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Let $c_0$ be the space of all sequences of scalars converging to $0$ with the supremum norm and $c$ be the space of all convergent sequences of scalars with the supremum norm. Is $c$ isometrically isomorphic $c \times c$? Here $c \times c$ is given the max norm: $||(\{x_n\},\{y_n\})||=\max (||(\{x_n\},||(\{y_n\}||)$. This must have appeared on MSE but I couldn't locate it. $c_0 \times c_0$ is obviously isometrically isomorphic to $c_0$ and there cannot be an isometric isomorphism of $c$ onto $c \times c$ that maps $c_0$ onto $c_0 \times c_0$ as seen by a consideration of codimensions. But I am unable to see if c isometrically isomorphic to $c \times c$

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    $\begingroup$ Is the isomorphism $(a_n) \mapsto \langle(a_{2n}) , (a_{2n-1})\rangle $? Why is it isometry? Why can't the same work for $c$? $\endgroup$ – Berci Mar 13 '18 at 8:54
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    $\begingroup$ If you want to ask about isometry you have to specify what norm on $c\times c$ you're considering. There are various equivalent norms - which one you take wouldn't matter for isomorphism but it matters a great deal for isometry. $\endgroup$ – David C. Ullrich Mar 13 '18 at 14:25
  • $\begingroup$ Just forgot to mention the norm. I have edited the question now. $\endgroup$ – Kavi Rama Murthy Mar 14 '18 at 4:51
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    $\begingroup$ @Berci the map you mentioned is an isometric isomorphism in the case of $c_0$ but it is not surjective in the case of $c$. The range consists of pairs of sequences with the same limit. $\endgroup$ – Kavi Rama Murthy Mar 14 '18 at 4:52
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Let $\alpha X$ denote the Alexandroff's compactification of a locally compact space $X$. Let $C(K)$ denote the space of continuous functions on a compact space $K$. By symbol $\cong_1$ we denote an isometric isomorphism of Banach spaces. Then, $$ c\cong_1 C(\alpha \mathbb{N})\\ c\times c\cong_1 C(\alpha \mathbb{N})\times C(\alpha \mathbb{N})\cong_1 C(\alpha \mathbb{N}\sqcup \alpha \mathbb{N}) $$ Suppose $c$ and $c\times c$ are isometrically isomorphic, then by Banach-Stone theorem two compact spaces $\alpha \mathbb{N}\sqcup \alpha \mathbb{N}$ and $\alpha \mathbb{N}$ are homeomorphic. Clearly, this is not the case since these spaces have two and one limit point respectively.

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