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I am reading the Functional Analysis book of Conway, one question from the book is find a closed subspace M of $l^{\infty}=l^{\infty}(\mathbb{N})$ with the property that $l^{\infty}/M$ is separable. I have found a solution for this but here is my question :

Is it true that every non-separable normed space $X$ always contains a closed (proper) subspace $M$ such that $X/M$ is linear isometric to a separable normed space whose dimension is infinite ? i.e, are there a map $A$ and a separable normed space $Y$ whose dimension is infinite, st: $A: X/M\to Y$ which is linear, onto, and preserve the distance?

(Edit: I already have an answer for the following question I am thinking a about $l^{\infty}$ : can it contain a closed proper subspace M that $l^{\infty}/M$ is isometric to $l^{1}=l^{1}(\mathbb{N})$?)

edited Jun 15 2011 at 5:46

Yemon Choi
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asked Jun 14 2011 at 21:58

Steven
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By the way, something seems to be wrong with your quote from Conway – Yemon Choi Jun 14 2011 at 22:30

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The answer is affirmative. Namely, take M as the kernel of any [non-trivial] continuous linear functional on X. – Ady Jun 14 2011 at 22:51

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In your first question, about merely getting a separable quotient, you want to require the subspace $M$ of $X$ to not only be a proper subspace but of infinite codimension. Otherwise that question is trivial. – Andreas Blass Jun 14 2011 at 22:52

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Steven, Ady has already mentioned that Question 1 in your current post is the Separable Quotient Problem; the answer is unknown but it is known to be yes for certain classes of Banach spaces. For a survey detailing the partial results known in the mid-1990s, see Mujica's survey Separable quotients of Banach spaces, Rev. Mat. Univ. Complutense Madrid 10 (1997), 299–330. A more recent result, that every Banach space isomorphic to a dual space has a separable quotient, has been shown my Argyros et al in Unconditional families in Banach spaces, Math. Ann. (2008) 341:15–38. – Philip Brooker Jun 15 2011 at 5:30

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Just as I was posting my comment Bill posted his answer (which contains more info than my comment, however maybe the survey article interests you; find it at mat.ucm.es/serv/revmat/vol10-2/vol10-2e.pdf ) – Philip Brooker Jun 15 2011 at 5:40

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Andreas Blass · Accepted Answer · 2011-06-15 15:53:11Z UTC

Your question is the famous "separable quotient problem", as Ady mentioned. From here on, "space" means "infinite dimensional Banach space". A space $X$ has a separable quotient provided $X^*$ has a reflexive subspace (obvious), a subspace isomorphic to $c_0$ (Rosenthal and me), or $\ell_1$ (Hagler and me). A result of PANDELIS DODOS, JORDI LOPEZ-ABAD and STEVO TODORCEVIC is that it is consistent with ZFC that if $X$ has density character at least $\aleph_\omega$ then $X$ has a separable quotient; see

http://arxiv.org/pdf/0805.1860.pdf

Every dual space has a separable quotient (Argyros, Dodos, Kanellopoulos):

http://users.uoa.gr/~pdodos/Publications/13-Unconditional.pdf

There are other striking things that I can't locate quickly.

Every non reflexive quotient of a $C(K)$ space contains a subspace isomorphic to $c_0$ (classical result of Pelczynski), so $\ell_1$ is not a quotient of $\ell_\infty$.

Jun Zhang · Answer 2 · 2011-06-16 01:56:23Z UTC

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Here seems to be another reference paper from Jorge MUJICA, who transfer this seperable quotient space problem to some other equivalent problems.

http://www.mat.ucm.es/serv/revmat/vol10-2/vol10-2e.pdf

answered Jun 16 2011 at 1:56

Jun Zhang
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Separable quotients of non-separable Banach spaces?

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