# Is the "hereditarily indecomposable" property separably determined?

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

• 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 Argyros-Tolias 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 Lopez-Adad $X_{\omega_1}$ that is reflexive non-separable and contains no UBS so it is HI saturated. That may be a candidate for a counterexample. Nov 21, 2021 at 16:05
• @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 Argyros-Tolias [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. Nov 21, 2021 at 18:06

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 $$X$$.