$C[0,1]$ 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
$C[0,1]$ is Banach-space isomorphic to $c_0(C[0,1])$.
Here,Banach-space isomorphism means a bounded invertible operator from $C[0,1]$ onto $c_0(C[0,1])$.
 A: There is a useful simple
Lemma. If $X\sim X\oplus X$, $Y\sim Y\oplus Y$, and each of $X,Y$ is isomorphic to a complemented subspace of another, then $X\sim Y$.
Proof. We have $X\sim Y\oplus A$, then $X\sim (Y\oplus Y)\oplus A=Y\oplus(Y\oplus A)=Y\oplus X$, analogously $Y\sim X\oplus Y$.
Now let $X=C([0,1])$, $Y=c_0(X)$. Property $Y\sim Y\oplus Y$ is clear, property $X\sim X\oplus X$ follows from $X\oplus X=C([0,1]\times \{0,1\})$ and Milyutin theorem. $X$ is clearly complemented in $Y$. At last, $Y$ is the space of functions on the compact space on the plane $K=\cup \{\frac1n \times [0,\frac1n]\}\cup (0,0)$, which are equal to 0 at $(0,0)$. Again by Milyutin theorem we see that $X\sim C(K)$, hence $Y$ is isomorphic to a hyperplane in the space isomorphic to $X$. Hyperplane is of course complemented.
This is maybe not a very good proof, since your claim may be used in the proof of Milyutin theorem (I do not remember). But if so, study the proof, it should contain this claim.
A: If you mean $c_0(X)=\lbrace (x_n)_{n\in\mathbb N} \in X^{\mathbb N}: \|x_n\|_X\to 0\rbrace$ with norm $\|(x_n)_n\|=\sup\lbrace \|x_n\|_X: n\in\mathbb N\rbrace$, then $c_0(C[0,1])$ and $C[0,1]$ are isomorphic (as explained by Fedor) but not isometrically isomorphic: The unit ball of $C[0,1]$ has extreme points but that of $c_0(C[0,1])$ does not.
