About the poset of subspaces of the vector space $\mathbb{R}$ over $\mathbb{Q}$.

Question

Let $L$ the set of all subspaces of the vector space $\mathbb{R}$ over $\mathbb{Q}$, ordered by the set strict inclusion: $V_1<V_2$ iff $\{x\in V_1 \Rightarrow x \in V_2$ and there exists $y \in V_2$ s.t. $y \notin V_1\}$.

Writing $V_1\le V_2$ if $V_1<V_2$ or $V_1=V_2$, , $(L,\le)$ is a partially ordered set, and I'm searching if it is a complete lattice, i.e. if

$ \forall S \subset L$ there exists a greatest lower bound $\bigwedge S $ and a least upper bound $\bigvee S $

Since $S$ can be a set finite or not and countable or not of subspaces $V_\alpha$ that can have uncountable basis I don't see how to show such elements ( if exist).

I will appreciate also any reference to sources where I can find a review of known results about the vector space $\mathbb{R}$ over $\mathbb{Q}$.

Answer 1

Any intersection of subspaces of a vector space $V$ is a subspace of $V$, so $\bigwedge S=\bigcap S$ for all $S\subseteq L$. Of course the union of subspaces need not be a subspace, but its linear span is, and is the smallest subspace containing the union: $\bigvee S=\operatorname{span}\bigcup S$ for each $S\subseteq L$. Thus, $L$ is a complete lattice.

Answer 2

If $P$ is a poset for which $\inf X$ exists for every $X\subseteq P$ (including $X=\varnothing)$, then $P$ is a complete lattice.