Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to prove that the open ball of radius 1 centered at the origin in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$. I believe the "shrinking map" from $\mathbb{R}^n$ to the ball given by $x \mapsto \dfrac{x}{1 + |x|}$ does the job, but I'm having trouble showing it's a homeomorphism, particularly the "continuous inverse" part. What's a good way to do this?

share|improve this question

1 Answer 1

up vote 13 down vote accepted

To figure out the inverse, note that if $|x|=\lambda$, then $$\left|\frac{x}{1+|x|}\right| = \frac{\lambda}{1+\lambda}.$$ Thus, given $y\in\mathbb{R}^n$ with $0\leq |y|\lt 1$, you want to find $\lambda$ such that $\lambda = (1+\lambda)|y|$. Letting $|y|=\mu$, we have $(1-\mu)\lambda = \mu$, or $\lambda = \frac{1}{1-\mu}$.

So the map you want for the inverse is $$y\longmapsto \frac{y}{1-|y|}.$$ Note that this is well-defined, since $0\leq |y|\lt 1$, so $0\lt 1-|y|\leq 1$. Also, the compositions are the identity: $$\begin{align*} x &\longmapsto \frac{x}{1+|x|}\\ &\longmapsto \left(\frac{1}{1- \frac{|x|}{1+|x|}}\right)\frac{x}{1+|x|} = \left(\frac{1+|x|}{1+|x|-|x|}\right)\frac{x}{1+|x|}\\ &= \vphantom{\frac{1}{x}}x.\\ y &\longmapsto \frac{y}{1-|y|}\\ &\longmapsto \left(\frac{1}{1 + \frac{|y|}{1-|y|}}\right)\frac{y}{1-|y|} = \left(\frac{1-|y|}{1-|y|+|y|}\right)\frac{y}{1-|y|}\\ &= \vphantom{\frac{1}{y}}y. \end{align*}$$ Now simply verify that both maps are continuous.

share|improve this answer
    
Thank you very much! –  DA1729 Feb 11 '12 at 22:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.