# I have to decide if $\ell^1\subset c_0$ is closed or not.

I was asked to decide if $\ell^1\subset c_0$ is closed or not, where $$\ell^1=\{(x_n)_{n\in\mathbb N}\subset\mathbb R:\sum_{n=0}^{\infty}|x_n|<\infty\}$$ $$c_0=\{ (x_n)_{n\in\mathbb N}\subset\mathbb R:\lim_{n\rightarrow\infty}x_n=0\}$$

In my opinion it is not closed so I want to prove my claim.

Here is what I did,

In order to show my claim I will try to show that the complement of $\ell^1$ can not be open.

Consider the sequence $(\frac{1}{n})_{n\in\mathbb N}\in c_0$. Clearly $(\frac{1}{n})_{n\in\mathbb N}\notin\ell^1$ since $\sum_{n=0}^{\infty}\frac{1}{n}=\infty$.

and define $B_{\epsilon}(\frac{1}{n}):=\{(x_n)_{n\in\mathbb N}\in c_0:\sup_n|\frac{1}{n}-x_n|<\epsilon\}$ for an arbitrary $\epsilon>0$.

Now consider the sequence $(\frac{\delta}{n^2})_{n\in\mathbb N}\in \ell^1$ for some $\delta$ strictly smaller $\epsilon$ i.e $\delta<\epsilon$.

Then we have,

$$\sup_n|\frac{1}{n}-\frac{\delta}{n^2}|=\sup_n|\frac{n^2-\delta n}{n^3}|=\sup_n|\frac{n-\delta}{n^2}|=\delta<\epsilon$$

This implies that $(\frac{\delta}{n^2})_{n\in\mathbb N}\in B_{\epsilon}(\frac{1}{n})\bigcap\ell^1\neq\emptyset\quad\forall \epsilon>0$

What finally yields that the complement of $\ell^1$ is not open and so $\ell^1$ is not closed.

Could someone look over it and tell if it is correct or not? Thank you!

• It seems off. For one, your $\sup$ in your last display is at least $|1-\delta|$ (take $n=1$). Instead, you can show that the sequence $(x_n)$ where $x_n=(1,1/2,1/3,\ldots,1/n,0,\ldots)$ converges to $(1/n)$ in $c_0$. This will show $\ell_1$ is not closed in $c_0$. – David Mitra Apr 4 '14 at 17:31
• @DavidMitra Thank you, of course the last sup is wrong and thank you for the very helpful hint! – Thorben Apr 4 '14 at 17:35

$\ell^1$ contains the sequences with finite support (i.e. which vanish eventually), and the set of such sequences is dense in $c_0$. Thus the closure of $\ell^1$ in $c_0$ is $c_0$. As noted in the OP, $\ell^1\neq c_0$ hence $\ell^1$ cannot be closed in $c_0$.