Timeline for Quotients of $\ell_\infty$ by separable subspaces
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2015 at 15:11 | vote | accept | M.González | ||
| Dec 10, 2013 at 19:02 | answer | added | Bill Johnson | timeline score: 16 | |
| Nov 16, 2013 at 11:44 | comment | added | M.González | @Bill Johnson: Thank you very much for your useful comments. | |
| Nov 15, 2013 at 19:35 | comment | added | Bill Johnson | Continuing the 2nd comment, it is then enough to assume that $M$ embeds into a separable conjugate space, for again by subspace homogeneity you can assume WLOG that $M$ is contained in a weak$^*$ closed separable subspace of $\ell_\infty$ and use the fact that if both $Y$ and $X/Y$ embed into $\ell_\infty$, then so does $X$. | |
| Nov 15, 2013 at 19:18 | comment | added | Bill Johnson | A second comment is that if $M$ is isomorphic to a separable conjugate space, then $\ell_\infty/M$ does embed into $\ell_\infty$. The reason is that then $M$ is isomorphic to a weak$^*$ closed subspace of $\ell_\infty$, so by the subspace homogeneity of $\ell_\infty$ you can assume that $M$ is weak$^*$ closed. | |
| Nov 15, 2013 at 16:16 | comment | added | Bill Johnson | Nice question. The only immediate comment I have is that $\ell_\infty/M$ does not embed into $\ell_\infty$ if $M$ is separable and $c_0$ embeds into $M$. WLOG [Lindenstrauss-Rosenthal] by the subspace homogeneity property of $\ell_\infty$ you can assume $c_0 \subset M$, and $c_0(R)$ embeds into $\ell_\infty/c_0$, and every operator from $c_0(R)$ into $\ell_\infty$ has separable range because weakly compact subsets of $\ell_\infty$ are separable since $\ell_\infty$ is the dual of a separable space. | |
| Nov 15, 2013 at 9:23 | history | edited | M.González | CC BY-SA 3.0 |
missing symbol
|
| Nov 15, 2013 at 7:50 | history | asked | M.González | CC BY-SA 3.0 |