Find a bijection from the closed disk to the open square

Question

This is a slightly specific question but I'm not sure how to even begin or go about doing it:

Find a bijection from a closed disk:

$D=\{(x,y)\in\mathbb{R}^2|x^2+y^2\leq 1\}$

To the open square:

$S = (0,1) \times (0,1)=\{(x,y)\in \mathbb{R}^2|0<x<1, 0<y<1\}$

Answer 1

1. If we define $\varphi$ through $\varphi(0,0)=(0,0)$ and $\varphi(x,y)=\frac{\max(|x|,|y|)}{\sqrt{x^2+y^2}}(x,y)$ then $\varphi$ maps the closed unit disk into the closed square $\max(|x|,|y|)\leq 1$ in a bijective way;
2. There is a bijective map $\gamma$ from $[-1,1]$ to $(-1,1)$: for instance, it is enough to enumerate the rational points of $[-1,1]$ as $q_0=-1,q_1=1,\ldots$, then map $q_n$ into $q_{n+2}$;
3. Let $\eta(x,y) = \left(\frac{\gamma(x)+1}{2},\frac{\gamma(y)+1}{2}\right)$.
   Then $\eta\circ\varphi$ maps $x^2+y^2\leq 1$ into $(0,1)\times(0,1)$ in a bijective way.

Of course there are some issues in dealing with continuous bijections since $x^2+y^2\leq 1$ is closed while $(0,1)\times(0,1)$ is open in the Euclidean topology.

Answer 2

You can do this in a sequence of steps. The only tricky one is the first.

1) Find a bijection from the closed unit disk to the open unit disk.

2) Find a bijection from the open unit disk to the square $(-1,1) \times (-1,1)$

3) Find a bijection from the square $(-1,1) \times (-1,1)$ to the square $(0,1) \times (0,1)$.

One bijection from the closed unit disk to the open unit disk is $$ f(x,y) = \left\{ \begin{array}{lc} (x/2,y/2) & x^2 + y^2 = 4^{-n},\ n \in \{0,1,2,\ldots\} \\ (x,y) & \text{otherwise} \end{array} \right.$$