Order preserving bijection between $(a,b)$ and $(c,d)$ to $(e,f)$ on $\Bbb Q$ I need to prove $((0,1)\cap\mathbb{Q})\cup((2,3)\cap\mathbb{Q}) $ is isomorphic to $(0,10)\cap\mathbb{Q}$. I have tried by looking for a bijective relation between sets and Yi think my best try has been a piecewise function:
\begin{equation}
f(x)=
\begin{cases}
       x, & 0\lt x\lt 1 \\
      9x-17, & 2\lt x\lt 3
\end{cases}
\end{equation}
The problem is $1$ belongs to $(0,10)$ but it is not on the image of $f$ so it doesn't work as a bijection. I'd really appreciate some help on this problem. Thanks
Edit: I need an order-preserving bijection.
 A: First, find an irrational "hole" in $(0,10)$, let's say $Z=\pi$.
Can you now construct an order-preserving bijection of $(0,1)\cap\mathbb Q$ to $(0,Z)\cap\mathbb Q$? For example:

*

*Make a (strictly) monotonically increasing sequence of rational numbers $a_i\in(0,1)$ such that $\lim_{n\to\infty}a_i=1$, e.g. $0.9, 0.99, 0.999,\ldots$.

*Also, make another (strictly) monotonically increasing sequence of rational numbers $b_i\in(0,Z)$ such that $\lim_{n\to\infty}b_n=Z$, e.g. $3, 3.1, 3.14, 3.141\ldots$.

*Now make the bijection piecewise-linear on: $(0,a_1]\to(0,b_1], [a_1,a_2]\to[b_1,b_2], [a_2,a_3]\to[b_2,b_3]$ etc.

Altogether, these linear mappings will provide an order-preserving bijection of $(0,1)\cap\mathbb Q$ to $(0,Z)\cap\mathbb Q$.
Similarly, construct an order-preserving bijection of $(2,3)\cap\mathbb Q$ to $(Z,10)\cap\mathbb Q$. (You will need strictly monotonically decreasing sequences of rational numbers converging to $2$ and $Z$ "from above".)
Finally, glue those two bijections together.
A: Let $a_n=1-\frac1n$ and partition $(0,1)$ as
$$(0,1)=\bigcup_{n=1}^\infty (a_n,a_{n+1}].$$
Likewise, let $b_n=2+\frac1n$ and partition
$$(2,3)=\bigcup_{n=1}^\infty [b_{n+1},b_n).$$
Next, fix a real number $\alpha$ between $0$ and $10$.
Let $\{c_n\}_n$ be a strictly increasing sequence of rational numbers that converge to $\alpha$ and starts with $c_1=0$.
Similarly, let $\{d_n\}_n$ be a strictly decreasing sequence of rational numbers that converge to $\alpha$ and starts with $d_1=10$.
Note that
$$ (0,10)=\bigcup_{n=1}^\infty (c_n,c_{n+1}]\;\cup\{\alpha\}\cup \;\bigcup_{n=1}^\infty [d_{n+1},d_n)$$
Now we can readily biject $(a_n,a_{n+1}]\to (c_n,c_{n+1}]$. This preserves rationality as all $a_n$ and all $c_n$ are rational. We can do the same with $[b_{n+1},b_n)\to [d_{n+1},d_n)$. Ultimately, this gives us an order-preseving bijection
$$\bigl((0,1)\cap \Bbb Q\bigr) \cup \bigl((2,3)\cap \Bbb Q\bigr)\to \bigl((0,10)\cap \Bbb Q\bigr)\setminus\{\alpha\}.$$
But we can ignore that spurious "$\setminus\{\alpha\}$" if we pick $\alpha$ to be irrational!
