Let $A$ be a directed set, and $\ell^\infty_A$ the (complex vector) space of all eventually bounded nets $A\to \mathbb{C}$. We can define the limit superior seminorm on $\ell^\infty_A$: $$ \vert\vert{(u_\alpha)}\vert\vert:= \inf_{\alpha\in A} \sup_{\beta\geq \alpha} |u_\beta|\ . $$ This is indeed finite, and we can check (unless I am mistaken) the inequality $\vert\vert{u+v}\vert\vert\leq \vert\vert{u}\vert\vert+\vert\vert v\vert\vert$ by an $\epsilon/2$ argument.

Now my questions are :

1) Is $\ell^\infty_A$ a complete seminormed space ?

2) Is the subspace $c_A$ of convergent nets $A\to \mathbb{C}$ a closed subset of $\ell^\infty_A ?$.

3) Is there a reference for these type of questions ? Is the space $\ell^\infty_A$ (for general directed set $A$) used or studied somewhere in the litterature?


1 Answer 1


Usually what we do with a seminormed space is factor out the null space, so that we get a norm on the quotient. If you do that with your space then you find that "eventually bounded" can be replaced by "bounded" because every eventually bounded net $u$ has null difference with a bounded net $u'$. (When $u$ is positive, let $u' = {\rm min}(u, \|u\|\cdot 1_A)$; for general $u$ take linear combinations.)

So what you get for the quotient space is just $l^\infty(A)$ modulo the nets which converge to 0. This is a version of a corona algebra --- that might be the keyword you want.

Edit: yes, of course I mean $l^\infty(A)$ modulo the nets in $l^\infty(A)$ which converge to 0.

  • $\begingroup$ Thank you for your answer. I suppose you denoted $\ell^\infty(A)$ the Banach space of bounded nets. So if I understand correctly, $\ell^\infty_A / c_0 = \ell^\infty(A) / c_0$ where $c_0$ is the subspace of nets that converges to 0 (the null space for the limit superior seminorm). And this equality is at the level of Banach spaces (since the right hand side is Banach, right ?). What are the consequences of this concerning questions 1 and 2 ? $\endgroup$
    – Jeremy
    Oct 25, 2012 at 23:43
  • $\begingroup$ Also, I am not sure why $c_0$ would be included in $\ell^\infty(A)$. Maybe you meant : $\ell^\infty_A/c_0 = \ell^\infty(A) /c_0'$ where $c'_0:=c_0\cap \ell^\infty(A)$ ? $\endgroup$
    – Jeremy
    Oct 25, 2012 at 23:47
  • $\begingroup$ @Jeremy: I hope the edit answers your last comment. The answers to questions 1 and 2 are both yes, this is a good exercise. $\endgroup$
    – Nik Weaver
    Oct 26, 2012 at 4:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.