Order isomorphism help

Question

Let $A=B=(0,1]$ and let $\{a_i\}_{i\in\mathbb{N}}\subset A$ and $\{b_i\}_{i\in\mathbb{N}}\subset B$ are be two sequences and let $a_i\le a_j$ iff $b_i\le b_j$. is there any order preserving embedding $f:A\to B$ such that $f(a_i)=b_i$?

Answer 1

If it is $b_i \leq b_j$, then in general such an embedding doesn't have to exist.

Suppose that all $a_i = 1/2$ and all $b_i = 1$. Then $f(1/2)=1$ and there's no room left for (say) $f(3/4)$.