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Dec 20, 2022 at 12:39 comment added HardyHulley Thanks @Mikael for the very comprehensive answer and to everyone else for their insights. I expected that the answer would be negative, but it is good to see the argument laid out clearly. I'm curious, though, about what happens with Banach spaces, such as $L^1(0,1)$, which have no predual.
Dec 15, 2022 at 8:34 history edited Matthew Daws CC BY-SA 4.0
Fix arXiv link
Dec 15, 2022 at 4:47 vote accept HardyHulley
Dec 14, 2022 at 23:54 comment added Mikael de la Salle I changed the phrasing to make this clear.
Dec 14, 2022 at 23:52 history edited Mikael de la Salle CC BY-SA 4.0
Changed isomorphic to isometrically isomorphic (cd Dirk Werner's comment).
Dec 14, 2022 at 23:47 comment added Mikael de la Salle Thanks to both for the clarification. I know that this is clear to you, but in case somebody less familiar read these comments, let me just point out that the mere fact that $\ell_1$ has two non-isometric preduals (eg $c$ and $c_0$) is enough to say that it admits two distinct weak-* topologies.and answer the question.
Dec 14, 2022 at 20:11 comment added Robert Furber Nonetheless, it is my favourite example where $\aleph_1$ occurs instead of $\mathfrak{c}$, there are $\aleph_1$ separable commutative C$^*$-algebras up to isomorphism (but $\mathfrak{c}$ up to isometry).
Dec 14, 2022 at 20:04 comment added Robert Furber We can get a predual of $\ell^1$ not isomorphic to $c_0$ by trying $C(\omega^{\omega}+1)$, though I don't know a simple way of proving the non-isomorphism, only the Szlenk index.
Dec 14, 2022 at 18:06 comment added Dirk Werner Actually, $c$ and $c_0$ are isomorphic, but not isometrically isomorphic.
Dec 14, 2022 at 15:09 history edited Mikael de la Salle CC BY-SA 4.0
Typos corrected
Dec 14, 2022 at 8:25 history answered Mikael de la Salle CC BY-SA 4.0