In many texts is noted the analogy between the transcendence degree of a field extension and the dimension of a vector space, so I'm tempting to use such analogy to better understand the structure of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, using the fact that $\mathbb{R}$ is also a field extension of $\mathbb{Q}$.
Given a subset $S \subset \mathbb{R}$, the $\Bbb Q$-vector space $V_S= \operatorname{span}_{\Bbb Q} (S)$, as a subspace of $\mathbb{R}$, has dimension $dim (V_S)$ that can be finite or infinite (denumerable or not).
We have also the extension field $\mathbb{Q}(S)$ that has a transcendence degree $Trd (\mathbb{Q}(S))$.
My question is if there exists some relation between these.
It seems to me that we can not have, in general: $dim (V_S)=Trd (\mathbb{Q}(S))$. A simple counter example being $S=\{\sqrt{2},\pi\}$. But there is some way to characterize the cases in which the identity is valid?
We can prove that $dim (V_S) \ge Trd (\mathbb{Q}(S))$ ? And there is some way to compare this two values when they are infinite?
I will appreciate also any reference to sources where I can find a review of known results about the vector space $\mathbb{R}$ over $\mathbb{Q}$.
One can prove that $$ Trd (\mathbb{Q}(S)) \aleph_0 + 1 \le \dim (\mathbb{Q}(S)) \le (Trd (\mathbb{Q}(S)) +1) \aleph_0 .$$ In particular $\dim (\mathbb{Q}(S)) = Trd (\mathbb{Q}(S)) \aleph_0$ if $Trd (\mathbb{Q}(S))$ is non-zero.
Let $S' \subset S$ be a transcendence basis of $\mathbb{Q}(S)$ then $S'$ is algebraically indepenent over the rationals and thus in particular $\{s^n \colon n>0, s \in S'\} \cup \{1\}$ is linearly independent over the rationals. Showing the one inequality.
In the other direction on gets $| \mathbb{Q}(S') | = \max\{\aleph_0, |S'|\}$ and $|\mathbb{Q}(S)| \le | \mathbb{Q}(S') | \ \aleph_0 $ as the extension $\mathbb{Q}(S)$ over $\mathbb{Q}(S')$ is algebraic.
In the same way, one can show that $$\dim V_S \ge Trd (\mathbb{Q}(S)).$$ A transcendence basis (over the rationals) is algebraically independent and thus in particular linearly independent (over the rationals).
However, in general this inequality will be strict. When $s$ is transcendental then setting $S_i = \{s^i \colon 1 \le i < n+1\}$, we get that $\dim V_{S_i} = i$ while $Trd (\mathbb{Q}(S_i)) = 1$.
So we get any finite dimension as well as a countable dimension with transcendence degree $1$ (we could even do it with transcendence degree $0$).
Yet our upper bound above shows that the discrepancy is at most countable since $V_S \subset \mathbb{Q}(S)$.
So, as soon as $ Trd (\mathbb{Q}(S)) \ge \aleph_0$ we have $ Trd (\mathbb{Q}(S)) = \dim V_S$.
(This uses AC, I guess)