Let $T$ be a transcendental basis of $\Bbb R$ and $A\subset T.$ For $B\subset\Bbb R$, by $\overline{\Bbb Q}(B)$ the sets of all $x\in\Bbb R$ that are algebraic over $\Bbb Q(B).$
Now, let $S\subset\Bbb R$, $S$ is not algebraically independent, and $S$ is a countable set such that $$\overline{\Bbb Q}(A\cup S)=\Bbb R$$
Notice that $A\cup S$ is not algebraically independent. So, I can not say $A\cup S$ is a transcendental basis. But we also know each algebraically independent set can be extended to a transcendental basis, using Zorn's lemma. Then, there exists an algebraically independent set, say $G$, such that $A\cup G$ is a transcendental basis of $\Bbb R$ over $\Bbb Q.$ Here is my question
$G$ will be a countable set. Is that right?
I think it is. Any idea.