Is $L$ order isomorphic to $\mathbf{R}$? 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 $\mathbf{R}$?
 A: 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.
A: 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.

