# $c_0[0,1]$ in $C(K)$

Let $K=[0,1]\times \{0,1\}$ be endowed with the topology arising from the lexicographic order on it. It is known that $K$ is compact, Hausdorff, first-countable and perfectly normal. Furthermore, the space $c_0[0,1]$ is a quotient of $C(K)$. Is $c_0[0,1]$ a (complemented) subspace of $C(K)$?

-
In Fabian-Habala et all the space $K$ is called "two arrow space", see p.634. You can find there also the result about the quotient, which is mentioned in the question. –  Martin Sleziak Nov 28 '11 at 18:03
Thanks. They prove that the dual $C(K)^*$ is weak*-separable, but the dual of $c_0[0,1]^*$ is not, hence $c_0[0,1]$ does not embed into $C(K)$. Is that right? –  Bakcyl Nov 28 '11 at 18:22
Double arrow space $K$ is hereditarily Lindelof, so it satisfies countable chain condition. By theorem 14.26 (ii) the space $C(K)$ can not contain $c_0(\Gamma)$, for uncountable $\Gamma$ and in particular $[0,1]$ since it is also uncountable.