# Finding a bijective function between an open disk and the open square

How can I find a bijective function between these two sets? $$\{(x,y)\in\mathbb{R}^2 \,|\, x^2+y^2<1\}, \quad (-1,1) \times (-1,1) .$$

I already thought of first writing between 2nd and set of real numbers, but then I find myself stuck with finding between the reals and the 1st set. Any help would be appreciated.

• think at the Polar coordinates.
– Surb
Commented Jun 30, 2015 at 6:34
• What do u mean with polar coordinates? Commented Jun 30, 2015 at 6:37
• $\{(x,y)\in\mathbb{R}^2 | x^2+y^2<1\}$ is an open disc with radius $1$, which in polar coordinates is $\{(\theta,r)\in{\Bbb R}^2| \theta\in[0,2\pi)\land r\in[0,1)\}$. However there are two problem you would have to solve: the angle range and radius range is both closed on one side and the subset $(\theta,r)\in\{[0,2\pi)\times\{0\}\}$ maps onto the single point $(x,y)=(0,0)$. Commented Jun 30, 2015 at 6:44

One domain is $\{\rho_1(x,y)<1\}$, the other is $\{\rho_2(x,y) <1\}$, where $\rho_1(x,y) =\sqrt{x^2+y^2}$, and $\rho_2(x,y) = \max(|x|,|y|)$. Find a map $F\colon (x,y) \mapsto (x',y')$ so that $\rho_1(x,y) = \rho_2(x',y')$. It should be linear on each line through the origin. One can take: $$F(x,y) = \frac{\rho_1(x,y)}{\rho_2(x,y)} \cdot (x,y)$$ that is \begin{eqnarray} F(x,y) = \frac{\sqrt{x^2+y^2}}{\max(|x|,|y|)} \cdot (x,y) \end{eqnarray} with inverse \begin{eqnarray} F^{-1}(x,y) = \frac{\max(|x|,|y|)}{\sqrt{x^2+y^2}} \cdot (x,y) \end{eqnarray}

$F$ provides a homeomorphism from the disk to the square.

• What about $(x,y) = (0,0)$? Commented Jun 30, 2015 at 9:13
• $(0,0)$ to $(0,0)$. Glad you asked. Commented Jun 30, 2015 at 9:20

Hint If we denote $$S := (-1, 1) \times (-1 ,1) \qquad \text{and} \qquad D := \{(x, y) : x^2 + y^2 = 1\},$$ we can easily write down a homeomorphism $f: \partial S \stackrel{\cong}{\to} \partial D$ between their boundaries in $\Bbb R^2$ by projecting points in $\partial S$ along rays from the origin; explicitly, this map is: $$f(x, y) := \frac{1}{\sqrt{x^2 + y^2}} (x, y).$$

Extending this map linearly on each ray from the origin determines a map $\Bbb R^2 \to \Bbb R^2$ that restricts to a bijection (in fact, a homeomorphism) $S \to D$.

• What do those deltas mean? Commented Jun 30, 2015 at 7:03
• They're not deltas---they're usually read, del, partial, or just d, and it's the same symbol that appears in the partial derivative notation $\frac{\partial f}{\partial x}$. For any topological space $X$ and subspace $A \subseteq X$, $\partial A$ denotes the boundary of $A$. In this case, $\partial S$ and $\partial D$ are the square and circle that are respectively the bound the given regions $S$ and $D$. Commented Jun 30, 2015 at 7:05
• Oh,but this it is from "naive" set theory, therefore i am not using any topology(not officially) . Commented Jul 1, 2015 at 7:41
• @user246608 It's just a convenient notation, it's rather faster to describe $\partial S$ this way than with so-called set-builder notation alone. The argument itself doesn't actually rely on any topology (even if the bijection we produce is compatible with the usual topologies on the involved spaces, i.e., is a homeomorphism between then). Commented Jul 1, 2015 at 10:03
• Oh, i think i get it now. Thank you. Commented Jul 1, 2015 at 10:05

Outline: Map the origin to itself. Now consider points $P$ in the disk with polar coordinates $(r,\theta)$, where $r\gt 0$ and $0\le \theta\le \frac{\pi}{4}$. Map $P$ to $\phi(P)$, where $\phi(P)$ has polar coordinates $(r\sec\theta,\theta)$.

Do the geometrically same thing for the remaining $7$ sectors of the disk.