Let $k$ be a field, and $E$ be an algebraic extension field of $k$. Suppose $E=k(\alpha_i)_{i \in I}$, where $\alpha_i$ is an algebraic element of $E$ over $k$, for each index $i \in I$.
I already know that if $I$ is a finite set, then the general form of an element of $E$ is the linear combination of products of $\alpha_i$'s.
My question: Does this hold even when $I$ is infinite?
More explicitly, is every element of $k(\alpha_i)_{i \in I}$ is contained in a finitely generated subextension $k(\alpha_{i(1)},\ \cdots,\ \alpha_{i(n)})$?
If the $\alpha_i$ are algebraic ( of degree $d_i$) then the field generated by them consists of all elements that are linear combinations of products of the form $$\alpha_{i_1}^{e_{i_1}}\cdots \alpha_{i_k}^{e^{i_k}}$$ with coefficients in $k$, where $i_1$, $\ldots$, $i_k$ are (distinct) from $I$, and $0\le e_{i_s}< d_{i_s}$, for all $1\le s \le k$. This equals the $k$ algebra generated by the $\alpha_i$'s, because the $\alpha_i$ are algebraic over $k$. Note that $I$ does not need to be finite.
The second equality ( about the union of finitely generated extensions) is true in general ( do not need the elements to be algebraic over $k$) and is a consequence of basic field theory. In fact, if you have a family $\mathcal{F}$ of subsets of $I$ such that every element of $I$ is contained in one of the subsets and the union of two elements of $\mathcal{F}$ is contained in another element of $\mathcal{F}$, then $$k(\alpha_i)_{i \in I}= \bigcup_{J \in F} k(\alpha_i)_{i \in J}$$ For instance, $\mathcal{F} = \textrm{ finite subsets of } I$ works OK. Note that this follows from the definitions, and similar results hold for vector spaces, groups, etc.