Is there a bijection from a bounded open interval of $\mathbb{Q}$ onto $\mathbb{Q}$?

Question

It is easy to create a bijection between two bounded open intervals of $\mathbb{R}$, such as:

$$ \begin{align} f : (a,b) &\to (\alpha,\beta) \\ x &\mapsto \alpha+(x-a)(\beta-\alpha). \end{align} $$

It is also possible to biject a bounded open interval of $\mathbb{R}$ onto the whole of $\mathbb{R}$, e.g.:

$$ \begin{align} f : (a,b) &\to \mathbb{R} \\ x &\mapsto \tanh^{-1} x. \end{align} $$

Consider now the set of rational numbers $\mathbb{Q}$. The bijection between two bounded open intervals still holds, but is it possible to biject: $$ f : (p,q) \to \mathbb{Q} $$

where $p,q\in\mathbb{Q}$ and $(p,q) = \{x\in \mathbb{Q} : p < x < q\}$?

Answer 1

The answer to this question: (Is there a bijection between $(0,1)$ and $\mathbb{R}$ that preserves rationality?) -- that is, $f(x) = \frac{1}{x} - 2$ for $(0,\frac{1}{2})$ and $f(x) = 2 - \frac{1}{x - 1}$ on $[\frac{1}{2},1)$ -- is a bijection from $(0,1)$ to $\Bbb R$ that preserves rationality; then $f|_{\Bbb Q}$ is a bijection from $(0,1)_{\Bbb Q}$ to $\mathbb{Q}$.

I'm not sure how this could be extended to general $(p,q)$, but this is an example of one function for one such interval.

EDIT: as Whacka has pointed out, simply by composing this function with a bijection from $(0,1)_{\Bbb Q}$ to $(p,q)_{\Bbb Q}$, this may be achieved. One such map is $x \mapsto (q-p)x + p$, whose inverse is $x \mapsto \frac{x - p}{q - p}$.

So in general, a bijection from $(p,q)_{\Bbb Q}$ to $\Bbb Q$ is:

$f(x) = \frac{q - p}{x - p} - 2,\;x \in (p, \frac{p+q}{2})$

$f(x) = 2 - \frac{q - p}{x - q},\;x \in [\frac{p+q}{2}, q)$

Answer 2

Cantor showed that any two countable densely ordered sets with no first or last element are order isomorphic. One can build the isomorphism using his Back and Forth Method.

If all we want is a bijection, we can more simply note that both sets are countably infinite.

Answer 3

Start: Pick any monotonic decreasing sequence $l_i$ of positive rationals converging to zero and any monotonic increasing sequence $r_i$ of positive rationals converging to $1$, with $l_1<r_1$. Consider

$$\begin{array}{lllll} \Bbb Q= & \cdots & \sqcup & [2,-1) & \sqcup & [-1,0) & \sqcup & [0,1) & \sqcup & [1,2) & \sqcup & [2,3) & \sqcup & \cdots \\ (0,1) = & \cdots & \sqcup & [l_3,l_2) & \sqcup & [l_2,l_1) & \sqcup & [l_1,r_1) & \sqcup & [r_1,r_2) & \sqcup & [r_2,r_3) & \sqcup & \cdots \end{array} $$

Note that $[a,b)\cong[c,d)$ are order-isomorphic for any rationals $a,b,c,d\in\Bbb Q$ with $a<b$, $c<d$.

All intervals are taken in $\Bbb Q$.

Answer 4

I know similar ideas already came up here, but the first thing I thought of was to construct the bijection $f:\mathbb R\longrightarrow(-2,2)$ given by $$ f(x)= \begin{cases} -2-\frac{1}{x}&x\in(-\infty,-1)\\ x&x\in[-1,1]\\ 2-\frac{1}{x}&x\in(1,\infty) \end{cases} $$ with explicit inverse $$ f^{-1}(x)= \begin{cases} -\frac{1}{x+2}&x\in(-2,-1)\\ x&x\in[-1,1]\\ \frac{1}{2-x}&x\in(1,2) \end{cases} $$ and then this could be composed with a linear bijection $g:(-2,2)\longrightarrow(p,q)$.