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:
\begin{equation} r(x)=x\;\mbox{ for all }\;x\in\partial{B}_1(0) \end{equation}
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.