Are $\Bbb{R}\times\Bbb{Q}$ and $\Bbb{Q}\times\Bbb{R}$ order-isomorphic?

Question

Consider the two following linear orders, $X=\langle \Bbb{R}\times\Bbb{Q}, <_\text{lex} \rangle$ and $Y=\langle \Bbb{Q}\times\Bbb{R}, <_\text{lex} \rangle$, where $<_\text{lex}$ is the lexicographic order on the cartesian product. Is it true that $X\cong Y$, that is, $X$ and $Y$ are order-isomorphic?

Intuition says that the assertion is false, as the similar orders $\Bbb{R}\times \Bbb{N}$ and $\Bbb{N}\times\Bbb{R}$ are non-isomorphic. Assume to the negation that $f:X\to Y$ is an order-isomorphism. I tried to consider the set $\{x\in\Bbb{Q}\mid x^2 \le 2\}\times\Bbb{R}\subseteq \Bbb{Q}\times\Bbb{R}$ and its upper closure, but couldn't really see if it leads anywhere. Any clues?

Answer 1

In $X$, $(0,0) < (0,1)$, and the interval between these pairs is countable infinite. $Y$ has no countably infinite intervals.

Answer 2

Hint

Indeed those two linearly ordered set are not isomorphic. The basis for the proof is that $\mathbb Q$ is countable while $\mathbb R$ is not. Suppose for the sake of contradiction that $\varphi$ is a linear isomorphism.

Take $(a,0) \lt (b,0) \in \mathbb R \times \mathbb Q$ and let $$(a_1, a_2)=\varphi((a,0)) \le (c_1,c) \lt (c_2,c) \lt \varphi((b,0)) =(b_1,b_2).$$

The contradiction is that between the inverse images of $(c_1,c), (c_2,c)$ under $\varphi$ there is a countable number of elements while there should be an uncountable number.