Lower Sections in Totally Ordered Set form Chain

Theorem
Let $\struct {S, \preceq}$ be a totally ordered set.

Let $\LL$ be a set of lower sections in $S$.

Then $\LL$ is a chain.

That is, $\LL$ is totally ordered by $\subseteq$.

Proof
Let $L, M \in \LL$.

Suppose that $M \nsubseteq L$.

Then:


 * $\exists x \in M: x \notin L$

Let $y \in L$.

Then since $\preceq$ is a total ordering, $x \preceq y$ or $y \preceq x$.

If $x \preceq y$, then since $L$ is a lower section: $x \in L$, a contradiction.

Thus $y \preceq x$.

Since $M$ is a lower section, $y \in M$.

Since this holds for all $y \in L$, $L \subseteq M$.

Hence, for all $L, M \in \LL$:


 * $M \subseteq L$ or $L \subseteq M$

That is, $\LL$ is a chain.