Is $c$ isomorphic to $c_0$?

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.

• Your map is not an injection; it sends any constant sequence to $0$. Also, your terminology could use some clarification: what is $k$? What does "isomorphic" mean, isomorphic as topological vector spaces? – Eric Wofsey Sep 22 '15 at 1:29
• Thank! but is possible find some $u$ such that $\alpha ||x_n||_{sup}\leq ||u(x_n)||_{sup}\leq\beta ||x_n||_{sup}$ ? – sti9111 Sep 22 '15 at 1:31
• @sti9111 - when someone asks you a question about what you mean, it is not helpful to simply restate what you've already asked without bothering to answer what was requested of you. We are not sure what you mean, so we can't properly answer you until you clarify it. – Paul Sinclair Sep 22 '15 at 1:44
• @PaulSinclair: In fairness to sti9111, I edited my comment to add the second half of it a minute after posting it and they may not have noticed. – Eric Wofsey Sep 22 '15 at 1:53
• My apologies then. Actually, I think I can address his question about the norms as is, and am currently writing it up. – Paul Sinclair Sep 22 '15 at 1:54

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 $$\alpha \le 0$$.
• More simply, instead of taking a sequence of positive numbers converging to $0$, just let $x_n=0$ for all $n$ (i.e., you are just looking at any nonzero constant sequence). – Eric Wofsey Sep 22 '15 at 2:21
• True - I was avoiding the $0$ sequence because I was thinking it would zero out the inequality, but I failed to notice that after adding $k$, that was no longer true. – Paul Sinclair Sep 22 '15 at 2:40
• Thanks! this say that for $u$ like before the numbers $\alpha \, \beta$ doesn´t exist, but is possible find some $u$ which is a surjection and this number exist? – sti9111 Sep 22 '15 at 19:28
• Well, since I still am not absolutely sure what $c$ and $c_0$ are, that is hard to say. I am guessing that $c$ is the set of all convergents sequences in $\Bbb K$, and $c_0$ are sequences that convege to 0, both supplied with the sup norm. Is this correct? – Paul Sinclair Sep 22 '15 at 22:22