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edited Feb 25, 2016 at 23:58

$c_0(C[0,1])$ is the $c_0$-direct sum of countably many $C[0,1]$.How to prove

$C[0,1]$ is Banach-space isomorphic to $c_0(C[0,1])$.

Here,Banach-space isomorphism means a bounded invertible operator from $C[0,1]$ onto $c_0(C[0,1])$.

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edited Feb 25, 2016 at 13:05

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asked Feb 25, 2016 at 12:41

$C[0,1]$ is Banach-space isomorphic to $c_0(C[0,1])$

$c_0(C[0,1])$ is the $c_0$-direct sum of countably many $C[0,1]$.How to prove

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