Is there an order preserving injection between unwell orders?

Question

Given arbitrary orders $(A,\le_{A}), (B,\le_{B})$. Is it obligatory for an injection from $A$ to $B$ or vice versa such that the order is preserved?

I have noticed that clearly with $\neg $AC this is untrue, because it would imply that cardinals are comparable in AC which is not true (Dedekind finite sets).

But given AC, I can't think of a way to construct such function or counter-example. The problem arises from inability to use the well ordering of the set to construct the function, because of order preservance. So does anyone have an idea how to solve this?

Answer 1

Try these two orders: $A := \{0,1,2,3,\dots\}$ and $B := \{\dots, -3, -2, -1, 0\}$.

Answer 2

Suppose that $X$ is a real closed field of cardinality $\kappa$, $Y$ is a real closed field of cardinality $k^{+}$, and $\Lambda > \kappa^{+}$ is an infinite cardinal. Let $A = B = X \cup \Lambda \cup Y$. When comparing two elements of $A$, two elements of $B$, or two elements of $\Lambda$ use the induced order. For convenience write $A = X_{A} \cup \Lambda_{A} \cup Y_{A}$ and $B = Y_{B} \cup \Lambda_{B} \cup Y_{B}$.

Suppose that $x \in X$, $y \in Y$, and $\lambda \in \Lambda$. In $A$ use the order $x <_{A} \lambda <_{A} y$. In $B$ use the order $y <_{B} \lambda <_{B} x$.

Suppose that $f : A \rightarrow B$ is an order preserving injection. A function $f$ can map the set $X_{A} \cup \Lambda_{A}$ into $Y_{B} \cup \Lambda_{B}$. Since $Y_{A}$ has infinite descending sequences and $| Y_{A} | > | X_{B} | $ there can not be an order preserving injection of $Y_{A}$ into $\lambda_{B} \cup X_{B}$.

Suppose that $g : B \rightarrow A$ is an order preserving injection. Since $Y_{B}$ has infinite descending sequences and $|Y_{B}| > |X_{A}|$, at least some elements of $Y_{B}$ have to be mapped into $X_{A}$. But there is not enough room to map $\Lambda _{B}$ into $Y_{A}$.