# Compact space $K$ without convergent sequence while $c_0$ is complemented in $C(K)$

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 $$c_0$$.

• "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. – YCor Jul 23 '19 at 20:58
• @YCor probably it means "contains a complemented copy of $c_0$". – Fedor Petrov Jul 23 '19 at 21:43
• @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. – YCor Jul 23 '19 at 22:04
• $c_0$ is the Banach space of sequences of reals that converges to 0, endowed with supremum norm. – Aligomez Jul 23 '19 at 22:07

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.