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

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!

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

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• @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. – Ben W Jan 2 at 19:22

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.