Is $C_0$ dense in $l^{\infty}$

Question

Is $C_0$ dense in $l^{p}$ with $1\leq p\leq \infty$ where $C_0=\{ (x_n): x_n\rightarrow 0, x_n\in R\}$. Well I think that if $p<\infty$ is true because by definition if i take $y=(y_n)\in l^p$ then $\sum (y_n)^p <\infty$ so $(y_n)^p \rightarrow 0$ imply $y_n \rightarrow 0$ then I can choose the same $y_n \in C_0\cap l^\infty$ such that $y_n \rightarrow y_n$ and this is the definition of density, for all $z$ in the big set exist one succession $z_n$ in the small set such that $z_n\rightarrow z$. But I don't know how to do with $p=\infty$. Please somebody can you help me? Thank you

Answer 1

$C_0$ is not dense in $l^{\infty}$. Actually it is a closed subset in $l^{\infty}$ and it does not have the constant sequence $1$ which is in $l^{\infty}$.

To show that it is closed, suppose that $((a_n)^{(k)})$ is a sequence of $C_0$ points congerges to $(x_n)$ in $l^{\infty}$. Then for every positive $\epsilon$, we can choose $k$ large enough so that $|(a_n)^{(k)}-x_n|<\epsilon$ for all $n$. Since $(a_n)^{(k)} \in C_0$, we can choose $N$ large enough so that $|(a_n)^{(k)}|<\epsilon$ whenever $n>N$. But then for each $n >N$ we have $|x_n|<2\epsilon$. Since $\epsilon$ were arbitrary, we have $x_n \in C_0$ which shows the desired closedness.

Thanks for a good question!