Union of Total Ordering with Lower Sections is Total Ordering

Theorem
Let $(Y, \preceq)$ be a totally ordered set.

Let $X$ be the disjoint union of $Y$ with the set of lower sets of $Y$.

Define a relation $\preceq'$ on $X$ extending $\preceq$ by letting:
 * $y_1 \preceq' y_2 \iff y_1 \preceq y_2$
 * $y \preceq' L \iff y \in L$
 * $L_1 \preceq' L_2 \iff L_1 \subseteq L_2$
 * $L \preceq' y \iff y \in Y \setminus L$

Then $\preceq'$ is a total ordering.

Proof
First note that by Lower Sets in Totally Ordered Set form Nest:
 * $\subseteq$ is a total ordering on the set of lower sets.

Also note that by Complement of Lower Set is Upper Set, the complement of each $\preceq$-lower set is a $\preceq$-upper set.

$\preceq'$ is reflexive
This follows immediately from the fact that $\preceq$ and $\subseteq$ are reflexive.

$\preceq'$ is transitive
There are eight possibilities to consider.

If $y_1 \preceq' y_2$ and $y_2 \preceq' y_3$, then $y_1 \preceq' y_3$ because $\preceq$ is transitive.

If $L_1 \preceq' L_2$ and $L_2 \preceq' L_3$, then $L_1 \preceq' L_3$ because $\subseteq$ is transitive.

If $y_1 \preceq' y_2$ and $y_2 \preceq' L$, then:
 * $y_1 \preceq y_2$ and $y_2 \in L$

Since $L$ is a lower set in $Y$:
 * $y_1 \in L$, so $y_1 \preceq' L$.

If $L \preceq' y_1$ and $y_1 \preceq' y_2$, then:
 * $y_1 \in Y \setminus L$ and $y_1 \preceq y_2$.

Since $Y \setminus L$ is an upper set in $Y$:
 * $y_2 \in Y \setminus L$, so $L \preceq' y_2$.

If $y \preceq' L_1$ and $L_1 \preceq' L_2$, then:
 * $y \in L_1$ and $L_1 \subseteq L_2$.

By the definition of subset:
 * $y \in L_2$, so $y \preceq' L_2$.

If $L_1 \preceq' L_2$ and $L_2 \preceq' y$, then:
 * $y \in Y \setminus L_2$ and $L_2 \supseteq L_1$

so $y \in Y \setminus L_1$, so $L_1 \preceq' y$.

If $y_1 \preceq' L$ and $L \preceq' y_2$ then
 * $y_1 \in L$ and $y_2 \in Y \setminus L$.

Since $L$ is a lower set, $y_2 \not\preceq y_1$.

Since $\preceq$ is a total ordering, $y_1 \preceq y_2$, so:
 * $y_1 \preceq' y_2$

If $L_1 \preceq' y$ and $y \preceq' L_2$, then:
 * $y \in L_2$ but $y \notin L_1$.

Thus $L_2 \not\subseteq L_1$.

Since $\subseteq$ is a total ordering on the lower sets, $L_1 \subseteq L_2$, so:
 * $L_1 \preceq' L_2$.

$\preceq'$ is antisymmetric
There are three cases.

If $y_1 \preceq' y_2$ and $y_2 \preceq' y_1$ then:
 * $y_1 \preceq y_2$ and $y_2 \preceq y_1$

Since $\preceq$ is antisymmetric, $y_1 = y_2$.

If $L_1 \preceq' L_2$ and $L_2 \preceq' L_1$ then:
 * $L_1 \subseteq L_2$ and $L_2 \subseteq L_1$

Since $\subseteq$ is antisymmetric, $L_1 = L_2$.

By the definition of $\preceq'$, it's impossible for $y \preceq' L$ and $L \preceq' y$, so the third case cannot occur.

Since $\preceq'$ is reflexive, transitive, and antisymmetric, it is an ordering.

$\preceq'$ is a total ordering of $X$ because $\preceq$ is a total ordering, the set of lower sets is a nest, and for any $y$ and $L$ either $y \in L$ or $y \in Y\setminus L$.