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Let $K$ be a compact Hausdorff infinite topological space and $C(K)$ the Banach space of continuous functions from $K$ in $\mathbb{R}$ with sup norm. It is known that $c_0$ is complemented in $C(K)$ for $K$ that contains infinite convergent sequence. I would appreciate if somebody let me know an example of $K$ without infinite convergent sequences such that $C(K)$ contains a complemented subspace isomorphic to $c_0$.

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  • $\begingroup$ "It is known that $c_0$ is complemented in $C(K)$ for $K$ that contains infinite convergent sequence" sounds confusing: what do you mean by $c_0$ in this context? Please explain, and give a reference to the "known" result. $\endgroup$ – YCor Jul 23 '19 at 20:58
  • $\begingroup$ @YCor probably it means "contains a complemented copy of $c_0$". $\endgroup$ – Fedor Petrov Jul 23 '19 at 21:43
  • $\begingroup$ @FedorPetrov I made this guess, but don't see how to find a copy of $c_0$ in $C(K)$, when $K$ has a closed subset consisting of an infinite converging sequence. Indeed those functions vanishing on it, if complemented, would yield a complemented copy of $c_0$. I don't see right away why it's complemented. $\endgroup$ – YCor Jul 23 '19 at 22:04
  • $\begingroup$ $c_0$ is the Banach space of sequences of reals that converges to 0, endowed with supremum norm. $\endgroup$ – Aligomez Jul 23 '19 at 22:07
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Take $K = \beta N \times \beta N$, the square of the Čech--Stone compactification of the integers. The space of continuous functions on that space is not a Grothendieck spaces. A space of continuous functions on a compact space is Grothendieck if and only if it doesn't contain complemented copies of $c_0$.

More generally you may take any infinite $K = L \times L$, such that $C(L)$ is a Grothendieck space.

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  • $\begingroup$ Thank you! Please, could you suggest me some bibliography for the proof of the last statement? $\endgroup$ – Aligomez Jul 24 '19 at 13:38
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    $\begingroup$ @Aligomez please check references here ams.org/journals/proc/1984-091-04/S0002-9939-1984-0746089-2/… $\endgroup$ – Tomasz Kania Jul 26 '19 at 20:46
  • $\begingroup$ @Aligomez, is my answer satisfactory? If not, please let me know what is missing $\endgroup$ – Tomasz Kania Aug 11 '19 at 8:36
  • $\begingroup$ Thank you, Tomek Kania for your answer and bibliography. $\endgroup$ – Aligomez Oct 11 '19 at 20:24

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