Lower Sections in Totally Ordered Set form Chain

Theorem
Let $\left({S, \preceq}\right)$ be a totally ordered set.

Let $\mathcal L$ be a set of lower sets in $S$.

Then $\mathcal L$ is a nest.

That is, $\mathcal L$ is totally ordered by $\subseteq$.

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

Suppose that $M \not\subseteq L$.

Then for some $x \in M$: $x \notin L$.

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 set: $x \in L$, a contradiction.

Thus $y \preceq x$.

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

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

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


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

That is, $\mathcal L$ is a nest.