears elsewhere too.

Theorem. Let $X$ be a Banach space. Then $X$ is separable iff the closed unit ball of $X^*$ is metrizable in the weak* topology (inherited from X*).

However, [M] only assumes $X$ is a normed space, not a Banach space. I've been through the proofs, and I can't see where completeness is being used.

Question 1. Is completeness of $X$ really necessary?

There is also a closely related theorem that appears in [D, p.305], [B, p.74], [DS, p.426]:

Theorem. Let $X$ be a Banach space. Then $X^{*}$ is separable iff the closed unit ball of $X$ is metrizable in the weak topology (inherited from X).

Question 2. Is completeness of $X$ really necessary?

No, in both caes. The dual does not "see" if $X$ is complete or not; the dual of a space and of its completion is the same. And separability and metrizability pass to subsets, so $X$ is separable/metrizable if and only if $\overline X$ is.
• But the weak$^*$ topology sees the completion: $\sigma(X^*,X)$ is strictly coarser than $\sigma(X^*,\tilde X)$. – Jochen Jun 29 '18 at 7:09
• But on the dual unter ball the weak topologies coincide: The ball is $\sigma(X^*,\tilde X)$-compact and hence the other weak topology can't be strictly coarser. – Jochen Jun 29 '18 at 19:55