Is $c$ isomorphic to $c_0$?

Question

I want to prove that $c(k)$ and $c_0(k)$ are isomorphic.
It is easy to find a linear map $u$ given by $u(\lbrace x_k\rbrace)=\lbrace x_k-\lim x_k\rbrace \in c_0(k)$, and show that this is a surjection. But I couldn't find the constants $\alpha ,\beta$ such that $\alpha \|x_n\|_{\sup}\leq\|u(x_n)\|_{\sup}\leq\beta\|x_n\|_{\sup}$, here $k$ is in $\mathbb{R}$ or $\mathbb{C}$.
Any help is welcome.

Answer 1

No, there is no $\alpha > 0$ satisfying the inequality for all sequences $\{x_n\}$. To see, this consider the case when the ${x_n}$ are positive real numbers converging to 0 and $k$ is a positive real constant, and for convenience, let $x = \|x_n\|_{\sup}$, then $\| x_n + k \|_{\sup} = x + k$. So your inequality becomes for $\{x_n + k\}, \alpha(x + k) \le x$. Therefore $(\alpha - 1)x +\alpha k \le 0$. By letting $k \gg x$, we see that this would require $\alpha \le 0$.