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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 :

  1. 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 $l^{1}=l^{1}(\mathbb{N})$?)

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    $\begingroup$ The answer is affirmative. Namely, take M as the kernel of any [non-trivial] continuous linear functional on X. $\endgroup$
    – Ady
    Commented Jun 14, 2011 at 22:51
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    $\begingroup$ 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. $\endgroup$
    – Andreas Blass
    Commented Jun 14, 2011 at 22:52
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    $\begingroup$ That's the 'Separable Quotient Problem'... $\endgroup$
    – Ady
    Commented Jun 14, 2011 at 22:58
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    $\begingroup$ 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. $\endgroup$
    – Philip Brooker
    Commented Jun 15, 2011 at 5:30
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    $\begingroup$ 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 ) $\endgroup$
    – Philip Brooker
    Commented Jun 15, 2011 at 5:40

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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 $\ell_\infty$.

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  • $\begingroup$ Small comment regarding the last part: would it be very slightly quicker to note that if a Banach space quotients onto $\ell_1$ then it has a complemented subspace isomorphic to $\ell_1$ (and then use Pelczynski's result about complemented subspaces of $\ell_\infty$)? $\endgroup$
    – Yemon Choi
    Commented Jun 15, 2011 at 5:46
  • $\begingroup$ It's the same proof, Yemon. Pelczynski proved that every non weakly compact operator from a $C(K)$ space preserves a copy of $c_0$. I don't think his proof simplifies if the operator is a projection (though I could be wrong about that). Quite a bit later Lindenstrauss proved (using Pelczynski's theorem, BTW) that every complemented subspace of $\ell_\infty$ is isomorphic to $\ell_\infty$. $\endgroup$
    – Bill Johnson
    Commented Jun 15, 2011 at 6:00
  • $\begingroup$ Thanks for the details Bill - as you have probably guessed, in my comment I meant Lindenstrauss when I said Pelczynski, and I was unaware that his proof used the result of Pelczynski which you mention. $\endgroup$
    – Yemon Choi
    Commented Jun 15, 2011 at 6:10
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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

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