# Showing that an open unit disk is homeomorphic to the surrounding space

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 $$-f$$.

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.

• 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 Feb 4, 2021 at 11:44
• @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. Feb 4, 2021 at 11:45
• 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 $v$? Feb 4, 2021 at 11:54
• 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. Feb 4, 2021 at 14:56