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
 A: 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.))
A: Remark: This still has some problems that need to be fixed
My attempt to answer the question: Let $S\subset\Bbb R.$ By $\bar{\Bbb Q}(S)$ denote the algebraic closure of $\Bbb Q(S)$ in $\Bbb R$, that is, $\bar{\Bbb Q}(S)$  is the set of $x\in\Bbb R$ that
are algebraic over $\Bbb Q(S)$. Also, recall that for $x\in\Bbb R$ there exist a finite set $E_x\subset T$ such that $x\in\bar{\Bbb Q}(E_x).$
Now, pick $a_1,a_2,\cdots, a_n\in\Bbb R,$ it is enought to show that there exist $r\in\Bbb R$ such that $r\not\in a_1 V+a_2 V+\cdots+a_n V.$
Let $C=\{a_1,a_2,\cdots, a_n\},$ then there exists a finite set  $E\subset T$ such that $C\subset\bar{\Bbb Q}(E).$ Notice that $V\cap T_0=\emptyset$ and since $E$ is a finite set. So, take $r\in T_0\setminus(V\cup E).$ Moreover, $r\cup V\cup E$ is algebraically independent. For contradition,  let $r\in a_1 V+a_2 V+\cdots+a_n V,$
but we have  $r\in a_1 V+a_2 V+\cdots+a_n V\subset\bar{\Bbb Q}(V\cup E)$ which is impossible since $r$ is algebraically independent of $V\cup E.$
