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Brouwer Fixed Point Theorem

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New member
Oct 15, 2013
19
Brouwer Fixed Point Theorem: Every continuous function from the closed ball B^n= {x∈R^n: abs(x)<1}
to itself has a fixed point.

Is anyone can help me to Prove the Brouwer Fixed point theorem for n = 1 using the fact: there is no retraction from the closed interval [-1,1] onto
the two point set {-1,1}.

also Assume that there is no retraction from the closed ballB^n= {x∈R^n: abs(x)<1} onto the sphere
S^n-1= {x∈R^n: abs(x)=1} Prove the
Brouwer Fixed Point Theorem.

Thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Brouwer Fixed Point Theorem: Every continuous function from the closed ball B^n= {x∈R^n: abs(x)<1}
to itself has a fixed point.

Is anyone can help me to Prove the Brouwer Fixed point theorem for n = 1 using the fact: there is no retraction from the closed interval [-1,1] onto
the two point set {-1,1}.
Proof by contradiction: Assume that $f:[-1,1]\to [-1,1]$ has no fixed point. Define $g(x) = 0$ if $f(x)>x$ and $g(x)=1$ if $f(x)<x$. Show that $g$ is a retraction.

also Assume that there is no retraction from the closed ballB^n= {x∈R^n: abs(x)<1} onto the sphere
S^n-1= {x∈R^n: abs(x)=1} Prove the
Brouwer Fixed Point Theorem.
Similar idea: if $f:B^n\to B^n$ has no fixed point, then for each $x\in B^n$ there is a well-defined line starting from $f(x)$ and going in the direction of $x$. Let $g(x)$ be the point where this line hits $S^{n-1}$, and show that $g$ is a retraction.