7
$\begingroup$

Let $K$ be a compact Hausdorff space. I wonder whether there are characterizations of $K$ such that $C(K)$ contains no copy of $l_{1}$. There are some compact Hausdorff spaces $K$ such that $C(K)$ contains no copy of $l_{1}$, for example, if $K$ is a countable compact metric space and has finite Cantor-Bendixson index. Are there more compact Hausdorff spaces $K$ such that $C(K)$ contains no copy of $l_{1}$?

Thank you!

$\endgroup$
2
  • $\begingroup$ no subspace isomorphic/isometric to $\ell^1$? $\endgroup$
    – YCor
    Commented Jun 10, 2016 at 23:06
  • $\begingroup$ no subspace isomorphic to $l_{1}$ $\endgroup$ Commented Jun 10, 2016 at 23:42

1 Answer 1

7
$\begingroup$

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 $C(K)$.

$\endgroup$
3
  • $\begingroup$ Could you tell me the reference about the above result you mentioned. $\endgroup$ Commented Jun 10, 2016 at 23:44
  • $\begingroup$ @DongyangChen, I think you will find this in H. Lotz, N. T. Peck and H. Porta, Semi-embeddings of Banach spaces, Proc. Edinburgh Math. Soc. (2) 22 (1979), no. 3, 233-240. $\endgroup$ Commented Jun 11, 2016 at 7:48
  • 2
    $\begingroup$ The result is due to Pelczynski and Semadeni: $$ $$ Pełczyński, A.; Semadeni, Z. Spaces of continuous functions. III. Spaces C(Ω) for Ω without perfect subsets. Studia Math. 18 1959 211–222. $\endgroup$ Commented Jun 11, 2016 at 8:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .