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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,

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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!

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$\ell^1$ 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 $c_0$.

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