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Tomasz Kania
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Yes, there is such characterisation. $C(K)$ contains no isomorphic copy of $\ell_1$ if and only if $K$ is scattered. Indeed, if $K$ is scattered then $C(K)^*$ is isometric to $\ell_1(K)$, so $C(K)$ cannot contain $\ell_1$, as then $C(K)^*$ would have contained a copy of $L_1$. Conversely, if $K$ is not scattered, then you may find a copy of $C[0,1]$ in $C(K)$.

Yes, there is such characterisation. $C(K)$ contains no isomorphic copy of $\ell_1$ if and only if $K$ is scattered. Indeed, if $K$ is scattered then $C(K)^*$ is isometric to $\ell_1(K)$, so $C(K)$ cannot contain $\ell_1$, as then $C(K)^*$ would have contained a copy $L_1$. Conversely, if $K$ is not scattered, then you may find a copy of $C[0,1]$ in $C(K)$.

Yes, there is such characterisation. $C(K)$ contains no isomorphic copy of $\ell_1$ if and only if $K$ is scattered. Indeed, if $K$ is scattered then $C(K)^*$ is isometric to $\ell_1(K)$, so $C(K)$ cannot contain $\ell_1$, as then $C(K)^*$ would have contained a copy of $L_1$. Conversely, if $K$ is not scattered, then you may find a copy of $C[0,1]$ in $C(K)$.

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Tomasz Kania
  • 11.3k
  • 2
  • 39
  • 75

Yes, there is such characterisation. $C(K)$ contains no isomorphic copy of $\ell_1$ if and only if $K$ is scattered. Indeed, if $K$ is scattered then $C(K)^*$ is isometric to $\ell_1(K)$, so $C(K)$ cannot contain $\ell_1$, as then $C(K)^*$ would have contained a copy $L_1$. Conversely, if $K$ is not scattered, then you may find a copy of $C[0,1]$ in $C(K)$.