# Establish Archimedean property of a vector-lattice

I am trying to find ways to prove the Archimedian property of a certain vector lattice and got stuck on the following type of problem.

I feel the statement below (or in fact weaker versions) should be provable, but I am not being very successful so far. I am posting this question here to get suggestions and/or counterexamples, thanks!

Statement: Let $$A$$ be a vector lattice (wikipedia entry) and let $$B\subseteq A$$ a subset of $$A$$ which:

1) B is closed under vector spaces operations (of $$A$$), i.e.: (i) if $$b_1,b_2\in B$$ then $$b_1 + b_2 \in B$$, and (ii) if $$b\in B$$ then $$r b\in B$$ for all $$r\in\mathbb{R}$$. In other words, $$B$$ is a partially orderd vector subspace of $$A$$, but $$B$$ is not necessarily a lattice.

2) $$B$$ generates $$A$$, i.e., every element $$a\in A$$ is expressible as a finite combination of meet and joins from $$B$$: $$a = \bigvee_i \bigwedge_j b_{i,j}$$.

3) $$B$$ is Archimedean, in the sense that it does not have infinitesimals (except $$0$$): for all $$b,b^\prime\geq0$$ in $$B$$, if for all $$n\in\mathbb{N}$$ it holds that $$nb \leq b^\prime$$, then $$b=0$$.

Under these assumptions it follows that $$A$$ is Archimedean as a vector lattice. I.e., for all $$a,a^\prime\geq0$$ in $$A$$, if for all $$n$$, $$n a \leq a^\prime$$ then $$a=0$$.

end of Statement

As I said, I have not been able to prove this so far. Perhaps there is some counterexample?

• Do you have an example motivating this question? – user458276 Mar 16 at 22:41
• Any examples would need to be infinite dimensional (see math.leidenuniv.nl/scripties/MasterJongen.pdf). Also, since you’d have to a finite combination of meets and joins of elements of $B$ to produce any elements of $A$, it seems unlikely that there are any finite proper subsets $B$ that satisfy (2). – user458276 Mar 17 at 0:41