Union of Total Ordering with Lower Sections is Total Ordering
Theorem
Let $\struct {Y, \preceq}$ be a totally ordered set.
Let $X$ be the disjoint union of $Y$ with the set of lower sections 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 Sections in Totally Ordered Set form Chain:
- $\subseteq$ is a total ordering on the set of lower sections.
Also note that by Complement of Lower Section is Upper Section, the complement of each $\preceq$-lower section is a $\preceq$-upper section.
Reflexivity
This follows immediately from the fact that $\preceq$ and $\subseteq$ are reflexive.
Thus $\preceq'$ is reflexive.
$\Box$
Transitivity
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 section 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 section 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 section:
- $y_2 \npreceq y_1$
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:
- $L_2 \nsubseteq L_1$
Since $\subseteq$ is a total ordering on the lower sections:
- $L_1 \subseteq L_2$
so:
- $L_1 \preceq' L_2$
Thus $\preceq'$ is transitive.
$\Box$
Antisymmetry
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 is impossible for $y \preceq' L$ and $L \preceq' y$, so the third case cannot occur.
Thus$\preceq'$ is antisymmetric.
$\Box$
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 sections is a chain
and:
- for any $y$ and $L$ either $y \in L$ or $y \in Y \setminus L$.
This article, or a section of it, needs explaining. In particular: Specify exactly why it follows from the above that $\preceq'$ is a total ordering You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
$\blacksquare$