Lower Sections in Totally Ordered Set form Chain
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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.
$\blacksquare$