Set of Division Subrings forms Complete Lattice

Theorem
Let $\left({K, +, \circ}\right)$ be a division ring, and let $\mathbb K$ be the set of all division subrings of $K$.

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

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

By Intersection of Division Subrings:
 * $\bigcap \mathbb S$ is the largest division subring of $K$ contained in each of the elements of $\mathbb S$.


 * The intersection of the set of all division subrings of $K$ containing $\bigcup \mathbb S$ is the smallest division subring of $K$ 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 division subrings of $K$.

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