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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?

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    Can you give us a more precise reference? Which lemma?2012-08-21
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    Lemma 7.23, page 151 (third edition). It is used in the proof of the change-of-variables theorem.2012-08-21
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    I see, I was reading the first edition.2012-08-21
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    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...2012-08-22
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    Thank you, but I was looking for a solution as elementary as possible, just because Rudin seems to suggest its existence..2012-08-22
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    In a different edition of Rudin, he assumes that $\|f(x)-x\| < \epsilon$ for all $\|x\| \le 1$, and then (maybe assuming $\epsilon < 1/2$ or something) it is indeed straightforward to prove. And if you try to prove the change-of-variables theorem, you should get this approximation by linear maps throughout the whole ball, not just on the boundary.2012-11-21
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    I'm thinking that there doesn't even exist a continuous and open map from $\bar B$ to $\mathbb R^n$. $\bar B$ is an open subset of itself, so if $f \colon \bar B \to \mathbb R^n$ were continuous and open, $f(\bar B)$ should be a nonempty, open, compact subset of $\mathbb R^n$. That's an impossible combination of attributes.2014-05-13
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    Never mind, guess I took it too literally. In the context of Theorem 7.24, it seems the domain of $f$ should actually be an open set in $\mathbb R^n$ containing $\bar B$.2014-05-13
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    The function $f(x)=(7\|x\|-6\|x\|^2)x$ shows the condition $\|x\|=1$ is as you say not enough.2015-01-30

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