Set of Subfields forms Complete Lattice

Theorem
Let $\left({F, +, \circ}\right)$ be a field, and let $\mathbb F$ be the set of all subfields of $F$.

Then $\left({\mathbb F, \subseteq}\right)$ is a complete lattice.

Proof
Let $\varnothing \subset \mathbb S \subseteq \mathbb F$.

By Intersection of Subfields:
 * $\bigcap \mathbb S$ is the largest subfield of $F$ contained in each of the elements of $\mathbb S$.


 * The intersection of the set of all subfields of $F$ containing $\bigcup \mathbb S$ is the smallest subfield of $F$ containing $\bigcup \mathbb S$.

Thus:
 * Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.


 * The supremum of $\mathbb S$ is the intersection of the set of all subfields of $R$.

Therefore $\left({\mathbb F, \subseteq}\right)$ is a complete lattice.