Countability of a Totally Ordered Set Prove or disprove the following statement:

If $X$ is a totally ordered set with the property that for every two elements $x$ and $y$ in $X$ such that $x<y$, there exists another element $z$ such that $x<z<y$, then $X$ is uncountable.
What about the converse of the statement?

 A: Let's try every variant of your statement.  Define the predicates


*

*U(S) means S is uncountable, 

*TO(S) means S is totally ordered, 

*D(S) means between any x

*for a predicate P, the predicate -P is the negation of that predicate.


We look for examples.  For each example we find, no one or two of the predicates listed can imply the negation of the third (since that would imply the nonexistence of the example).


*

*U(S), TO(S), D(S): You observe that $\Bbb{R}$ is an example.

*U(S), TO(S), -D(S):  Henno Brendsma observes that $[0,1] \cup [2,3]$ is an example with $x = 1, y = 2$.  (In fact, chopping any open interval out of a U+TO set would work.)

*U(S), -TO(S), D(S): $\Bbb{R} \times \{a,b\}$, partially ordered by "$(s,t) < (u,v) \iff s < u$" is not totally ordered.  $(x,a)$ and $(x,b)$ are incomparable for all $x \in \Bbb{R}$.

*U(S), -TO(S), -D(S): $\left( [0,1] \cup [2,3] \right) \times \{a,b\}$ as ordered above works.

*-U(S), TO(S), D(S): Henno Brandsma observed that $\Bbb{Q}$ is an example.

*-U(S), TO(S), -D(S): $\Bbb{Q} \cap \left( [0,1] \cup [2,3] \right)$ is an example.

*-U(S), -TO(S), D(S): $\Bbb{Q} \times \{a,b\}$, ordered as above, is an example.

*-U(S), -TO(S), -D(S): $\left( \Bbb{Q} \cap \left( [0,1] \cup [2,3] \right) \right) \times \{a,b\}$, ordered as above, is an example.


Conclusion:  No variety of converse or contrapositive of the given statement is a valid statement about a universe containing $\Bbb{Q}$ and $\Bbb{R}$.  (... and "0, 1, 2, 3", I suppose.  But it would be a bizarre universe where we did not know that $\Bbb{Q}$ and/or $\Bbb{R}$ contained all four of these elements.)
A: The statement is false: the rationals have this property.
And there are uncountable linear orders without this: just $\omega_1$ as an ordinal. Or simpler: $[0,1] \cup [2,3]$ in the inherited order of the reals. There is then a gap between $1$ and $2$.
