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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 $\mathbf{R}$?

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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.

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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.
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