I cannot figure out how I would show $R\setminus Q$ is not order isomorphic to $R$.
Both sets have the same cardinality, so I've tried to show the order preserving property doesn't work, but I'm having a lot of difficulty doing so.
Thank you for any help
On
Suppose there is an order-isomorphism $f : \mathbb{R} \setminus \mathbb{Q} \to \mathbb{R}$.
Consider the sequence $(x_n)_{n \ge 1}$ in $\mathbb{R} \setminus \mathbb{Q}$ defined by $x_n=1-\frac{\sqrt{2}}{n}$ for all $n \in \mathbb{N}$. This is a bounded increasing sequence, and hence $(f(x_n))_{n \ge 1}$ is also a bounded increasing sequence in $\mathbb{R}$. Let $y$ be the least upper bound for $(f(x_n))_{n \ge 1}$, which exists by completeness of $\mathbb{R}$, and let $x = f^{-1}(y)$.
It is easy to verify that $x$ is a least upper bound for $(x_n)_{n \ge 1}$, contradicting that $(x_n)_{n \ge 1}$ has no least upper bound in $\mathbb{R} \setminus \mathbb{Q}$.
For $i\in \{1,2\}$ let $<_i$ be a linear order on $A(i)$ and let $T(i)$ be the topology on $A(i)$ induced by $<_i.$ Suppose $f:A(1)\to A(2)$ is a bijection such that $\forall x,y\in A(1)\;(x<_1y\implies f(x)<_2f(y).$ Then $f$ is a homeomorphism with respect to the topologies $T(1), T(2).$
Therefore, with $A(1)=\Bbb R$ and $A(2)=\Bbb R$ \ $\Bbb Q,$ and with $<_1$ and $<_2$ being the usual (arithmetic ) order, the existence of an order-isomorphism $f:\Bbb R\to \Bbb R$ \ $\Bbb Q$ would imply that the ordered spaces $A(1)=\Bbb R,$ $A(2)=\Bbb R$ \ $\Bbb Q$ are homeomorphic. But $\Bbb R$ is a connected space and $\Bbb R$ \ $\Bbb Q$ is $not$ a connected space, so no such $f$ exists.