Cardinality of order preserving functions from totally ordered set with dense subset

Question

I know that in the category of continuous functions, if $X$ and $Y$ are Hausdorff topological spaces and $D\subseteq X$ is a dense subset of $X$, then the cardinality of the continuous functions from $X$ to $Y$ is at most $\vert Y^D\vert$. Can we obtain a similar result for ordered spaces with respect to order preserving functions $A\to B$ where $D\subseteq A$ is dense, in the order sense, and saying there are at most $\vert B^D\vert$ order preserving functions? For example if we assume that $A$ is the completion of $D$ or perhaps something stronger?

Answer 1

No. For instance, consider order-preserving maps $f:\mathbb{R}\to\mathbb{R}\times\mathbb{R}$ where $\mathbb{R}$ has its usual order and $\mathbb{R}\times\mathbb{R}$ has its lexicographic order. For any function $g:\mathbb{R}\to\mathbb{R}$, the map $f(x)=(x,g(x))$ is order-preserving. So the number of order-preserving maps $\mathbb{R}\to\mathbb{R}\times\mathbb{R}$ is $(2^{\aleph_0})^{2^{\aleph_0}}=2^{2^{\aleph_0}}$ which is greater than $|(\mathbb{R}\times\mathbb{R})^\mathbb{Q}|=2^{\aleph_0}$ even though $\mathbb{Q}$ is dense in $\mathbb{R}$.