Let $a:=[0,1)\times\Bbb R$ and $b:=[0,1)\times\Bbb Q$ and let $\prec_a$ and $\prec_b$ denote their respective antilexicographic orders. Are $(a,\prec_a)$ and $(b,\prec_b)$ similar?
$\underline{\boldsymbol{\text{definition 1}:}}$ Partially ordered sets $(a,\prec_a)$ and $(b,\prec_b)$ are said to be similar if there is an order-preserving bijection $f:a\to b$ with an order-preserving inverse $f^{-1}.$
$\underline{\boldsymbol{\text{definition 2 }:}}$ Subset $b$ of a partially ordered set $a$ is said to be convex if, for any $x,y\in b$ and $z\in a,$ $$x<z<y\implies z\in b.$$
$\underline{\boldsymbol{\text{definition 3 }:}}$ Partially ordered set $(a,<)$ is set to be complete if every subset of $a$ bounded from above has a least upper bound in $a$.
$\underline{\boldsymbol{\text{definition 4 }:}}$ Let $(a,\prec_a)$ and $(b,\prec_b)$ be two partially ordered sets. We define the antilexicographic order on the product $a\times b$ as
$(x_1,y_1)<(x_2,y_2):\iff y_1\prec_b y_2\text{ or } (y_1=y_2\text{ and } x_1\prec_a x_2)$
Convexity and completeness are properties invariant under a similarity so, a complete convex subset of $[0,1)\times\Bbb R$ should be mapped to a complete convex subset of $[0,1)\times\Bbb Q.$ I believe the only such subsets in both of $a$ and $b$ are segments $[(x,z),(y,z)]$ for $x,y\in[0,1), x\le y$ and $z\in\Bbb R$ or $\Bbb Q$ because the segments go zig-zag, yet, for example $[0,1)\times\{0\},$ which is bounded from above, doesn't have a supremum in either of the sets as we can choose an arbitrarily small $x>0$ so that $(0,x)$ is an upper bound for $[0,1)\times\{0\}.$
However, since $\Bbb R$ is uncountable, I couldn't figure out how to obtain a countable partition of $a$ into such subsets that would map to suitable subsets in $b$ and it seems in $b$ such partitions would be countable, but I wasn't able to write my thoughts down any further.
Any advice would be greatly appreciated.
Best Answer
Answer in the follow up to Bof's comment
Let's consider $(x_1,y), (x_2,y)\in [0,1)\times\Bbb R=a.$
If $c\subseteq a$ is dense in $a,$ then there is $(z_1,z_2)\in c$ such that $(x_1,y)\prec_a(z_1,z_2)\prec_a(x_2,y).\space \implies$ $x_1<z_1<x_2$ and $z_2=y.$
As Dave L. Renfro pointed out, there are uncountably many copies of $[0,1)$ in $a,$ therefore, a $c$ dense in $a$ is necessarily uncountable. Hence $a$ doesn't have a countable subset dense in it.
On the other hand, $b$ has, for example $[0,1)\times\Bbb Q\cap\Bbb Q\times\Bbb Q.$
Existence of a subset dense in a set is a property invariant under a similarity, so $a$ and $b$ cannot be similar.