Why $F(S)=\cup_{r=1}^{\infty} F(\alpha_1, \dots ,\alpha_r)$ instead of $F(S)=F(\alpha_1, \dots ,\alpha_r)$? I'm reading Lang's Undergraduate Algebra, here:



I don't understand: Why $F(S)=\cup_{r=1}^{\infty} F(\alpha_1, \dots ,\alpha_r)$ instead of $F(S)=F(\alpha_1, \dots ,\alpha_r)$? In a previous section of the book, there was:



This made me think (mistakenly, I guess) that all fields $F(\alpha_1),F(\alpha_1 , \alpha_2)\dots $ were already contained in $F(\alpha_1, \dots ,\alpha_r)$. I can't see why that doesn't hold.
 A: (2) We definitely have
\begin{equation} \label{inc}
F(\alpha_1) \subseteq F(\alpha_1, \alpha_2) \subseteq F(\alpha_1, \alpha_2, \alpha_3) \subseteq \cdots \subseteq F(\alpha_1, \alpha_2, \ldots \alpha_r) \subseteq \cdots. \tag{$\ast$}
\end{equation}
For instance, $F(\alpha_1)$ is the smallest field containing $F$ and $\alpha_1$. Since $F(\alpha_1, \alpha_2)$ contains both $F$ and $\alpha_1$, then $F(\alpha_1) \subseteq F(\alpha_1, \alpha_2)$.
(1) The problem is that $S$ can be infinite. Lang has presumably defined $F(\alpha)$. Since we can iterate this, we can define $F(\alpha, \beta) := F(\alpha)(\beta)$ and more generally
$$
F(\alpha_1, \alpha_2, \ldots, \alpha_r) := F(\alpha_1, \alpha_2, \ldots, \alpha_{r-1})(\alpha_r)
$$
for any finite collection of elements $\alpha_1, \ldots, \alpha_r$. However, this doesn't work as is to define $F(S)$ for an infinite set $S$. If $S = \{\alpha_1, \alpha_2, \ldots\}$ is a countably infinite set, Lang points out that we can still define $F(S)$ as a sort of "limit" of the previous idea. Since we have the inclusions (\ref{inc}) that you mentioned, we can define $F(S)$ to be the union of all these fields obtained by adjoining finitely many elements:
$$
F(\alpha_1, \alpha_2, \ldots, ) := \bigcup_{r=1}^\infty F(\alpha_1, \ldots, \alpha_r) \, .
$$
In general we can define $F(S)$ as a direct limit of the fields $F(A)$ where $A$ ranges over all finite subsets of $S$. Meaning that, given finite subsets $A$ and $B$ of $S$, if $A \subseteq B$, we have a natural inclusion map $F(A) \hookrightarrow F(B)$. Since $A \cup B$ is still a finite set and $F(A) \hookrightarrow F(A \cup B)$ and $F(B) \hookrightarrow F(A \cup B)$, then this system is directed, so we can form the direct limit and set
$$
F(S) := \varinjlim_{\substack{A \subseteq S\\ A \, \text{ finite}}} F(A) \, .
$$
