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 $(0,1)?$

The answer to my original question has an inverse which is not "rational-continuous", but I think a nicer solution exists.

Answer 1

Let $T_1:=\mathbb{Q}\cap(0,1]$ and $T_2:=\mathbb{Q}\cap (0,1)$ be given the subspace topologies. Your question essentially translates to asking whether $T_1$ and $T_2$ are homeomorphic.

Firstly, observe that if $(a,b)$ and $(c,d)$ are two intervals, then $(a,b)\cap \mathbb{Q}$ and $(c,d)\cap \mathbb{Q}$ are homeomorphic under the subspace topologies. To sketch the argument, one may construct an order preserving bijection between the above sets by enumerating (using a bijection to $\mathbb{N}$) both sets and assigning inductively. Note that this argument uses the axiom of dependent choice. Since the subspace topologies coincide with the respective order topologies, this map also turns out to be a homeomorphism.

Let $\{\alpha_i\}_{i=1}^{\infty}$ be a strictly sequence of irrational numbers in $(0,1)$ converging to $1$. Similarly, let $\{\beta_j\}_{i=1}^{\infty}$ and $\{\gamma_k\}_{i=1}^{\infty}$ be strictly increasing and strictly decreasing sequences respectively of irrational numbers in $(0,1)$, both converging to $\frac{1}{2}$. Let $\alpha'$ be an irrational number such that $0<\alpha'<\alpha_1$. We define a function $\phi:T_1\rightarrow T_2$ as per the following conditions:

1. $\phi:1\mapsto \frac{1}{2}$
2. $\phi$ maps $(0,\alpha')\cap \mathbb{Q}$ homeomorphically onto $(0,\beta_1)\cap \mathbb{Q}$ and $(\alpha',\alpha_1)\cap \mathbb{Q}$ homeomorphically onto $(1,\gamma_1)$
3. $\phi$ maps $(\alpha_{2i},\alpha_{2i+1})\cap \mathbb{Q}$ homeomorphically onto $(\beta_i,\beta_{i+1})\cap \mathbb{Q}$ and $(\alpha_{2i-1},\alpha_{2i})\cap\mathbb{Q}$ homeomorphically onto $(\gamma_i,\gamma_{i+1})\cap \mathbb{Q}$

It may be checked that $\phi$ is be a homeomorphism.