Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

Intuitively, it seems like the statement is true, but I have no idea how to prove it. This is just something I thought up of when thinking about how the Real numbers were uncountable. Thank you.

share|cite|improve this question
What about the dyadic rationnals? Are they uncountable? – nombre Jan 24 at 12:10
Heck, what about the rationals? – BrianO Jan 24 at 13:42
Superheck, what about a singleton set $X$? – Hagen von Eitzen Jan 24 at 21:14

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 $2$.

share|cite|improve this answer

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.