Definition 1.1. Let $F$ be an extension field of $K$, and $S$ a subset of $F$. Then $S$ is said to be
- algebraically dependent over $K$ if for some positive integer $n$ there exists a nonzero polynomial $f\in K[x_1,\dots,x_n]$ such that $f(s_1,\dots, s_n) = 0$ for some distinct $s_1,\dots, s_n \in S$;
- algebraically independent over $K$ if $S$ is not algebraically dependent over $K$.
Also, we need to recall the following lemma.
Lemma 1. Let $E$ be an extension field of the field $F$ and suppose that $a, b\in E$ are algebraically independent over $F$. Then $F(a)\cap F(b) = F$.
Now, Let $B$ be a non-empty non-singleton algebraically independent over $\Bbb Q.$ I have the following claim:
Claim For any non-empty non-singleton $C \subset \Bbb Q$ and $r\in\Bbb R$, we have $rC \subsetneq B.$
Proof of the claim: Consider two cases for $r$. If $r\in \Bbb Q,$ then it is clear $rC\subsetneq B$ since $\Bbb Q\cap B=\emptyset.$ Otherwise, $r\subset\Bbb R\setminus\Bbb Q.$ For $q,q'\in C,q\neq q'$, then there exist $x_q,x_{q'}\in B$ such that $rq=x_q$ and $rq'=x_{q'}$. But this means that $$r\in\Bbb Q(x_q)\cap\Bbb Q(x_{q'})=\Bbb Q.\tag{1}$$ Note that equation $(1)$ follows by Lemma $1$ .This is a contradiction since $r\not\in\Bbb Q.$ Lastly, if $x_q=x_{q'}$, then we have $r(q-q')=0$ which is again a contradiction since $q\neq q'$ and $r\in\Bbb R\setminus\Bbb Q.$ This finishes the proof of the claim.
Question: Is that right?
Any help or comments will be appreciated greatly.