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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 $Y_{A}$.