Is it true that a Banach space $X$ is hereditarily indecomposable if every separable closed subspace of $X$ is hereditarily indecomposable?

$\begingroup$ I don't know, but I think the answer should be yes. All HI spaces embed into $\ell_\infty$ and so they can't be too large. The proof of this fact is short and uses a characterization of HI spaces due to Milman. It is the Memoirs of ArgyrosTolias that I don't have with me. Maybe one can use that characterization to prove that the HI property is separable determined? In the other direction, there is a space of Argyros, Todocevic and LopezAdad $X_{\omega_1}$ that is reflexive nonseparable and contains no UBS so it is HI saturated. That may be a candidate for a counterexample. $\endgroup$– Kevin BeanlandNov 21, 2021 at 16:05

$\begingroup$ @KevinBeanland Thanks for your comment. For the record, Milman's characterization is given in Proposition 1.1, and the embedding is proved in Proposition 1.3 in the Memoirs of ArgyrosTolias [dx.doi.org/10.1090/memo/0806] I can see at this moment that these beautiful results would be overkill to answer a perhaps simple question. $\endgroup$– Onur OktayNov 21, 2021 at 18:06
1 Answer
If $X$ is not HI, then there exist a decomposable closed subspace $Y\subseteq X$. Let $Y=Y_1\oplus Y_2$ be a decomposition, let $Z_1\subseteq Y_1$ and $Z_2\subseteq Y_2$ be separable closed subspaces, and let $Z=Z_1\oplus Z_2$. Clearly $Z$ is a separable decomposable closed subspace of $X$.

$\begingroup$ This is perhaps obvious, however it took a rookie (that would be me) a couple of days to see it. $\endgroup$ Nov 21, 2021 at 18:01


1$\begingroup$ @KevinBeanland It is encouraging to know that I had a company in this island of no see, although I was there for a long period of time and my company only for a few minutes. Luckily I built the Titanik (selfsarcasm alert!) and saved the humanity from a difficult place. $\endgroup$ Nov 21, 2021 at 18:51