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 $C(K)$?
$c_0[0,1]$ in $C(K)$
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general-topology
functional-analysis
banach-spaces
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0Thanks. 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? – 2011-11-28
1 Answers
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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.