0
$\begingroup$

I was browsing throughout the literature and I found the following fact:

  • Every $\sigma$-Dedekind complete Riesz space $E$ which admits a strictly positive functional $f$ is Dedekind complete.

I believe I found this result a couple of years ago in some paper or a book but I can't remember where or when. I would really appreciate it if someone can give me a reference or provide a proof.

I can prove this result in the case when the functional is also $\sigma$-order continuous.

Proof in this case: Let $\{x_\lambda\}_{\lambda}$ be an increasing net that is bounded by $y$ in $E$. We may also assume that the net is closed under taking finite suprema. Then there exists $$S=\sup_{\lambda} f(x_\lambda).$$ Let us choose an increasing subsequence $\{f_n\}_{n\in\mathbb N}$ with $$S=\lim_{n\to\infty} f(x_n).$$ Due to the fact that $E$ is $\sigma$-Dedekind complete, there exists $x=\sup_{n\in\mathbb N}x_n.$ Since $f$ is $\sigma$-order continuous, we have $f(x)=S.$ The following $$x_n\leq x_n\vee x_\lambda \nearrow x\vee x_\lambda$$ together with $\sigma$-order continuity of $f$ implies $f(x_\lambda \vee x)=S=f(x).$ Here I also applied that $f(x_n\vee x_\lambda)\leq S.$ Since $f$ is strictly positive, we have $x=x_\lambda\vee x$ so that $x_\lambda\leq x.$ This implies that $x$ is an upper bound for our net. If $x_\lambda \leq z\leq x$ for all $\lambda$, then positivity of $f$ implies $f(z)=S$, so that, again by its strict positivity we have $x=z.$ This finishes the proof.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I think you can deduce what you want from Section 12 in "Positive Operators" by Aliprantis and Burkinshaw. $\endgroup$
    – Bill Johnson
    Feb 3, 2016 at 16:41
  • $\begingroup$ Thank you. I checked Section 12 and the most promising result in this direction is Theorem 12.9. (the original version of the book) and Theorem 4.9 (in the reprinted version). However, the whole section is devoted to Banach lattices. I am not sure if I am still missing something $\endgroup$
    – Marko
    Feb 3, 2016 at 17:06

1 Answer 1

Reset to default
1
$\begingroup$

Proof. By theorem [1] $E$ has the countable sup property, now theorem [2] completes the proof.

[1] If a Riesz space $E$ admits a strictly positive linear functional, then $E$ is Archimedean and has the countable sup property. read

[2] Every $\sigma$-Dedekind complete vector lattice having the countable sup property is super Dedekind complete. book

Improve this answer
$\endgroup$

Your Answer

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

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