Let $l^1 = \left\{ (x_i)_{i\in\mathbb{N}} \mid \sum_{i =1}^{\infty} | x_i |<\infty \right\}$, the space of all sequences whose associated series converge absolutely.

Let $A= \{\sum_{i = 0}^{\infty} | x_i |\leq1\}$

(Note: $A$ is the unit ball with respect to $d_1$ where $d_1((x_i)_{i\in\mathbb{N}},(y_i)_{i\in\mathbb{N}}) = \sum_{n = 0}^{\infty} | x_i - y_i |$.)

Let $d_\infty((x_i)_{i\in\mathbb{N}},(y_i)_{i\in\mathbb{N}}) = sup_{i\in\mathbb{N}}| x_i - y_i |$.

I want to show $A$ is closed with respect to the topology induced by $d_\infty$. (I see how to show it's closed with respect to $d_1$)

Suppose we have a sequence $(x^n)_{n\in\mathbb{N}}$ of members of $A$ converging to $y\in l^1$. (Each $x^n \in A$, so this is a sequence of sequences). I will try to show $lim_{n\rightarrow \infty} \sum_{i=1}^\infty|x_i^n|=\sum_{i=1}^\infty|x_i|$. Here's my attempt.

For each $n$, we have: $$|(\sum_{i=1}^\infty|x_i^n|-\sum_{i=1}^\infty|x_i|)|=\sum_{i=1}^\infty(|x_i^n|-|x_i|)|\leq\sum_{i=1}^\infty||x_i^n|-|x_i||\leq \sum_{i=1}^\infty|x^n_i -x_i|$$ $$=\sum_{i=1}^{k}|x^n_i-x_i| + \sum_{i=k +1}^\infty|x^n_i-x_i|$$ for any $k$. (We may want k to depend on n). We can make $\sum_{i=1}^{k}|x^n_i-x_i|$ small by the convergence of $(x^n)$ to $y$ with respect to $d_\infty$. And given a particular $n$, we can make $\sum_{i=k_n +1}^\infty|x^n_i-x_i|$ small by taking $k_n$ large. However, the latter is not good enough. Given $\epsilon>0$, it seems we need to find $k$ such that For All sufficiently large $n$, $\sum_{i=k_n +1}^\infty|x^n_i-x_i|<\epsilon/2$. I don't see how to do this.

  • $\begingroup$ Note that $$\sup_{i\in\mathbb N}|x_i-y_i|\leqslant\sum_{i\in\mathbb N}|x_i-y_i|, $$ so if a sequence converges in $\ell^1$ it also converges in $\ell^\infty$. $\endgroup$
    – Math1000
    Oct 6, 2017 at 9:27
  • $\begingroup$ @Math1000 I'm not seeing how that helps since the hypothesis is convergence in $l^\infty$ not convergence in $l^1$, and the former does not imply the latter. $\endgroup$
    – Smithey
    Oct 6, 2017 at 18:40

1 Answer 1


The problem with your proof is that it does not use the structure of the set. It's as if you try to prove that every $\ell^1$-closed set is $\ell^\infty$-closed, which is false. (In fact, the converse is true: a smaller norm, such as $\ell^\infty$, induces weaker topology on the same space.)

Let's work with the complement and show it's open. Suppose $\|x\|_1>1$. Then there exists $n$ such that $|x_1|+\dots+|x_n| >1$. Let $\epsilon = (|x_1|+\dots+|x_n| -1)/n$. Then for any $y$ with $\|y-x\|_\infty<\epsilon$ we have $$ \|y\|_1 \ge |y_1|+\dots+|y_n| > |x_1|+\dots+|x_n| - n\epsilon = 1 $$ which proves that the complement of unit ball is open in $\ell^\infty$ norm.


You must log in to answer this question.

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