Let $E/F$ be a field extension.
Then $E/F$ is called normal iff there exists $\mathscr{A}\subset F[X]\setminus\{0\}$ such that $\forall f\in\mathscr{A}$, $f$ splits over $E$ and $E=F(\{\alpha\in E: \exists f\in \mathscr{A} f(\alpha)=0\}$
Usually normal extension is defined for algebraic extensions, so that the standard definition of normal extensions is "the above definition + algebraic extension".
However, doesn't the above definition (for arbitrary field extension) imply that the extension is algebraic?
That is, isn't "algebriac" condition superfluous in the standard definition of normal extension?
Yes. The elements of $E$ algebraic over $F$ form a subfield of $E$, call it $L$, that contains $F$. Since $E=F(\{\alpha\})$ where $\{\alpha\}$ is a set of algebraic elements, $E$ is contained in $L$. Thus $E=L$ is algebraic over $F$.