Seeking an isomorphism from $\langle\mathbb{Q},=,<\rangle\to\langle\mathbb{Q}\cap((0,1)\cup(2,3)),=,<\rangle$

Question

As the question suggests, I am seeking an isomorphism from $$\langle\mathbb{Q},=,<\rangle\to\langle\mathbb{Q}\cap((0,1)\cup(2,3)),=,<\rangle.$$ I know for example that $$\frac{x}{2+2|x|}+\frac{1}{2}$$ is an isomorphism from $\langle\mathbb{Q},=,<\rangle\to\langle\mathbb{Q}\cap(0,1),=,<\rangle$. Also, I know that $$\frac{x}{2+2|x|}+\frac{5}{2}$$ is an isomorphism from $\langle\mathbb{Q},=,<\rangle\to\langle\mathbb{Q}\cap(2,3),=,<\rangle$. However, I'm not able to think of an obvious isomorphism from $\langle\mathbb{Q},=,<\rangle\to\langle\mathbb{Q}\cap((0,1)\cup(2,3))$.

My textbook defines a structure isomorphism as an injective function which preserves relations among elements. So if $x=y$ and $g$ is a bijection, then $g(x)=g(y)$. Similarly, $x<y$ implies $g(x)<g(y)$. Is such a construction even possible?

Answer 1

Let's define first an order-isomorphism from $(-\infty,\sqrt2)\cap\Bbb Q$ to $(0,1)\cap\Bbb Q$ and also one from $(\sqrt2,\infty)\cap\Bbb Q$ to $(2,3)\cap\Bbb Q$. The same idea works for both, so I'll just describe the first. Putting these two order-isomorphisms together gives you the one you seek.

Let $(a_n)_{n=-\infty}^\infty$ be a strictly increasing doubly infinite sequence in $(-\infty,\sqrt2)\cap\Bbb Q$ with $\lim_{n\to\infty}a_n=\sqrt2$ and $\lim_{n\to-\infty}a_n=-\infty$. Likewise, Let $(b_n)_{n=-\infty}^\infty$ be a strictly increasing doubly infinite sequence in $(0,1)\cap\Bbb Q$ with $\lim_{n\to\infty}b_n=1$ and $\lim_{n\to-\infty}b_n=0$. Now define $f:(-\infty,\sqrt2)\cap\Bbb Q \to(0,1)\cap\Bbb Q$ so that $f$ is maps each $[a_n,a_{n+1}]\cap\Bbb Q$ linearly to $[b_n,b_{n+1}]\cap\Bbb Q$.