I'm asked to build an order-perserving function from $<\mathbb{Q},<_{\mathbb{Q}}>$ to $<B,<_{B}>$, where $B=\{(p,q)|p<q\land p,q\in\mathbb{Q}\}$ and $<$ is defined $(p,q)<(t,l)\iff t>q$.
So my proof is this: We can prove that $<$ on $B$ is dense and has no maximum nor minimum element. Moreover, $\mathbb{Q}$ is countable, so we can use Cantor's "back and forth" method, using only the forth part. Let us first define enumeration of $\mathbb{Q}$ - $A=\{q_{n}|n\in\mathbb{N}\}$.
So we want to inductively define $F=\{f_{n}|n\in\mathbb{N}\}$ where for each $n$, $f_{n}$ satisfy the following: $1$. $domf_{n}\subset A$ and $Imgf_{n}\subset B$, and both are finite. $2$. $q_{n}\in domf_{n}$ $3$. $f_{n}$ is order-perserving. $4$. $m<n\rightarrow f_{m}\subseteq f_{n}$
Finally, we define $f=\bigcup_{n\in\mathbb{N}}f_{n}$, where $domf=A$, $Imgf\subseteq B$ and $f$ is order-perserving.
Let's define $f_{0}=\emptyset$, and assume that for every $m<n$, $f_{m}$ is defined. Let's define $f_{n}$: if $q_{n}\in domf_{m}$, we define $f_{n}=f_{m}$. Else, we first define $X_{0}=\{q_{m}|q_{m}<q_{n}\land q_{m}\in domf_{m}\}, X_{1}=\{q_{m}|q_{m}>q_{n}\land q_{m}\in domf_{m}\}$. Since $q_{n}\notin domf_{m}$, $X_{1}\cup X_{0}=domf_{m}$. Since we assume each $f_{m}$ is finite, we can find maximum in $X_{0}$ that we'll denote $q_{m}$ and minimum in $X_{1}$ that we'll denote $q_{M}$. Now - if neither $X_{0}$ nor $X_{1}$ are empty, because $B$ is a dense order we can find $b\in (f(q_{m}),f(q_{M}))$ (the interval in not empty) and define $f_{n}=f_{m}\cup \{<q_{n},b>\}$. Otherwise if one of them is empty, we can use the fact that $B$ has neither maximum nor minimum, and again find such $b$ that fits.
Does my prove work? I'm still uncomfortable with such constructions.