Timeline for Is this property an isomorphic characterization of $\ell_1(\Gamma)$?
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10 events
| when toggle format | what | by | license | comment | |
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| May 24, 2018 at 14:53 | comment | added | M.González | Thank you for the comments. I was trying to show that a separably projective space $E$ is isomorphic to $\ell_1(\Gamma)$. The case $E$ separable is easy, and the question tries to know what happens if $E$ has the separable complementation property. | |
| May 23, 2018 at 16:17 | comment | added | Bill Johnson | From my paper with Odell (numdam.org/article/CM_1974__28_1_37_0.pdf) one can deduce that Manuel's questions have positive answers if you replace $\ell_1$ by $\ell_p$, $1<p<\infty$. The argument does not work for $\ell_1$; maybe that is why Manuel asked this question. | |
| May 23, 2018 at 15:46 | comment | added | Bill Johnson | Thanks, Tomek. I am glad that I did not try to check out my wrong guess! | |
| May 23, 2018 at 14:32 | comment | added | Tomasz Kania | @BillJohnson, however there is a compact space $K$ for which Bill's idea works. Indeed, Marciszewski constructed an Eberlein compact space $K$ such that $C(K)$ contains an uncomplemented copy of $c_0(\Gamma)$ (See Remark 2 in impan.pl/en/publishing-house/journals-and-series/…) As $K$ is Eberlein, $C(K)$ is WCG, hence it has the separable complementation property. In particular, every copy of $c_0$ in $C(K)$ is complemented. | |
| May 23, 2018 at 8:27 | comment | added | Tomasz Kania | @BillJohnson, actually it will be complemented. See Theorem 1.1 in londmathsoc.onlinelibrary.wiley.com/doi/epdf/10.1112/… | |
| May 22, 2018 at 1:15 | comment | added | Bill Johnson | Let $X=c_0(\omega_1)$ and $Y=C(\omega_1)$. Clearly every separable subspace of $X$ is contained in a larger separable subspace of $X$ that is 2-complemented in $Y$. Is $X$ complemented in $Y$? My guess is that standard arguments give a negative answer, but I have not tried to check it out. | |
| May 21, 2018 at 19:37 | comment | added | Bill Johnson | I am not sure. Maybe Lewis and Stegall proved that such a space must embed into $\ell_(\Gamma)$ (because $E$ has the Radon-Nikodym property). Even if that is correct, there is still the problem of seeing that $E$ is complementably embeddable into $\ell_1(\Gamma)$. Probably no one has ever looked at this. $$$$ Does there exist $X\subset Y$ s.t. every separable subspace of $X$ is contained in a separable subspace of $X$ that is complemented in $Y$, yet $X$ is not complemented in $Y$? | |
| May 21, 2018 at 17:35 | comment | added | M.González | And what happens replacing $(1+\varepsilon)$ by a fixed constant $C$ with $1<C<\infty$ in both places? | |
| May 21, 2018 at 17:12 | comment | added | Bill Johnson | Yes. $E$ is $\cal{L}_1$-$(1+\epsilon)$ for all $\epsilon >0$, hence is isometrically isomorphic to $L_1(\mu)$ for some measure $\mu$, and $\mu$ must be purely atomic since, e.g., your hypothesis implies that $E$ has the Schur property. | |
| May 21, 2018 at 16:35 | history | asked | M.González | CC BY-SA 4.0 |