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Show that a subspace of $c_{0}$ with the norm induced by $c_{0}$, can’t be isomorphic to $l_1$.

Any idea or hint? I think have to show that the given subspace has a certain property that $l_1$ no has (or the reverse) but I can’t understand which property... thank you!

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closed as off-topic by Namaste, Saad, Holo, José Carlos Santos, TheSimpliFire Jan 3 at 8:46

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  • $\begingroup$ @mechanodroid had a great answer. I just feel like adding that since $c_0$ is subprojective, if $c_0$ contained a copy of $\ell_1$ then $\ell_1$ would contain a copy of $c_0$. However $\ell_1$ has the Schur property whereas $c_0$ does not. $\endgroup$ – Ben W Jan 2 at 19:22
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The dual space of $\ell^1$ is isometrically isomorphic to $\ell^\infty,$ which is nonseparable.

However, the dual of $c_0$ is isometrically isomorphic to $\ell^1,$ which is separable. Hence the dual of any subspace of $c_0$ is separable as well.

This argument actually shows that there is not a linear homeomorphism between $\ell^1$ and a subspace of $c_0$, let alone an isometric isomorphism.

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