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$c_0(C[0,1])$ 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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  • $\begingroup$ Is Banach-space isomorphic the term you want? Ordinarily I think of "Banach-space isomorphic" as synonymous with isometrically isomorphic. $\endgroup$
    – Todd Trimble
    Commented Feb 25, 2016 at 15:37
  • $\begingroup$ @ToddTrimble I think (but I don't claim to speak for everyone) that "Banach-space isomorphic" is in contrast to "Banach-algebraically isomorphic" (both objects are Cstar algebras, so if one just said "isomorphic", one might think this is isomorphism of Cstar algebras) $\endgroup$
    – Yemon Choi
    Commented Feb 25, 2016 at 17:02
  • $\begingroup$ @ToddTrimble As I think was discussed on the nLab a while ago, "isometrically isomorphic" is not usually used as the default by people who work with Banach spaces - if we want isometric we usually feel obliged to explicitly mention it $\endgroup$
    – Yemon Choi
    Commented Feb 25, 2016 at 17:04
  • $\begingroup$ @YemonChoi (Referring to your second comment) I defer to your judgment here, although I've seen "linearly homeomorphic" also used to indicate the weaker notion of isomorphism (and FWIW, I personally believe that would be preferable language). $\endgroup$
    – Todd Trimble
    Commented Feb 25, 2016 at 17:22
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    $\begingroup$ In Banach space theory, isomorphism means linear homeomorphism. This is completely standard. $\endgroup$ Commented Feb 26, 2016 at 4:29

2 Answers 2

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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.

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  • $\begingroup$ @FedorPetrov.What is "A" in the proof of the lemma above? $\endgroup$ Commented Feb 25, 2016 at 13:40
  • $\begingroup$ Complement subspace. $\endgroup$ Commented Feb 25, 2016 at 13:50
  • $\begingroup$ @FedorPetrov.Thank you for help. I am confused at the statement of your lemma. Do you mean that if $X\sim X\oplus X, Y\sim \oplus Y$, "Then" each of $X,Y$ is isomorphic to a complemented subspace of another? $\endgroup$ Commented Feb 25, 2016 at 14:04
  • $\begingroup$ No, it is another assumption. Sorry, I forgot to formulate the conclusion, fixed now. $\endgroup$ Commented Feb 25, 2016 at 14:10
  • $\begingroup$ oh, I see. thank you. $Y=\{f\in C(K)| f(0,0)=0\}$. I can chek this statemet by building an isometric isomorphism. But is there any easier way to see this? $\endgroup$ Commented Feb 25, 2016 at 14:14
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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.

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