# $\ell_1$ dense in $c_0$?

This may be a silly question, but here goes. To ensure clarity, $\ell_1$ is the space of absolutely summable sequences, and $c_0$ the space of bounded sequences with limit 0. So we know that $\ell_1\subset c_0$ by basic principles. My question is: is $\ell_1$ when equipped with the sup-norm dense in $c_0$?

Here is my thought, and I would appreciate a comment on correctness or if something went wrong:

Let $\xi\in c_0$ and write $\xi=\{\xi_1,\xi_2,\xi_3,\dots\}$. Now define $P_n:\ell_1\to c_0$ by $$P_n(\eta)=\{\eta_1,\eta_2,\dots,\eta_n\}$$ So if $\xi\in c_0$, we can say $$\xi=\underset{n\to\infty}{\lim}P_n\xi$$

So does this get us all of $c_0$?

A typical example would be the harmonic sequence $\{1, 1/2, \dots, 1/n,\dots\}$. This is in $c_0$ but not $\ell_1$, but taking finite pieces of this sequence at a time guarantees us to remain in $\ell_1$, and we can approximate the sequence in $c_0$ as the limit of elements of $\ell_1$.

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Your proof for the harmonic sequence generalizes. –  dls Mar 28 '12 at 23:41
$c_{00}$ is dense in $c_0$, so any set containing $c_{00}$ is dense in $c_0$. –  user16299 Mar 29 '12 at 2:50
@Yemon Will the same reasoning always hold for sums of Banach spaces? It seems like the same arguments will work to show that the $\ell_1$ sum of Banach spaces is dense in the $c_0$ sum of Banach space, and so on. Or is there an example out there where this would break down? –  Keaton Mar 29 '12 at 16:04
@keaton I think it should work, just as you say –  user16299 Apr 1 '12 at 0:18

## 1 Answer

Yes. Let $(x_n)\in c_0$ and take $\epsilon>0$. Since $x_n\to 0$, we have some $N$ such that $n\geq N\implies |x_n|<\epsilon$. Define the sequence $(y_n)$ by $y_n=x_n$ for $n<N$ and $y_n=0$ for $n\geq N$. Clearly $(y_n)\in \ell^1$, and for any $n$, $|x_n-y_n|\leq \epsilon$, hence $\sup\limits_{n}|x_n-y_n|\leq \epsilon$.

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