# In $(\mathbb{R}^n,\varepsilon_n)$ prove the unit open ball and $Q=\{x \in \mathbb{R}^n| | x_i| <1, i=1,…,n \}$ are homeomorphic

I have the following example in my lectures notes:
In $$(\mathbb{R}^n,\varepsilon_n)$$ , where $$\varepsilon_n$$ is the usual euclidean topology or the one induced from it. Let $$B=B(O,1)$$ be the open ball with center the origin $$O$$ and radious $$1$$ and $$Q=\{x \in \mathbb{R}^n| | x_i| <1, i=1,...,n \}$$ the n-dimensional cube of center $$O$$ and side $$2$$ (or in other words the open ball of center $$O$$ and radious $$r$$ with rispect to the metric $$d_{L^{\infty}}$$) We want to prove $$B$$ is homeomorphic to $$Q$$

Let $$x=(x_1,...,x_n) \in \mathbb{R}$$, we have that $$\|x\|_{L^{\infty}}=d_{L^{\infty}}(x,O)=max\{|x_i|| i=1,...,n\}$$. Let $$k(x)=\frac{\|x\|_{L^{\infty}}}{\|x\|}$$, and let's consider the mapping $$f:(\mathbb{R}^n,\varepsilon_n) \rightarrow (\mathbb{R}^n,\varepsilon_n)$$ defined by: $$f(x)=\begin{cases} k(x)x & \text{if } x \neq 0, \\ 0 &\text{if } x=0\end{cases}$$

Now, $$f$$ is a homeomorphism from $$(\mathbb{R}^n,\varepsilon_n)$$ to itself such that $$f(Q)=B$$, that is $$f$$ is open

In fact $$1/\sqrt(n) \leq k(x) \leq 1$$ and $$k(\alpha x)=k(x)$$ for each $$x \neq O$$ and $$\alpha \in \mathbb{R}^n\setminus\{0\}$$, then $$\lim_{x \to O}f(x)=O$$ and $$f$$ is continuous over $$\mathbb{R}^n$$

Furthermore the mapping $$g(x)=\begin{cases} x\frac{1}{k(x)} & \text{if } x \neq 0, \\ 0 &\text{if } x=0\end{cases}$$

is defined and continous over $$\mathbb{R}^n$$ and $$g\circ f=f \circ g = I_{\mathbb{R}^n}$$, then $$g=f^{-1}$$ and f is a homeomorphism. Besides $$\|f(x)\|=\|x\|_{L^{\infty}}$$ and $$\|g(x)\|_{L^{\infty}}=\|x\|$$, so $$f(Q)=B$$

I need some help filling in the details . To prove f is an homeomorphism I have to prove it is continuous, bijective and open.

(1) continuity.

The relation $$1/\sqrt(n) \leq k(x) \leq 1$$ comes from the equivalence of the $$L^p$$ metrics:

I know $$\|x\|_{L^{\infty}} \leq \|x\|_{L^p} \leq n^{1/p}\|x\|_{L^{\infty}}$$ for $$p=2$$ and $$\|x\|_{L^2}=\|x\|$$ I get $$1/\sqrt(n) \leq k(x) \leq 1$$. Also, the expresion $$k(\alpha x)=k(x)$$ is straightforward, but then I can't figure out how to show $$\lim_{x \to O}f(x)=O$$

(2)injectivity For the injectivity I tried the ususal approach: $$f(x_1)=f(x_2)$$, so $$\frac{\|x_1\|_{L^{\infty}}x_1}{\|x_1\|} = \frac{\|x_2\|_{L^{\infty}}x_2}{\|x_2\|}$$ but then I don't know how to get to $$x_1=x_2$$

(3) surjectivity For the surjectivity I need to show the equation $$\frac{\|x\|_{L^{\infty}}x}{\|x\|}=y$$ has a solution for any $$y \in \mathbb{R}^n$$, but I can't figure out how to solve for $$x$$

(4)openness

I don't get how $$\|f(x)\|=\|x\|_{L^{\infty}}$$ and $$\|g(x)\|_{L^{\infty}}=\|x\|$$ manage to conclude $$f(Q)=B$$

I was trying to do like this: By definition of image $$f(Q)=\{y|f(x)=y, x \in Q\}$$ I take $$x \in Q \iff\|x\|_{L^{\infty}}<1 \iff$$ $$\|x\|_{L^{\infty}}= \|\frac{1}{k(x)}y\|_{L^{\infty}}=\frac{1}{k(x)}\|y\|_{L^{\infty}}<1$$ $$\implies \|y\|_{L^{\infty}} and I am getting nowhere, I should be getting to something like $$\|y\|<1$$ but instead I have the infinity norm there

Any help would be much appreciated

• What is $\varepsilon_n$ in $(\mathbb R^n,\varepsilon_n)$ in your problem statement? – kimchi lover Jul 7 '20 at 15:18
• @kimchi lover Its the usual euclidean topology – J.C.VegaO Jul 7 '20 at 15:20
• Thanks. Edit your post to say so. You have $\varepsilon_n, \|\cdot\|$ and $\|\cdot\|_{L^2}$, all meaning the same thing. May I suggest you replace them all with $\|\cdot\|_2$, and so on? – kimchi lover Jul 7 '20 at 15:21
• @kimchi lover all right, $\varepsilon_n$ is the euclidean topology. I think others denote like $\tau_e$, for other two you are right – J.C.VegaO Jul 7 '20 at 15:27
• @kimchi lover Actually , I did it on purpose to see the conection of $p$ with the inequality in which it appears as an exponent: $\|x\|_{L^{\infty}} \leq \|x\|_{L^p} \leq n^{1/p}\|x\|_{L^{\infty}}$, then I did put $\|x\|_{L^2}=\|x\|$, – J.C.VegaO Jul 7 '20 at 15:29

Continuity

$$\lim_{x\rightarrow 0}\|k(x)x\|=\lim_{x\rightarrow 0} \frac{\|x\| \ \|x\|_{L^\infty}}{\|x\|}=0$$

Injectivity and surjectivity

I would suggest the following aproach:

Prove that: $$x\in Q$$ implies that $$f(x)\in B$$ $$x\in B$$ implies that $$g(x)\in Q$$

Then prove that $$g$$ is the inverse of $$f$$, i.e. that $$f(g(x))=x$$ and $$g(f(x))=x$$.

This is enough to show that $$f$$ is a bijection from $$Q$$ to $$B$$.

• Not sure how $x\in Q$ implies that $f(x)\in B$ , $x\in B$ implies that $g(x)\in Q$ together wih having the inverse implies f is bijective – J.C.VegaO Jul 7 '20 at 17:51
• Functions with an inverse are bijective. math.stackexchange.com/q/1215365/15624 – Angela Pretorius Jul 8 '20 at 4:04
• ok, and what about these: " $x\in Q$ implies that $f(x)\in B$ , $x\in B$ implies that $g(x)\in Q$" what are they for? – J.C.VegaO Jul 8 '20 at 8:50