I seek for a short and rigorous way to extend an embedding of a partial order into $\mathbb{Q}$

Question

Let $(X,\leq ) $ be a partial order and $A$ be a proper subset of $X$ and $t\in X\setminus A$. Knowing that there exists an order-preserving map $\varphi :A\rightarrow \mathbb{Q}$ whose range is finite, I want to show that this map can be extended to $A\cup \{ t\}$ still preserving the order. The problem is that I could not show it in a rigorous way. I had to use expressions like “if [some formal statement] doesn’t hold, then check the next one”, “continuing this process…” , which makes the proof ugly. Is there any suggestion of formal argument for that?

Answer 1

Rigour and ugliness are not mutually incompatible. But:

Put $A_{< t} = \{a \in A \mathrel: a < t\}$, and analogously for $A_{> t}$. Since $\varphi(A)$, and so its subsets $\varphi(A_{< t})$ and $\varphi(A_{> t})$, are finite, and since the order on $\mathbb Q$ is total, there are a maximum element $b_-$ of $\varphi(A_{< t})$, and a minimum element $b_+$ of $\varphi(A_{> t})$. Fix $a_- \in A_{< t}$ and $a_+ \in A_{> t}$ such that $b_\pm = \varphi(a_\pm)$, so that $a_- < t < a_+$.

Assuming order-preserving means strictly order-preserving (i.e., $a_1 < a_2 \implies \varphi(a_1) < \varphi(a_2)$ for all $a_1, a_2 \in A$), then $b_- < b_+$. By the density of the order on $\mathbb Q$, there is some $c \in \mathbb Q$ such that $b_- < c < b_+$. Then we may extend $\varphi$ to $A \cup \{t\}$ by putting $\varphi(t) = c$.

If you do not require strict order preservation (and so assume only that $a_1 \le a_2 \implies \varphi(a_1) \le \varphi(a_2)$ for all $a_1, a_2 \in A$), then the analogous argument concludes only that $b_- \le b_+$, and we may choose, say, $\varphi(t) = b_-$.