monotone bijection between ordered set which is not a order-isomorphism

Question

What would be an example of a bijection $f : (A,<_A) \to (A,<_A)$ such that that $\forall x,y \in A$ if $x<y$ then $f(x)<f(y)$ but $f$ is not an order-isomorphism? I thought of different bijection from $\mathbb{Z}$ to $\mathbb{Z}$ but it fails.

Answer 1

Consider

$$A_1=\{a_1,a_2,\ldots\}, A_2=\{b_1,b_2,\ldots\}, A_3=\{c_1,c_2,\ldots\} \text{ and } A = A_1 \cup A_2 \cup A_3, $$

and

$$<_A = \{(b_n,c_n)\;|\;n=1,2,\ldots\}.$$

So all the $a_n$ are incomparable to anything, and each $b_n$ is less than the correspondig $c_n$, and that's all there is for comparable elements.

Now consider the following $f: A \rightarrow A$:

$$ f(x)=\begin{cases} a_{n-2}, & \text{if } x=a_n, n\ge 3, \\ c_1, & \text{if } x=a_2, \\ b_1, & \text{if } x=a_1, \\ b_{n+1}, & \text{if } x=b_n, n\ge 1,\\ c_{n+1}, & \text{if } x=c_n, n\ge 1.\\ \end{cases} $$

The $b_n$ and $c_n$ get shifted up an index, the "empty" place of $b_1,c_1$ is taken by $a_1, a_2$ and the $a_n$ get shifted down by 2 indices, to fill the "hole" left by $a_1,a_2$.

So $f$ is a bijection from $A$ onto itself. It's also order preserving, because we only need to check what happens to $f(b_n)$ compared to $f(c_n)$, and we have

$$f(b_n)=b_{n+1} <_A c_{n+1} =f(c_n),$$

which is exactly what we need to prove for order preservation.

But $f$ is not an order isomorphism, because $a_1$ and $a_2$ are incomparable, but we have $f(a_1)=b_1 <_A c_1=f(a_2)$.

This concludes the proof that the given $f$ is a monotone bijection (order preserving bijection), but not an order isomorphism.

Answer 2

Here is the minimal example of a monotone bijection between posets that is not an isomorphism of posets.

Let $P$ be the poset with two elements $x$ and $y$, neither of which is $\le$ the other. Let $P'$ be the poset with two elements $x$ and $y$ obeying $x \le y$. So, $P$ and $P'$ have the same underlying set. The identity function from $P$ to $P'$ is a monotone bijection that is not an isomorphism of posets, because its inverse is not monotone.

Up to isomorphism, every example can be obtained as follows: take a set and make it into a poset in two different ways, say $P$ and $P'$, by creating two partial orderings $\le_P$ and $\le_{P'}$, with the property that

$$ x \le_P y \implies x \le_{P'} y $$

but not

$$ x \le_{P'} y \implies x \le_{P} y $$