# Brouwer fixed point theorem and completeness axiom.

I want to show that the completeness axiom of the real numbers is equivalent to the Brouwer fixed point theorem (in $\mathbb{R}^2$), i.e., without loss of generality:

For all nonempty set $A\subset (0,1)$, such that $A$ is bounded above implies that $A$ has supreum iff any continuos mapping $f:\overline{B_{1}(0)}\rightarrow\overline{B_{1}(0)}$ has at least one fixed point.

First of all, the "Brouwer fixed point theorem" is equivalent to "there is no continuous mapping $r:\overline{B_{1}(0)}\rightarrow \partial{B}_1(0)$" such that:

$$r(x)=x\;\mbox{ for all }\;x\in\partial{B}_1(0)$$

Now, suppose that $A\subset (0,1)$ has no supremum, then:

$$r(x)=\left\{\begin{array}{l|l} \frac{x}{|x|}\hspace{1cm}\mbox{ if }|x| \mbox{ is an upper bound of A}\\ (1,0) \hspace{1cm}\mbox{ if }|x| \mbox{ isn't an upper bound of A}\end{array}\right.$$

It's easy verify that r is continuos and $r(x)=x$ for all $x\in\partial{B}_{1}(0)$. I don't know how to show the other implication. ¿Can someone give me an idea?

Thank you.

• You could look at the traditional proofs of Brouwer fixed point theorem. Invariably, they will relate to some consequence of the supremum axiom. – Theo Bendit Nov 14 '17 at 5:51
• Your $r$ doesnt look to be continuous. – user99914 Nov 14 '17 at 5:51
• JohnMa , Why?, I think that it is continuos. Because of definition of supremum. – Andrés Felipe Nov 14 '17 at 6:00
• @AndrésFelipe Let say $A = (0,1/2)$. Then at the circle $|x| = 1/2$, $r(x)$ is non-constant, but $r(x)$ is constant on $|x|<1/2$. – user99914 Nov 14 '17 at 6:14
• To be clear, you are asking how to prove the Brouwer fixed point theorem assuming the completeness axiom? The completeness axiom implies your ordered field is isomorphic to the usual real numbers, so this just means you want to prove the ordinary Brouwer fixed point theorem... – Eric Wofsey Nov 14 '17 at 6:40