Let $$F \subseteq K=F(A)$$ be an algebraic field extension, where $A$ is a subset of $K \setminus F$, for any $\beta \in F(A)$, show that there exists finitely many elements $\alpha_1,\alpha_2,...,\alpha_n$ in $A$, s.t. $$\beta \in F(\alpha_1,\alpha_2,...,\alpha_n) \; .$$
I have no idea about this question.
It suffices to show $$ F(A) = \bigcup_{\alpha_1,...,\alpha_n \, \in A, n\in \mathbb{N}} F(\alpha_1,...,\alpha_n) \; . $$ It is easy to check that $$ \bigcup_{\alpha_1,...,\alpha_n \, \in A, n\in \mathbb{N}} F(\alpha_1,...,\alpha_n) $$ is a field containing $F$ and $\alpha_1,...,\alpha_n$, and $$ \bigcup_{\alpha_1,...,\alpha_n \, \in A, n\in \mathbb{N}} F(\alpha_1,...,\alpha_n) \subseteq F(A) \; , $$ since $F(A)$ is the smallest field containing $F$ and $\alpha_1,...,\alpha_n$, then we obtain the equation.