Let $\mathbb{F}$ be an arbitrary ordered field. We define:
- $\textbf{Topological Connectedness:}$ If $\mathbb{F}=A \cup B$ where A,B are nonempty and open, then $A \cap B \neq \phi$.
- $\textbf{Order Completeness :}$ If $S \subset \mathbb{F}$ is bounded above, then $\exists \ c \in \mathbb{F}$ that is an upperbound of S and for every upperbound $b$ of S, we have $c \leq b.$
- Cut Property: If $A,B \subset \mathbb{F} \ (A \cap B = \phi \ , \ A \cup B=\mathbb{F} \ , \ A<B)$, then
$\exists \ c \in \mathbb{F} \ (x \in A \implies x \leq c,
\ x \in B \implies c \leq x)$
By $A<B$, I mean every element of $A$ is less than every element of $B$.
It can be shown that 2. $\iff$ 3. So we proceed as follows,
"$\implies$"
We prove the contrapositive by the way of Cut Property.
Let $A$ and $B$ be sets satisfying the hypothesis of the Cut Property but violating its conclusion.
There exists no $c$ such that everything less than $c$ is in $A$ and everything greater than $c$ is in $B$.
Then for every $a \in A$ there exists $a' \in A$ with $a < a'$. And for every $b \in B$ there exists
$b' \in B$ with $b' < b$.
(How?, by choosing $c=max(A)$ or $max(B)$).
Using this it can be shown that both $A$ and $B$ are open(How?, by choosing $\epsilon=a'-a$), that contradicts the topological connectivity of $\mathbb{F}$.
My question: Is the converse true?