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.


2 Answers 2


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.

  • $\begingroup$ how we can show the continuity part? $\endgroup$
    – vqw7Ad
    Sep 24, 2019 at 21:19
  • $\begingroup$ $|f(x)-f(y)|\le \|x-y\|_\infty$. $\endgroup$
    – Jochen
    Sep 25, 2019 at 7:15
  • $\begingroup$ @Jochen , why this inequality holds... $\endgroup$
    – Hamilton
    Oct 3, 2021 at 13:18
  • $\begingroup$ @Beginner Because the l inf norm is defined to be the supremum of the entire series, i.e the greatest difference two points of the sequence can have. $\endgroup$ Feb 20, 2022 at 20:29

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.


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