Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Let $L$ be a linearly ordered set which is equinumerous to $\mathbf{R}$, and $L$ is order dense,which means that for every $x,y\in L$,if $x<y$ ,then there is a $z$ such that $x< z < y$. And $L$ has no first and no last element. Is it true that $L$ is order isomorphic to $\mathbf{R}$?

share|cite|improve this question
You can find in model theory textbooks that the theory of dense linear orders with no endpoints is $\aleph_0$-categorical but not $\kappa$-categorical for $k>\aleph_0$. – Seirios Jan 25 '13 at 18:22
up vote 2 down vote accepted

No. Let $\Bbb P$ be the irrationals in their usual order. Then $\Bbb R$ and $\Bbb P$ have the same cardinality, and both are dense linear orders without endpoints, but $\Bbb R$ is order-complete, and the irrationals are not, so they are not order-isomorphic.

share|cite|improve this answer

Brian Scott's answer is excellent, but maybe it's worth noticing the following too:

  • Say you have a copy of $\mathbb Q$ followed by a copy of $\mathbb R$. Or you can intersperse lots of copies of $\mathbb Q$ and $\mathbb R$, and get a linearly ordered set satisfying all the stated conditions, but not order-isomorphic to $\mathbb R$.
  • Look at the plane whose points are $(x,y)$, which $x,y$ real, in lexicographic order. It's not order-isomorphic to $\mathbb R$.
  • Look at the set of all countable ordinals. That's an uncountable set. Between each such ordinal and the next, put a copy of the interval $(0,1)$. Then the cardinality of that ordered set is $\aleph_1 2^{\aleph_0}=2^{\aleph_0}$. But it's not order-isomorphic to $\mathbb R$ since the latter has no subset that's order-isomorphic to the set of all countable ordinals. Every subset of $\mathbb R$ that's well-ordered in the usual order on $\mathbb R$ is at most countable.
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.