Brouwer's fixed-point theorem and the intermediate value theorem?

Question

In one dimension, Brouwer's fixed-point theorem (BPFT) can be proved easily based on the Intermediate Value Theorem (IVT). Is the inverse also true? I.e, is it possible to prove the IVT directly from the BFPT?

Answer 1

Yes. Let $f\colon [a,b]\to \mathbb R$ be continuous, and assume, without loss of generality, that $f(a)<f(b)$. Let $t\in \mathbb R$ with $f(a)<t<f(b)$. Assume that $t$ is never attained by $f$.

Construct $g\colon [a,b]\to \mathbb R$ by:

• $g(x)=a$ if $f(x)>t$;
• $g(x)=b$ if $f(x)<t$.

We don't have to define $g$ on points in which $f(x)=t$, since $f$ never equals $t$ in $[a,b]$.

Then $g$ is a continuous function whose image is contained in $[a,b]$. By the assumption on $t$, $g(a)=b$, and $g(b)=a$, and $g(x)\ne x$ for all $x\in (a,b)$ (by construction of $g$). In other words, $g$ has no fixed-point, contradicting Brouwer's fixed-point theorem.