TITLE: Is it possible to construct an infinite subset of $\Bbb R$ that is not order isomorphic to any proper subset of itself?
QUESTION [23 upvotes]: Is it possible to construct an infinite subset of $\Bbb R$ that is not order isomorphic to any proper subset of itself?

REPLY [18 votes]: The answer is yes in ZFC. We can construct a dense infinite set
$A\subset\mathbb{R}$ such that the only order-preserving map
$f:A\to A$ is the identity. In particular, $A$ is not
order-isomorphic with any proper subset of itself.
To see this, note first that any order-preserving map
$f:B\to\mathbb{R}$ defined on a dense set $B\subset\mathbb{R}$ can
be extended to a total order-preserving map $\bar
f:\mathbb{R}\to\mathbb{R}$ defined on the closure of $B$, by
defining $\bar f(x)=\sup_{y\leq x, y\in B}f(y)$. Further, note
that any such monotone map will have at most countably many points
of discontinuity, since every discontinuity will be a jump
discontinuity. Thus, there are precisely continuum many such
order-preserving functions $\mathbb{R}\to\mathbb{R}$, since any
one of them is determined by countably much information about
their values on a countable dense set and the information about
what their values are on the countably many points of
discontinuity.
We may therefore enumerate all order-preserving functions
$f_\alpha:\mathbb{R}\to\mathbb{R}$ in a sequence of length
continuum, $\alpha\lt\mathfrak{c}$.
Let's now build the set $A$ by a transfinite process, making
promises at each stage about some reals being definitely in $A$
and other promises about keeping some reals out of $A$, in such a
way that we kill off $f_\alpha$ at stage $\alpha$ as a possible
order-preserving map from $A$ to $A$. We may begin at stage $0$ by
placing all the rational numbers into $A$, so that it will
definitely be dense. Suppose we have carried out our process up to
stage $\alpha$, and $f_\alpha$ is the next non-identity
order-preserving map $\mathbb{R}\to\mathbb{R}$ presented for our
consideration. Since $f_\alpha$ is order-preserving and not the
identity, it must be that there is an interval $(a,b)$ with
$(f(a),f(b))$ disjoint from $(a,b)$. Since we've made fewer than
continuum many promises so far, there must be an $x\in (a,b)$ such
that we've made no promises about $x$ or $f_\alpha(x)$. In this
case, we place $x$ into $A$ and promise to keep $f_\alpha(x)$ out
of $A$. This will prevent $f_\alpha$ from being an
order-isomorphism of $A$ to a proper subset of $A$.
The end result is that $A$ is dense, but is strongly rigid in the
sense that there is no non-identity order-preserving map from $A$
to $A$. In particular, $A$ is not order-isomorphic with any proper
subset of itself.