Let a strict linear order $C = (V, <)$, be an irreflexive and transitive relation < defined on $V$, and call a section of $C$ a partition of $V$ into two sets $A, B$, such that $x < y$, whenever $x \in A$ and $y \in B$. The strict linear order $C$ is called continuous if all sections of the $C$ are such that either 1. or 2. hold:
- $A$ has a maximal element (i.e, $\exists z \in A (\forall x \in A (x < z))$) and $B$ doesn't have a minimum element (i.e, $\neg \exists z \in B (\forall x \in B (z < x))$).
- $A$ doesn't have a maximal element and $B$ has a minimum element.
I have read that a fundamental example of a continuous strict linear order is the real numbers under the relation of inequality. I can't see how any partition of the reals into two sets either satisfies condition 1. or condition 2. above. Can anyone explain this?
Under the required conditions, you have $A\cup B = \mathbb R$ with $A\cap B=\varnothing$, and $a<b$, whenever $a\in A$ and $b \in B$.
It follows that $\bigvee A = \bigwedge B$ (the supremum of $A$ is the infimum of $B$). Call this element $c$.
If $c \in A$, then the partition satisfies the first condition.
Otherwise, $c \in B$ and the partition satisfies the second condition.