I was wondering that can I show that an open unit disk in $\mathbb{R}^n$ is homeomorphic to the surrounding space $\mathbb{R}^n$ by arguing that all of the half-axes of the space are homeomorphic to the half-axes of the unit disk. By half-axes of the space I refer to the two parts, $(-\infty, 0], [0, \infty)$ of each coordinate axis, and by half-axes of the open unit disk I refer to the intervals $(-1, 0], [0, 1)$ along all of the coordinate axis of the surrounding space.

So essentially I would argue that by some univariate functions $f$ and $-f$ all $[0, 1) \approx [0, \infty)$ and $(-1, 0] \approx (-\infty, 0]$. Then the mapping $g(x_1, x_2,\dots,x_n)$ from the $n$-dimensional open disk to the $\mathbb{R}^{n}$ could be constructed by considering the sign of the arguments, so that if $x_i \in [0, 1)$ then the $i$th coordinate of the image of $g$ is determined by $f$, and if $x_i \in (-1, 0]$ then it is determined by $-f$.


2 Answers 2


Map $x \in \Bbb R^n$ to $f(x)=\frac{1}{1+\|v\|}\cdot v$, where $\cdot$ is scalar multiplication and the norm is the standard Euclidean one. So every vector is scaled down (in the same direction) so that its norm becomes

$$\left\| \frac{v}{1+\|v\|} \cdot v \right\| = \left(\frac{1}{1+\|v\|}\right)\|v\| = \frac{\|v\|}{1+\|v\|} < 1$$

and so $f$ maps $\Bbb R^n$ into the open unit ball $B(0,1)$.

$f$ is clearly continuous as the norm as a function is continuous and division in $\Bbb R$ and scalar multiplication are too.

$f$ is invertible as we can solve $f(tv)=v$ for $v \in B(0,1)$ uniquely and it's easy to see we get $f^{-1}(v)=\frac{1}{1-\|v\|}\cdot v$ which is also continuous (and well-defined on $B(0,1)$).


You say "if $x_i\in[0,1)$". But not all such coordinates combined give a point in an open disk (assuming the standard Euclidean norm). For example for $n=2$ the point $(0.9,0.9)$ does not belong to the unit disk. Your approach would be ok for $(-1,1)^n$, but not for unit disk.

The simpliest method for $B\to\mathbb{R}^n$ homeomorphism is to map $v\mapsto f(\lVert v\rVert)\cdot v$ where $f:[0,1)\to[0,\infty)$ is a homeomorphism, say $f(x)=\frac{\pi}{2}\cdot\tan(x)$. That's because it has a simple inverse: $v\mapsto f^{-1}(\lVert v\rVert)\cdot v$. As a nice bonus: this approach works with any norm.

  • $\begingroup$ Why is $(0.9, 0.9)$ not part of the open unit disk? By unit disk I am referring to this: en.wikipedia.org/wiki/Unit_disk $\endgroup$ Feb 4, 2021 at 11:44
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    $\begingroup$ @Qwaster with the Euclidean norm we have $\lVert (0.9,0.9)\rVert \approx 1.27 > 1$. You can draw the open unit disk on the plane and see that the $(-1,1)^2$ square is bigger. $\endgroup$
    – freakish
    Feb 4, 2021 at 11:45
  • $\begingroup$ You are right! My bad. While I believe your homeomorphism map, could you explain the intuition behind it? We take vector $v$ with a length little bit under one (since it is from the unit disk) and then scale it by the a scalar determined by the mapping $f$ and the norm of the vector $v$? $\endgroup$ Feb 4, 2021 at 11:54
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    $\begingroup$ You are correct. The intuition is similar to stretching $[0,1)$ to $[0,\infty)$, except that in $\mathbb{R}^n$ you take a line connecting $0$ to $v$ and stretch it to infinity. $\endgroup$
    – freakish
    Feb 4, 2021 at 14:56

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