Continuous bijection from $(0,1]$ to $(0,1)$ on rational numbers.

Question

Recently, I have been dealing with generalizing notions using the real numbers by using the rationals instead. 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 from the rationals in $(0,1]$ to the rationals in $(0,1)?$

Answer 1

Define $f\colon (0,1]\cap \mathbb{Q}\to (\pi/4,1+\pi/4)\cap \mathbb{Q}$ by:

$$f(t)=1+t,\qquad t<\pi/4,$$ $$ f(t)=t, \qquad t>\pi/4.$$

Then $f$ is continuous and bijective.

It remains to define a continuous bijective function $(\pi/4,1+\pi/4)\cap \mathbb{Q}\to (0,1)\cap \mathbb{Q}$.

However this can be done piecewise linearly on intervals:

Let $a_n\in\mathbb{Q}$, $n\in \mathbb{Z}$ be an increasing sequence, with:$$ \lim_{n\to-\infty} a_n=\pi/4,\qquad \lim_{n\to\infty} a_n=1+\pi/4,$$ and let $b_n\in\mathbb{Q}$, $n\in \mathbb{Z}$ be an increasing sequence, with:$$ \lim_{n\to-\infty} b_n=0,\qquad \lim_{n\to\infty} b_n=1.$$

Then simply map each interval $[a_n,a_{n+1}]$ linearly to $[b_n,b_{n+1}]$.