# Subset of sequence space closed

Is the set $E:=\{(x_n)_{n\in \mathbb{N}} \in \ell^{\infty}\ |\; x_i \in \mathbb{C}, \lim_{n \rightarrow \infty } x_n = 0 \}$ closed in $\ell^{\infty}$ equipped with $\lVert (x_n)_{n\in \mathbb{N}} \lVert_{\infty}=\sup_{n\in\mathbb{N}} |x_n|$?

I've come to the following solution: Let $(x_n)_{n\in\mathbb{N}} \in \ell^{\infty}\setminus E$. So $\exists\, \varepsilon >0$ such that $\forall N \in \mathbb{N} \quad \exists m>N$ with $|x_m|>\varepsilon$. So for $(y_n)_{n\in\mathbb{N}} \in B_{\epsilon /2}((x_n)_{n\in \mathbb{N}})\quad \forall N\in\mathbb{N}\quad \exists m>N$ with $|y_m|>\epsilon /2$ and hence $(y_n)_{n\in\mathbb{N}} \in\ell^{\infty}\backslash E$. So $\ell^{\infty}\backslash E$ is open and $E$ is closed.

Is this solution correct?

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That looks good to me. Also, you may want to include some absolute values around the $x_n$ in your definition of $\|\cdot\|_\infty$. – froggie May 18 '12 at 15:31
I added the absolute values! – Julian May 18 '12 at 15:40

If $x=(x_n)_{n\in\Bbb N}\notin E$, it's not convergent to $0$ and we can find $\varepsilon>0$ and $A\subset\Bbb N$ infinite such that $|x_n|\geq 2\varepsilon$ for all $n\in A$ (but it's not necessarily true for $n$ large enough). The ball $B(x,\varepsilon)$ is contained in $E^c$ since $|y_n|\geq \varepsilon$ for all $n\in A$ and $y=(y_n)_{n\in\Bbb N}\in B(x,\varepsilon)$.