Open ball does not have fixed point

Question

How we can prove that the open ball in $R^n$ does not have fixed point property (by algebraic topology concepts)?
I know $D^n$ -closed ball in $R^n$- has fixed point property by Brouwer's theorem, but can't see how to find the function which has not a fixed point in the aforementioned case.

Would be grateful for your help.

How about the one-dimensional case? E.g. a fixed-point-free map from $(0,1)$ into itself? — hardmath, Jun 12 '16 at 08:06
In $\mathbb{R^n}$, $x \to \dfrac{x-1}{2}$. Basically a translation followed by a contraction. — Hmm., Jun 12 '16 at 08:16
@Soumya : The map you've defined has a fixed point in $\mathbb{R}^n$. — hardmath, Jun 12 '16 at 12:43
But it doesn't have a fixed point in $D^n$, which is what Brouwer would've promised. — Hmm., Jun 12 '16 at 12:45
Possible duplicate of Show that any non-empty, non-compact convex subset of $\mathbb{R}^n$ lacks the fixed point property — Moishe Kohan, Jun 13 '16 at 18:23

score 7 · Accepted Answer · answered Jun 12 '16 at 08:07

7

$\mathbb R^n$ does not have fixed point property (think of translation). The open ball in $\mathbb R^n$ is homeomorphic to $\mathbb R^n$, so it does not have fixed point property either.

answered Jun 12 '16 at 08:07

score 0 · Answer 2 · answered Jun 12 '16 at 08:14

0

Consider the open ball of radius $\frac 1 2$ around $(\frac 1 2, 0, 0, \ldots, 0)$, and the function $f$ which takes the $(n-1)$-dimensional cross-section at first coordinate $x$ to that of first coordinate $x^2$.

answered Jun 12 '16 at 08:14

David Schneider-Joseph

393

Excuse me I can't understand the $(n-1)$-dimensional cross-section; could you a little more explain? Thanks. – Aref Jun 12 '16 at 10:22

Open ball does not have fixed point

2 Answers2