# Extension of order-preserving bijection from rationals to reals.

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?

• Dedekind cuts${}$? Commented Dec 19, 2018 at 18:07

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 $$x
• $$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.