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?

  • 3
    $\begingroup$ That looks good to me. Also, you may want to include some absolute values around the $x_n$ in your definition of $\|\cdot\|_\infty$. $\endgroup$
    – froggie
    May 18, 2012 at 15:31
  • $\begingroup$ I added the absolute values! $\endgroup$
    – Julian
    May 18, 2012 at 15:40

1 Answer 1


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 $y=(y_n)_{n\in\Bbb N}\in B(x,\varepsilon)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .