Let $X$ be a Banach space. Is the following implication valid?

$$ (X,w) \textrm{ is hereditarily Lindelöf}~ \Rightarrow X^*~ \textrm{is separable} $$

The converse is clearly true, since the closed unit ball is relatively weak star second countable.

Def. A topological space $X$ is hereditarily Lindelöf if every subspace $Y\subseteq X$ is Lindelöf.

  • $\begingroup$ could you sketch why $X^\ast$ separable (strongly?) implies that $X$ is weakly hereditarily Lindelöf? $\endgroup$ – Henno Brandsma Jun 5 '18 at 13:00
  • $\begingroup$ @HennoBrandsma: Combination of two points get the result: (1) Let $Y$ be separable Banach space then, the closed unit ball of $Y^*$ is weak-star second countable. (2) The relative weak star topology of $X^{**}$ on $X$ is just the weak topology. $\endgroup$ – Ali Bagheri Jun 5 '18 at 14:50

Under CH there exists an example of a non-metrizable compact scattered Hausdorff space $K$ such that the Banach space $X=C(K)$ endowed with the weak topology is hereditarily Lindelof. The non-metrizability of $K$ implies that the Banach space $X=C(K)$ is not separable and then the dual $X^*$ is not separable as well. This example is due to Kunen and is described in the survey paper of Negrepontis in "Handbook of Set-Theoretic Topology" (1984).

  • $\begingroup$ Is $C(K)$ hereditarily Lindelöf in the weak topology? $\endgroup$ – Henno Brandsma Jun 6 '18 at 18:23
  • $\begingroup$ @HennoBrandsma Yes, see Theorems 7.3 and 7.4 in the Negrepontis survey. $\endgroup$ – Taras Banakh Jun 9 '18 at 19:01
  • $\begingroup$ interesting that $C_p(K)$ and $C(K)$ (in the traditional weak topology) are the same when $K$ is scattered compact. I really have to look at Zemadeni again.. I looked at $C(X)$ for another one of Kunen's example: his compact $L$-space. Nice area, IMHO. $\endgroup$ – Henno Brandsma Jun 10 '18 at 22:36
  • $\begingroup$ @HennoBrandsma It seems that you are right: the weak topology and the topology of pointswiese convergence are different on $C(K)$ even for infinite scattered $K$. What is true that they coincide on bounded subsets of $C(K)$. But this is sufficient for our purposes. $\endgroup$ – Taras Banakh Jun 11 '18 at 6:58

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