Is c isometrically isomorphic $c \times c$?

Question

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 $c \times c$

Answer 1

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.