# If there are order preserving injections between two countable sets, will there be an order preserving bijection?

During my Topology class the following question was brought up:

Question: Suppose that $A$ and $B$ are ordered countable sets. If there are two order preserving injections $f:A \to B$ and $g:B \to A$, is there an order preserving bijection between $A$ and $B$?

I do know that one can construct a bijection using two injections, it is a classic problem. However, I couldn't adapt the proof for order preserving injections.

False or not, I was hoping to find a proof for the following preposition:

Proposition: Every densely totally ordered countable set without lower or upper bound is isomorphic to $\mathbb{Q}$.

• The Proposition is true (Cantor proved it). Its usual proof now uses the "back and forth" or "zig-zag" method, which came later -- see en.wikipedia.org/wiki/Cantor's_back-and-forth_method. Re your Question: by "ordered" do you mean totally ordered? Apr 4, 2016 at 16:34
• Note that while the answer to your question is 'no' (see below), it is true if you require well-orders instead of merely linear orders. Apr 4, 2016 at 16:52
• I didn't mean totally ordered, I was just wondering as stated during my class. For the proposition, thank you for the hyperlink. I read a little and it is the same proof used in class. I may take a look for others uses of the method. Thank you. Apr 4, 2016 at 17:20

It appears that your conjecture as stated is false. Indeed, let $A=(0,1)\cap \Bbb Q$ and $B=\Bbb N \cup A$, $f:A\to B$ be the identity mapping, $g:B\to A$ be defined by $$g(b)=\begin{cases} \frac b2\ ;\ b\in A \\ 1-\frac 1{b+1}\ ;b\in \Bbb N \end{cases}$$ , then both $f$ and $g$ are order preserving injections. However, $A$ and $B$ have different order type so there cannot be any order isomorphism between them.
• A simpler example: consider $(0, 1)\cap\mathbb{Q}$ versus $[0, 1]\cap\mathbb{Q}$. Also, another source you may be interested in is Rosenstein's book "Linear orderings". Apr 4, 2016 at 16:52
• Or even simpler $(0, 1)\cap\mathbb{Q}$ vs. $(0, 1]\cap\mathbb{Q}$, $g$ being $g(x)=x/2$ :) Apr 4, 2016 at 17:05