Set of Division Subrings forms Complete Lattice

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

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

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

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


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

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