Extension of order-preserving bijection from rationals to reals.

Question

If $f:\mathbb{Q}\rightarrow\mathbb{Q}$ is order-preserving bijection. Prove that $f$ can be extended to an order-preserving homeomorphism $F:\mathbb{R}\rightarrow\mathbb{R}$.

Attempt for Proof:The inverse of the given function is also order preserving and bijective. Lets define the extension first. Given a real number $x$, pick a sequence of rationals converging to $x$ from below, call them $a(n)$. Similarly, pick $b(n)$ that converges from above. Then we look at the images of these points. Now we use the fact that $f$ preserves order and conclude (how?) that there is a unique number between all $f[a(n)]$ and $f[b(n)]$. Define it to be the image of x.

Next we would need to prove such an extension is continuous and continuous inverse?

Answer 1

My first idea was to prove that $f$ is locally uniformly continuous to extend it by completeness of $\mathbb{R}$ (using this). But even though is easy to prove that $f$ is continuous I didn't figure out how to prove that it is uniformly continuous.

Here is another approach using Lord Shark the Unknown natural idea.

Given $x\in \mathbb{R}$ define $F:\mathbb{R}\rightarrow \mathbb{R}$ by $$F(x):=\sup \left ( f((-\infty,x) \cap \mathbb{Q})\right )$$ Then

• $F\mid_\mathbb{Q}=f$: This is just the identity $f(-\infty,a)=(-\infty,f(a))$ for $a\in \mathbb{Q}$.
• $F$ is monotone: This came from $x<y \implies f((-\infty,x) \cap \mathbb{Q})\subseteq f((-\infty,y) \cap \mathbb{Q})$
• $F$ is continuous: Take $x\in \mathbb{R}$ and $\varepsilon>0$. Take $a\in (f(x),f(x)+\varepsilon)\cap \mathbb{Q}$ an define $b=f^{-1}(a)$. Then $\delta=b-x$ is such that $f(x,x+\delta)\subseteq (f(x),f(x)+\varepsilon)$. This prove upper semicontinuity at $x$, similarly we can prove the lower semicontinuity at $x$ and so $f$ is continuous.

Now we can define in a similar way for $g:=f^{-1}$ the function $$G(x):=\sup \left ( g((-\infty,x) \cap \mathbb{Q})\right )$$ and prove the three properties above.

As $F\circ G\mid_\mathbb{Q}=G\circ F\mid_\mathbb{Q}=\text{Id}_\mathbb{Q}$ we deduce that $G=F^{-1}$ by continuity and so $F$ is a monotone homeomorphism.