# Prove the set of sequences $c_0$ which converge to zero in $l_{\infty}$ is closed.

Prove the set of sequences which converge to zero in $$l_{\infty}$$ is closed.

Let $$x_n(k)\rightarrow x(k)$$ as $$n\rightarrow\infty$$. With $$x_n(k)\in c_0$$ and $$x(k)\in l_{\infty}$$.

Let $$\varepsilon>0$$. Then there exists an $$N>0$$ such that $$\parallel x_N-x\parallel_{\infty}:=\sup_{k\in\mathbb{N}}|x_N(k)-x(k)|\leq\varepsilon.$$

Then we have, \begin{align} |x(k)| &= |x(k) - x_N(k) + x_N(k)| \\\\ &\leq |x_N(k) - x(k)| + |x_N(k)| \\\\ &\leq \varepsilon + |x_N(k)|\rightarrow \varepsilon\;\; \text{as}\;\; k\rightarrow\infty. \end{align}

Therefore since $$\varepsilon$$ was chosen arbitrarily we can conclude that $$x(k)\rightarrow0$$ and thus that $$x(k)\in c_0$$

Can someone check my work on this? It seems too slick and painless to be correct.

One more proof: $f:\ell_\infty \to \mathbb R$, $x\mapsto \lim\sup |x_n|$ is continuous so that $c_0= f^{-1}(\lbrace 0\rbrace)$ is closed.
• $|f(x)-f(y)|\le \|x-y\|_\infty$. Sep 25, 2019 at 7:15
The solution is correct. Just to beef up this post, I'll sketch a slightly different proof: the complement of $l_0$ is open.
If $x\notin l_0$, let $r=\frac12\limsup_{k\to\infty} |x(k)|$. If $\|x-y\|\le r$, then $$\limsup|y(k)| \ge \limsup_{k\to\infty} |x(k)|-r =r$$ hence $y\notin l_0$.
By the way, this is the first time I see notation $l_0$ used for this subspace; all sources I know use $c_0$. I think $l_0$ is prone to confusion.