Continuous bijection from $[0,1]$ to $(0,1)$ on rational numbers with a continuous inverse.

Question

A function $f\colon\Bbb Q\to\Bbb Q$ is rational-continuous at a rational number $\alpha$ iff for any $\varepsilon > 0,$ there exists a corresponding $\delta$ based on $\alpha$ and $\varepsilon$ such that if $|\alpha-x| \leq \delta, |f(\alpha)-f(x)| \leq \varepsilon.$ Is there a "rational-continuous" bijection with a "rational-continuous" inverse from the rationals in $[0,1]$ to the rationals in $(0,1)?$

In this question it was shown that we can get such a bijection from $(0,1]$ to $(0,1).$ Such a bijection from $[0,1]$ to $(0,1]$ is acceptable as an answer.

Answer 1

In fact there is a much more general result: any two countably infinite metric spaces without isolated points are homeomorphic. Note that "homeomorphic" is a much weaker condition than "isometric" - a homeomorphism (= continuous function with continuous inverse) is allowed to distort distances substantially.

The proof is by a back-and-forth argument (see the wiki page for the ur-example, or this old answer of mine for a slightly messier example that might help motivate the construction below), relying heavily on the countability of the spaces in question. Let $A,B$ be countably infinite metric spaces without isolated points. For simplicity, let's assume that they each have diameter $1$ (this can be arranged via rescaling if necessary). Let's define a partial homeomorphism from $A$ to $B$ as a tuple $$(f, R,S)$$ with the following properties:

• $f$ is a bijection between some finite subset $A_0$ of $A$ and some finite subset $B_0$ of $B$ (let $n=\vert A_0\vert=\vert B_0\vert$);

• $R$ is a set of "forward continuity rules" - precisely, $R$ is a function from $A_0\times\{2^{-i}: i<n\}$ to $\mathbb{R}_{>0}$ such that

  • $f$ "agrees with $R$ so far" in the sense that for all $i<n$ and all $x,y\in dom(f)$ with $d_A(x,y)<R(x, 2^{-i})$ we have $d_B(f(x),f(y))<2^{-i}$, and

  • $R$ "avoids potentially nasty overlaps in the future" in the sense that for all $i,j<n$ and all $x,y\in dom(f)$, if $$R(x, 2^{-i})+R(y, 2^{-j})\ge d_A(x,y)$$ then $2^{-i}+2^{-j}\ge d_B(f(x),f(y))$.

• $S$ is a set of "backward continuity rules" - precisely, $S$ is a function from $B_0\times\{2^{-i}: i<n\}$ to $\mathbb{R}_{>0}$ such that

  • $f^{-1}$ "agrees with $S$ so far" in the sense that for all $i<n$ and all $x,y\in ran(f)$ with $d_B(x,y)<R(x, 2^{-i})$ we have $d_A(f^{-1}(x),f^{-1}(y))<2^{-i}$, and

  • $S$ "avoids potentially nasty overlaps in the future" in the sense that [exercise - basically, just "inverseify" the corresponding condition for $R$].

Given partial homeomorphisms $\mathfrak{f}=(f,R,S),\mathfrak{g}=(g,U,V)$, say that $\mathfrak{f}$ extends $\mathfrak{g}$ iff $f$ extends $g$, $R$ extends $U$, and $S$ extends $V$ (remember that a function $\alpha$ extends a function $\beta$ iff $dom(\alpha)\supseteq dom(\beta)$ and $\alpha\upharpoonright dom(\beta)=\beta$). It's then easy (if tedious) to prove the following:

$(*)\quad$ For any partial homeomoprhism $\mathfrak{f}=(f,R,S)$, any $a\in A$, and any $b\in B$, there is a partial homeomorphism $\mathfrak{g}=(g,U,V)$ such that $\mathfrak{g}$ extends $\mathfrak{f}$, $a\in dom(g)$, and $b\in ran(g)$.

We can then iterate $(*)$ to build a genuine homeomorphism from $A$ to $B$. Basically, fix enumerations $(a_i)_{i\in\mathbb{N}},(b_i)_{i\in\mathbb{N}}$ of $A,B$ respectively, and repeatedly apply $(*)$ to build a sequence of partial homeomorphisms $\mathfrak{f}_i=(f_i,R_i,S_i)$ ($i\in\mathbb{N}$) such that for each $i$ we have that $\mathfrak{f}_{i+1}$ extends $\mathfrak{f}_i$ and $a_i\in dom(f_i)$ and $b_i\in ran(f_i)$. Then the "union of the approximations" $$f=\bigcup_{i\in\mathbb{N}}f_i$$ is a genuine homeomorphism from $A$ to $B$.