set theory similarity between sets : Are $\{0,1\}×\mathbb R$ and $\mathbb N×\mathbb R$ similar?

Question

Are $\{0,1\}×\mathbb R$ and $\mathbb N×\mathbb R$ similar ? With antilexicographic order $( (a,b)<(c,d)\iff( b < d)\vee (b=d\text{ and }a <c))$ . Two sets are similar if there exists a bijection between them which keeps the order in sets. I tried by looking at invariants of similarity ( minimum, maximum, density, if ther are uncountable or countable) but they don't differ in anything I can think of. There is a characterisation that a set is similar with $\mathbb R$ if it is continuous ( every subset of $X$ has a supremum in $X$ ), has a countable subset which is dense and doesn't have a min or max. Maybe that would work, but are these sets continuos?

Answer 1

Here's an invariant which distinguises between these two sets.

In an ordered set $S$, define an subset $C \subset S$ to be convex if for all $x<y<z \in S$, if $x,z \in C$ then $y \in C$.

Any order preserving bijection takes convex sets bijectitvely to convex sets. To be more precise, if $f : S \to T$ is an order preserving bijection between two ordered sets, and if $C \subset S$ is a subset, then $C$ is convex if and only if $f(C)$ is convex. (Perhaps you might also call $C$ an "interval").

In the ordered set $\mathbb{N} \times \mathbb{R}$, there exists a convex subset of any finite cardinality $n$, namely the set $$\{(1,0),(2,0),(3,0),....,(n,0)\} $$ So, any ordered set which is similar to $\mathbb{N} \times \mathbb{R}$ must also have a convex set of any given finite cardinality $n$.

However, in the ordered set $\{0,1\} \times \mathbb{R}$, every finite convex subset has cardinality $\le 2$, for the following reason. Suppose that $(a_1,b_1) < (a_2,b_2) < (a_3,b_3)$ are all elements of the same convex set $C$. The elements $a_1,a_2,a_3 \in \{0,1\}$ cannot all be different. So, one of two things happens (possibly both): either

1. $a_1=a_2$ and $b_1<b_2$;

or

2. $a_2=a_3$ and $b_2 < b_3$.

In case 1 there are infinitely many elements in $\{0,1\} \times \mathbb{R}$ between $(a_1,b_1)$ and $(a_2,b_2)$; and in case 2 there are infinitely many elements between $(a_2,b_2)$ and $(a_3,b_3)$. In either case, $C$ is infinite.