# Is an open $n$-ball homeomorphic to $\mathbb{R}^{n}$?

I read on Wikipedia the following claim:

Any open topological $n$-ball is homeomorphic to the Cartesian space $\mathbb{R}^{n}$.

No reason or proof was given. Can someone explain? I did try looking at other SE questions and didn't find anything I could understand.

Thanks!

We will show when we have a normed space $E$ and $r>0$, $E$ and its subset $B_r(0) = \{x \text{ }|\text{ }\|x\| < r\}$ are homeomorphic. $($Wikipedia gives a pretty okay primer on normed spaces, Normed vector space, if one has not seen them before.$)$

Definition. A base, or basis for the topology on a topological space, is a collection $\mathcal{F}$ of open sets such that for each open set $U$ is a union of sets in $\mathcal{F}$. When a base for the topology on a space is fixed, we ofen refer to elements of $\mathcal{F}$ as basic open sets.

$($Often a topological space is defined by giving a set and a basis for the topology $($or a sub-basis, which is a collection of sets whose finite intersections form a basis$)$. In any metric space, balls are a basis for the topology, but for some purposes not the most convenient one. And some topological spaces are not metric spaces.$)$

Consider the following basis for the topology on a normed space. For each set $U$ which is open as a subset of $\{x \in E\text{ }|\text{ }\|x\| = 1\}$, and for each ordered pair of positive reals $(a, b)$ the sets $\{tu\text{ }|\text{ }a < t < b,\, u \in U\}$, together with balls centered at $0$, form a basis. Indeed, it is easy to see that every ball is a union of such sets, and every such set is a union of balls. Also, the same collection of sets, except with the restriction $b \le r$ is a basis for the topology on the open ball of radius $r$. Now, notice that the map$$f: B_r(0) \to E,\text{ }f(x) = {x\over{r - \|x\|}}$$ is a bijection from the ball of radius $r$ to all of $E$, as we can explicitly write down the inverse $$g: E \to B_r(0),\text{ }g(y) = {{ry}\over{1 + \|y\|}}.$$ Finally, we see that both $f$ and $g$ have the property that the inverse image of every basic open set is a basic open set, with respect to the bases we picked for $E$ and its ball of radius $r$. $U$ open implies $U$ is a union of basic open sets, so $f^{-1}(U)$ is a union of basic open sets, so $f^{-1}(U)$ is open. Thus $f$ is continuous, and $g$ is continuous by the same argument, so $f, g$ are homeomorphisms. $($This argument can be summarized as "a bijection that induces a bijection of bases is a homeomorphism."$)$

In the interests of completeness, here are some details for showing $f$ is one-to-one and onto.

$f(x)$ is one-to-one.

Assume there are $x_1, x_2 \in B_r(0)$ such that $f(x_1) = f(x_2)$. We have$${{x_1}\over{r - \|x_1\|}} = {{x_2}\over{r - \|x_2\|}} \implies {{\|x_1\|}\over{r - \|x_1\|}} = {{\|x_2\|}\over{r - \|x_2\|}} \implies \|x_1\|(r - \|x_2\|) = \|x_2\|(r - \|x_1\|)$$$$\implies \|x_1\| = \|x_2\| \implies {{x_1}\over{r - \|x_1\|}} = {{x_2}\over{r - \|x_1\|}} \implies x_1 = x_2.$$

$f(x)$ is onto.

Let $x \in E$, and let $x' = rx/(1 + \|x\|)$. Consider$$f(x') = f\left({{rx}\over{1 + \|x\|}}\right) = {{rx}\over{1 + \|x\|}}\left(r - \Bigg\|{{rx}\over{1 + \|x\|}} \Bigg\|\right)^{-1} = {{rx}\over{1 + \|x\|}}\left( {r\over{1 + \|x\|}}\right)^{-1} = x.$$Because$$\Bigg\|{{rx}\over{1 + \|x\|}} \Bigg\| = {{r\|x\|}\over{1 + \|x\|}} < r$$we have that $x' \in B_r(0)$. Hence $f(x)$ is onto.

• Wow, that was really helpful, Kevin! Thanks a lot! Apr 6 '15 at 5:36

Assume the ball is of radius $r$ and centered at the origin. A homeomorphism is given by sending vectors $v$ in the ball to $$\frac{v}{r-\|v\|}$$ To see that this is bijective, let $v'\in\mathbb{R}^n$. If $$v'=\frac{v}{r-\|v\|}$$ Then $\|v'\|=\dfrac{\|v\|}{r-\|v\|}$, so $v$ is the vector with magnitude $\dfrac{r\|v'\|}{1+\|v'\|}$ pointing in the same direction as $v'$ if $v'$ is nonzero, of which there is exactly one, and $v$ is inside the ball. Furthermore, $0\mapsto 0$ and $0$ is the only vector that maps to $0$. The map is continuous since scalar multiplication and the map $v\mapsto \dfrac{1}{r-\|v\|}$ are continuous, and the inverse is $v'\mapsto \dfrac{rv'}{1+\|v'\|}$, which is also continuous.

Let $f: [0,1) \to \mathbb{R}$ be given by $f(x) = {x \over 1-x}$. (We have $f^{-1}(x) = {x \over 1+x}$.)

Then let $h:B(0,1) \to \mathbb{R}^n$ be given by $h(x) = f(\|x\|){x \over \|x\|}$ (except for $h(0) = 0$), and $h^{-1}(x) = f^{-1}(\|x\|){x \over \|x\|}$ (similarly, $h^{-1}(0) = 0$).

The mapping $\phi_{x_0,r}(y) = x_0+ry$ is a homeomorphism between $B(0,1)$ and $B(x_0,r)$ ($\phi_{x_0,r}^{-1} = {1 \over r} \phi_{-x_0,1}$).

It is not necessary to confine ourselves to open balls that are centered at the origin. Let $$X:= R^n$$, $$Y:= B_{r}(\mu) = \{x\in\mathbb R^n \mid \lvert\lvert x-\mu\rvert\rvert < r\}$$. Then $$f: X\rightarrow Y, x\mapsto \frac{r\left( x-\mu\right)}{1+\lvert\lvert x-\mu \rvert\rvert}$$ is a homeomorphism with inverse $$f^{-1}: Y\rightarrow X, x\mapsto \frac{x}{r-\lvert\lvert x\rvert\rvert} + \mu.$$

For center $$c$$ and radius $$r>0:$$ For $$\|v-c\| let $$f(v)=(v-c)\tan (\|v-c\|\pi/2r).$$