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Let $\Gamma$ be an infinite set. Then every $(x_i)_{i\in\Gamma}\in \ell_1(\Gamma)$ has at most a countable number of components $x_i\neq 0$. As a consequence, every separable subspace $M$ of $\ell_1(\Gamma)$ is contained in an isometric copy of $\ell_1$ which is $1$-complemented in $\ell_1(\Gamma)$. Indeed, there exists a countable subset $I$ of $\Gamma$ such that for each $(x_i)\in M$, $x_i=0$ for $i\in \Gamma\setminus I$, thus $M$ is contained in $\ell_1(I)\subset \ell_1(\Gamma)$.

QUESTION: Suppose that $E$ is an infinite dimensional Banach space such that, for every $\varepsilon>0$, each separable subspace $M$ of $E$ is contained in a $(1+\varepsilon)$-isometric copy of $\ell_1$ in $E$ which is $(1+\varepsilon)$-complemented in $E$.

Is $E$ isomorphic to $\ell_1(\Gamma)$ for some $\Gamma$?

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    $\begingroup$ 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. $\endgroup$
    – Bill Johnson
    Commented May 21, 2018 at 17:12
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    $\begingroup$ And what happens replacing $(1+\varepsilon)$ by a fixed constant $C$ with $1<C<\infty$ in both places? $\endgroup$
    – M.González
    Commented May 21, 2018 at 17:35
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    $\begingroup$ @BillJohnson, actually it will be complemented. See Theorem 1.1 in londmathsoc.onlinelibrary.wiley.com/doi/epdf/10.1112/… $\endgroup$
    – Tomasz Kania
    Commented May 23, 2018 at 8:27
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    $\begingroup$ @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. $\endgroup$
    – Tomasz Kania
    Commented May 23, 2018 at 14:32
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    $\begingroup$ Thanks, Tomek. I am glad that I did not try to check out my wrong guess! $\endgroup$
    – Bill Johnson
    Commented May 23, 2018 at 15:46

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