Does bijectivity imply the same order type?

Question

Question: Given an ordered set A which must have bijective correspondences with set B, there must be a correspondence that preserve order?

I know it's wrong. For example, 1.the dictionary order on $\mathbb{Z}_+\times\mathbb{Z}_+$, 2.$(x_0,y_0)<(x_1,y_1)$ if either $x_0+y_0<x_1+y_1$, or $x_0+y_0=x_1+y_1\,and\,y_0<y_1$. They must have bijective correspondence because they have same amount of elements, but are in different order type.

But I feel uncomfortable with that because we all have an intuition that for two groups with same amount of elements( For example, if I can match 1~11, 2~12, 3~13....9~19, 10~20 for sets {1, ... ,10} {11, ... ,20}, I can actually match 1 with 12 with other every elements in {1, ... ,10} matches with elements in {11, ... ,20} ), we can build random correspondence(some correspondence preserves order in this case) between them.

What's wrong with this intuition.

Answer 1

Everything is wrong with that intuition. In fact, it doesn't work if your sets have more than two elements!

Look at the structure of divisibility, $m\mid n\iff\exists k(k\cdot m=n)$. So for example, $2\mid 4$ and $4\mid 8$, but $4\nmid 3$. Then $\{1,\dots,10\}$ with $|$ is very different from $\{11,\dots,20\}$ with $|$. The latter is an antichain, whereas the former has nontrivial comparisons (as we pointed out: $1\mid 2\mid 4\mid 8$).

But you might want to say, I didn't sign up for preserving divisibility. I only wanted to preserve a linear order! But that's the point. Bijections don't have to preserve any "natural structure".

For that matter, setting $f\colon\{0,1\}\to\{2,3\}$ by $f(0)=3$ and $f(1)=2$ also does not preserve the order. It flips it. Of course those two are isomorphic, and you could argue there that that's fine. But that only means that a function $f\colon\Bbb{Z\to N}$ needs to be an isomorphism with respect to some structure, not all structure. And that's true, given any bijection from $X$ to $Y$ we can use it to define a structure on $X$ (or on $Y$) such that $f$ is an isomorphism.

And I mean, what's the order on $\{x,y\}$ when I only tell you that $x\neq y$? Did I tell you what are these objects and how they relate to each other? Not all orders are linear.

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Remember: almost any naive intuition that you're coming with is wrong, in almost any field of mathematics. That is why we have definitions. Work with the definitions, step by step, and very very carefully, and you'll develop the correct definitions.

Or to quote John von Neumann, "in mathematics you don't understand things, you just get used to them."

Answer 2

$\mathbb{Z}$ and $\mathbb{N}$ are bijective but there's no order-preserving map between them. Loosely speaking, this is to do with the fact that $\mathbb{N}$ has a starting number, but $\mathbb{Z}$ doesn't.

$\mathbb{Z}$ and $\mathbb{Q}$ are bijective but there's no order-preserving map between them. This has to do with $\mathbb{Q}$ being dense, whereas $\mathbb{Z}$ isn't.

Some sets of cardinality $|\mathbb{R}|$ also have same property: I don't think there is an order-preserving map between $f: [-1,0] \cup [1,2] \to [0,1]$ because $f(0) < f(1)$, which means that for $f(0) < b < f(1)$ there is no $a \in [-1,0] \cup [1,2]$ such that $f(a) = b$.

I'm not sure how to characterise all these results, but I'm sure it can be done.

Anyway, hopefully I have shown that it is clear that your intuition that "Given an ordered set $A$ which is bijective to set $B$, there must be a correspondence that preserves order" is incorrect. I understand why you have this intuition, but if it's incorrect then it's incorrect.