2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.