Definition. $\bar{\Bbb Q}(S)$ denotes an 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 $\Bbb Q(S).$
It seems that I have trouble with the notation of algebraic closure since I could not figure this question out. Here is my question
Q. Let $T$ be a transcendental basis and let $x\in\Bbb R$ such that $x\in\bar{\Bbb Q}(S\cup\{t\})$, $S\subset T.$ Now, let $y\in\Bbb R$ such that $y\not\in\bar{\Bbb Q}(E)$ and $|E|\leq |S|+1$, where $E\subset T.$ Then $y\neq x.$
I was thinking to do it by contradiction. Assume $y=x$. This means, $y\in \bar{\Bbb Q}(S\cup\{t\})$. By definition, we have $y$ is algebraic over $\Bbb Q(S\cup\{t\}).$ But I got stuck here what should I do next? Thank you in advance.