About linear space generated by algebraically independent set

Let $T$ be a transcendental basis of $\Bbb R$ over $\Bbb Q$ with cardinlity continuum , a maximal algebraical independent set, and $ V$ be a linear space over $\Bbb Q$ generated by $T\setminus T_0$, where $T_0\subset T$ is a countable infinite. I am trying to show the following (if it is possible)

$a_1 V+a_2 V+\cdots+a_n V\neq \Bbb R$ for every $a_1, a_2,\cdots,a_n\in\Bbb R.$

I think it is true. Notice that $V\neq \Bbb R$ since $\text{co-dim}(V)>0,$ where $ \text{co-dim}$ denots to co-dimension, see codimension. My guess is that co-dimension might help. Another way to think about by contradition. To do this, pick $r\in\Bbb R$ then there exists $a_1, a_2,\cdots,a_n\in\Bbb R$ and $v_1, v_2,\cdots,v_n\in\Bbb R$ such that $r=a_1v_1+a_2v_2+\cdots+a_nv_n,$ but I did not see the contradition. However, I couldn't move forward in either appraoch.

Any help will be appreciated greatly

Solution 1:

Let $k=\mathbb{Q}(T\setminus T_0)$ be the subfield of $\mathbb{R}$ generated by $T\setminus T_0$. Then $T_0$ is algebraically independent over $k$, and in particular linearly independent over $k$. So, the dimension of $\mathbb{R}$ as a vector space over $k$ is infinite. Now if you had $a_1 V+a_2 V+\cdots+a_n V= \Bbb R$ then in particular $a_1,\dots,a_n$ would span $\mathbb{R}$ as a vector space over $k$, since $V\subset k$. This is impossible since $\mathbb{R}$ is infinite dimensional over $k$.

(In fact, the result would hold even if $T_0$ were empty, i.e. if $V$ were the entire linear span of $T$ over $\mathbb{Q}$. This is because $\mathbb{R}$ is infinite-dimensional as a vector space over $\mathbb{Q}(T)$ as well. For instance, this follows from the fact that for any $t\in T$ and any odd $n\in\mathbb{N}$, there is an $n$th root of $t$ in $\mathbb{R}$, which is the root of a degree $n$ irreducible polynomial $x^n-t$ over $\mathbb{Q}(T)$. (This irreducibility can be proven using Eisenstein's criterion, for instance.))