Timeline for Finding closed subspaces whose sum isn't closed
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2022 at 14:42 | comment | added | S Argyros | I do not claim that I have proved that this is a counterexample. it seems to me that if someone wants to think then the simplest case is the subspaces generated by uncountable many branches. | |
| Dec 27, 2022 at 14:28 | comment | added | S Argyros | We define the space $X$ through a norming set $W$. We consider a partial order $\preceq $ which induces a dyadic tree on $N$ .Denote by $s,t$ the finite segment of the tree and by $\sigma , \tau $ its branches. Then W is the minimal subset of $c_{00} $ containing all finite segments of the tree and it is closed in $(1/2, S^M )$ operation.This means that it includes $1/2 \sum_{i=1} ^{n}f_i $ with $\{f_i\}$ belond to $W$, have disjoint support and the min of their support is a Schreier set. The space $X^*$ is generated by all singletons and all branches. | |
| Dec 27, 2022 at 13:17 | comment | added | S Argyros | Bill,I agree with you that we should look for a negative answer. I thought a little in this direction and I think that the following is a good candidate. First there are James Tree spaces $X$ not containing $\ell_1$, with a boundedly complete basis, non separable dual such that $X^*/X_* $ is isomorphic to $c_0 (2^N )$. I believe that every non separable subspace of $X^*$ intersects the subspace $X_*$. I shall add in a next comment some more informations on the space X and its dual. There are such examples in the literature like the non separable HI in the Memoirs paper.But there are simpler | |
| Dec 27, 2022 at 0:58 | comment | added | Bill Johnson | Spyros, Lindenstrauss proved that every subspace of a reflexive space has a quasi-complement (actually, something more general than that), so any example that gives a negative answer to your question must be non reflexive. | |
| Dec 26, 2022 at 11:06 | comment | added | S Argyros | After thinking a little more I understood that there is no non separable extension of Mazur theorem since there are non separable HI spaces. However the question in my previous comment remains open. | |
| Dec 25, 2022 at 10:38 | comment | added | S Argyros | It seems to me that the answer to the question could be positive if there exists a nonseparable extension of Mazur theorem. Is there something like this? | |
| Dec 25, 2022 at 10:31 | comment | added | S Argyros | I return to the main step of answering the question. Namely for any $V_0$ closed subspace of $V$ such that space $V/ V_0 $ is of infinite dimension then there exists a closed inf. dim. subspace $W$ such that $V_0 \cap W =0 $.My question is the following. Assume that the space $V/V_0 $ is non separable.Is it possible the subspace $W$ be chosen be nonseparable. For a separable $W$ , as Bill explained , it is enough to find a Schauder basic sequence $(x_n)$ such that their biorthogonals $(f_n)$ belong to $ V_{0} ^ {\perp} $. For this we can also use a variant of the proof of Mazur Thm. | |
| Dec 24, 2022 at 22:58 | history | edited | Nik Weaver | CC BY-SA 4.0 |
added 57 characters in body
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| Dec 24, 2022 at 16:04 | comment | added | Nik Weaver | Nice! Thank you, Bill and Spyros. | |
| Dec 24, 2022 at 16:03 | vote | accept | Nik Weaver | ||
| Dec 24, 2022 at 16:03 | history | edited | Nik Weaver | CC BY-SA 4.0 |
added 184 characters in body
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| Dec 24, 2022 at 9:20 | comment | added | S Argyros | Thanks Bill. It was clear to me that the critical point in my previous comment was to determine the subspace 𝑊 . Your explanation is clear and more correct from what I had in mind. | |
| Dec 23, 2022 at 23:20 | history | answered | Bill Johnson | CC BY-SA 4.0 |