Quotient Riesz space given by a filter $\mathscr{F}$ is Archimedean iff $\mathscr{F}$ is closed under countable intersections.

Question

The following is given as Example 1XE in Fremlin's Topological Riesz Spaces and Measure Theory.

Let $X$ be a nonempty set, $E = \mathbb{R}^X$, and $\mathscr{F}$ be a filter on $X$. Let $$F = \{x \in E: \{t: x(t)=0 \} \in \mathscr{F} \}.$$

The quotient Riesz space $E/F$ is Archimedean iff $\mathscr{F}$ is closed under countable intersections.

I am having trouble proving this. In fact, I haven't been able to make any progress at all, which means I must be failing to grasp something very basic. I am not looking for a complete answer, just a hint to get me started.

I suspect my gap in understanding has to do with Riesz quotient spaces. All I really know about them is from Fremlin's 14G, where he shows that $E/F$ can be given a Riesz space structure. In particular, he shows that the positive cone $P$ of $E/F$ is given by the canonical map $\phi: E \to E/F$ by $P = \{\phi(x): x \in E^+ \}$, and that $\phi(x) \vee 0 = \phi(x^+)$.

For completeness, Fremlin's definition is that $E$ is Archimedean if $x,y \in E$ and $ny \leq x$ for all $n \in \mathbb{N}$ imply $y \leq 0$.

Answer 1

First you need to check that $F$ is an ideal. This is the easy part and it follows from the filter properties. I will write one as an example: Let $x, y\in E$ such that $|x|\leq |y|$ and $y\in F$. Then $\{t: y(t)=0\}\subseteq \{t: x(t)=0\}$ and the first belongs in $\mathscr{F}$ by assumption. So the second one also belongs in $\mathscr{F}$, because $\mathscr{F}$ is closed under supersets. So $x\in F$. Similarly you can show that $F$ is a subspace.

Then we have the Archimedean property. There is a very useful theorem which characterizes the Archimedean property on quotient spaces:

Definition: A sequence $(x_n)_n $ in a Riesz space is called relatively uniformly convergent to $x$, if there exist a $u>0$ and a strictly decreasing sequence of positive scalars $(\varepsilon_n)_n$ with $\varepsilon_n\rightarrow 0$, such that $|x_n-x|\leq \varepsilon_n u$, for every $n$.

A subset $A$ is uniformly closed if for every relatively uniformly convergent sequence of $A$, the limit is in $A$.

Theorem (Luxemburg-Moore): Let $A$ be an ideal of a (not necessarily Archimedean) Riesz space $E$. Then $E/A$ is Archimedean if and only if $A$ is a uniformly closed ideal of $E$.

Its proof is an easy exercise and you can find it, for example, in Aliprantis - Positive Operators, pg. 100.

Using this theorem, your question becomes:

Show that $F$ is uniformly closed if and only if $\mathscr{F}$ is closed under countable intersections.

• Suppose that $\mathscr{F}$ is closed under countable intersections and let $(x_n)_n$ be a sequence in $F$ that converges uniformly to $x\in E$. Then for some $u>0$, $(\varepsilon_n)_n\downarrow 0$,$$|x_n-x|\leq \varepsilon_n u, \ \ \ \forall n\in \mathbb{N}.$$ Since each $x_n$ is in $\mathscr{F}$, each of the sets $$A_n=\{t: x_n(t)=0\}$$ belongs in $\mathscr{F}$ and so does $A=\bigcap_{n=1}^\infty A_n$. If $t\in A$, then $$|x(t)|\leq \varepsilon_n u(t)+ |x_n(t)|=\varepsilon_n u(t) \rightarrow 0,$$ so $A\subseteq \{t:x(t)=0\}$, $A\in \mathscr{F}$ and therefore $x\in F$. So $F$ is uniformly closed.

• Suppose that $F$ is uniformly closed and let $(A_n)_{n\in\mathbb{N}}$ be a sequence with $A_n\in \mathscr{F}$, for every $n\in\mathbb{N}$. We will show that $A=\bigcap_{n=1}^\infty A_n \in \mathscr{F}$. Without loss of generality we can assume that $(A_n)_{n\in\mathbb{N}}$ is additionally decreasing. If not, just set $B_n=\bigcap_{i=1}^nA_n$. Now we need to define a sequence $(x_n)_{n}$ and a $x$ such that $x_n\rightarrow x$ uniformly and $A=\{t: x(t)=0\}$. I am sure you can fill in the details for the rest of the proof. I tried to define these functions as: $$ x_n(t)= \begin{cases} 1, \text{ if } t\notin A_n,\\ 0, \text{ if } t\in A_n, \end{cases} $$ $$ x(t)= \begin{cases} 1, \text{ if } t\notin A,\\ 0, \text{ if } t\in A, \end{cases} $$ but I didn't prove it in detail. If they don't work, you need to modify them accordingly.