Is there a continuous bijection from $\mathbb{Q}$ to $\mathbb{Q^*}$ I am trying to find an explicit function but I don't know if that is even possible.
Thanks
 A: Let's start by writing down an explicit bijection from $\Bbb Q\cap(a,b]$ to $\Bbb Q\cap(c,d]$, for any real numbers $a,b$ and rational numbers $c>a$ and $d>b$. Choose decreasing sequences $b = b_0 > b_1 > b_2 > \cdots$ and $d = d_0 > d_1 > d_2 > \cdots$ of rational numbers such that $b_n\to a$ and $d_n\to c$. Then let $f\colon (a,b]\to(c,d]$ be the piecewise linear function joining the points $(b_n,d_n)$, that is:
$$
f(x) = d_n + \frac{d_n-d_{n-1}}{b_n-b_{n-1}} (b_n-x) \quad\text{for } x\in [b_n,b_{n-1}].
$$
Since this line has rational slope, $f$ is a bijection between $\Bbb Q\cap (b_n,b_{n-1}]$ and $\Bbb Q \cap (d_n,d_{n-1}]$. The fact that $f$ is a bijection between $\Bbb Q\cap(a,b]$ to $\Bbb Q\cap(c,d]$ follows from the fact that $\bigcup_{n=1}^\infty (b_n,b_{n-1}] = (a,b]$ and $\bigcup_{n=1}^\infty (d_n,d_{n-1}] = (c,d]$.
Now for the stated problem: fix an irrational number $y$ and two rational numbers $n < 0 < p$ with $|n|,|p|>|y|$. Use the above process to write down a bijection from $\Bbb Q\cap(0,p]$ to $\Bbb Q\cap(y,p]$, and the same process (multiplied through by $-1$) to write down a bijection from $\Bbb Q\cap[n,0)$ to $\Bbb Q\cap[n,y)$. The union of these two bijectinos with the obvious bijections from $(p,\infty)$ to itself and $(-\infty,n)$ to itself forms a bijection from $\Bbb Q^*$ to $\Bbb Q$.
A: Any two countable densely ordered sets with no first or last element are order isomorphic. One can build such an order isomorphism using the Back and Forth Method of Cantor.
