# Open map which “almost fixes” the boundary of an open ball

We have a continuous function $f:\bar{B}\to\mathbb{R}^n$, where $\bar{B}=\{x\in\mathbb{R}^n:\|x\|\le 1\}$, such that if $\|x\|=1$ then $\|f(x)-x\|<\epsilon$, for a fixed $\epsilon\in(0,1)$. We have to prove that $B(0,1-\epsilon)\subseteq f(\bar{B})$.

This appears as a lemma in Rudin's Real and Complex Analysis. The author claims that it is possible to prove it without Brouwer's fixed point theorem, under the additional hypothesis that $f$ is open.

So far I've only observed that the problem reduces to showing that $f(\bar{B})\cap B(0,1-\epsilon)$ is not empty. Any ideas?

• Can you give us a more precise reference? Which lemma? – Siminore Aug 21 '12 at 15:19
• Lemma 7.23, page 151 (third edition). It is used in the proof of the change-of-variables theorem. – Mizar Aug 21 '12 at 15:38
• I see, I was reading the first edition. – Siminore Aug 21 '12 at 15:39
• If $f(\overline{B}) \subset \mathbb{R}^n \setminus B(0,1-\varepsilon)$, then you are essentially mapping a ball into a domain with a hole. I guess this is impossible, by comparing homologies... – Siminore Aug 22 '12 at 8:20
• Thank you, but I was looking for a solution as elementary as possible, just because Rudin seems to suggest its existence.. – Mizar Aug 22 '12 at 19:31

Assume that $f(\bar{B})∩B(0,1−ϵ)$ is empty. Since $f$ is open $f(B)$ is open. Since $f$ is continuous, $f(\bar{B})$ and the boundary of $f(\bar{B})$ (which is equal to the the boundary of it's complement) are compact sets. Since $0$ and $\infty$ are both in the complement of $f(\bar{B})$ it must be disconnected, which is a contradiction.