Positive Basis of a Riesz Space

Question

Does a Riesz space in general always has a positive basis? ae. if $E$ is a Riesz space, can we assume that there exists a set $B\subset E$ such that it is a basis of the vector space $E$ and for every $v\in B$, $v>0$?

Proving for the special case $E=\mathbb{R}^n$ is quite easy to prove. Is that true in general?

Edit

Digging in a little bit shows that if $E$ has a finite dimension, it has a positive basis. This is followed from the following theorem:

Suppose that $X$ is a vector lattice of finite dimension $n$. If $X$ is Archimedean ordered then it is (vector lattice) isomorphic with $\mathbb {R} ^{n}$ under its canonical order. Otherwise, there exists an integer $k$ satisfying $2 \le k \le n$ such that $X$ is isomorphic to $\mathbb {R} _{L}^{k}\times \mathbb {R} ^{n-k}$ where $\mathbb {R} ^{n-k}$ has its canonical order, $\mathbb {R} _{L}^{k}$ is $\mathbb {R} ^{k}$ with the lexicographical order, and the product of these two spaces has the canonical product order.

Since $\mathbb{R}^n$ always has a positive basis ($\left\{e_i\right\}$ for example), we can deduce that $E$ has a positive basis (by the inverse isomorphism).

Now, the question is: is it also true for infinite dimension Riesz space?

Answer 1

So, it seems like the answer is YES.

Proof:

Let $S=\left\{B\subseteq E\mid B\text{ is independent and all its items are positive}\right\}$. We know that a chain of independent sets has a maximum value (basic proof in linear algebra). Therefore, using Zorn's lemma, we know that $S$ has a maximum value via the $\subseteq$ relation. We'll mark it as $B'$

We want to prove that this is actually spanning $E$. We'll assume by contradiction that it is not, therefore there is $v\in E$ such that $v\notin \operatorname{span}(B')$, which means $v$ is independent of the elements of $B'$. It is known that: $$v=v^+-v^-$$ such that $v^+,v^-\ge 0$ (This is true for all $v\in E$ when $E$ is a Riesz space). If both $v^+,v^-\in \operatorname{span}(B')$, than $v\in \operatorname{span}(B')$, which is a contradiction. Therefore, $v^+\notin \operatorname{span}(B')$ or $v^-\notin \operatorname{span}(B')$, but if that's the case we know that $B'\cup\left\{v^+\right\}\in S$ or $B'\cup\left\{v^-\right\}\in S$, which is in contrast to the maximality of $B'$. Therefore, $\boxed{\operatorname{span}(B')=E}$