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If $c(\mathbb{N})$ is the space of all real-valued convergent sequences, $c_0(\mathbb{N})$ the subspace of sequences converging to $0$, I want to show that there is no norm so that $c_0$ is dense in $c$. I can show that $c_0$ is closed in the infinity norm, but not all norms are equivalent on $c_0$.

My idea is to show that $c_{00}$ is dense in $c_0$ for any norm, so if $c_0$ is dense in $c$ for some norm, so is $c_{00}$. I'm stuck trying to derive a contradiction from this, and I would appreciate any ideas!

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I do not think that you can prove this proposition for all norms.

For a $x=(x_0,x_1,...)\in c(\mathbb{N})$ define the following norm $$||x||:=\sum_{k=0}^\infty \frac{|x_k|}{2^k}$$

Then we can approximate $x\in c(\mathbb{N})$ by a sequence $y_j \in c_{00}(\mathbb{N})$ via setting $y_j :=(x_0,...,x_j,0,0,0....)$.

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    $\begingroup$ @MaoWao, convergent $\implies$ bounded. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 9 '19 at 20:27
  • $\begingroup$ Sorry, I misread it as all sequences. You are of course right. I will delete the misleading comment. $\endgroup$ – MaoWao Mar 10 '19 at 9:33

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