Corollary 1

— Let E be an extension of K and A a subset of E consisting of elements that are algebraic over K. Then $K(A)$ is algebraic over K and we have $K[A] = K(A)$.

For each $x \in K(A)$ belongs to a field $K(F)$, where F is a finite subset of $A$ ($V$, p. 11, Cor.); now $K(F)$ is algebraic over $K$ and equal to $K[F]$ by Th. 2, hence $x$ is algebraic over $K$ and $K(A) = K[A]$.