By Fixed Point Theorem, I know that it deals with closed interval, for example $[0,1]$. And this theorem will be false if $[0,1]$ is replaced by $(0,1)$. A counter-example is $f:(0,1)\rightarrow (0,1)$ and $f(x)=x/2$.
I have no idea to show the only intervals are the closed intervals. How to show that it is "ONLY"?
Best Answer
It suffices to analyse the cases $(0,1)$, $(0,1]$ and $[0,1]$, since any other interval (including the $\infty$ cases) is homeomorphic to one of those, and the "fixed point property" is invariant under homeomorphism.*
The case $[0,1]$ follows easily from the intermediate value theorem applied to $f(x)-x$.
To see that $(0,1)$ and $(0,1]$ do not satisfy it, take your function $f(x)=x/2$.
*To see this, let $A$ be a topological space with the FPP and $B$ an homeomorphic topological space, where the homeomorphism is $f:A \to B$. Let $g: B \to B$ be a continuous map. Then $f^{-1}gf$ is a continuous map from $A$ to $A$, hence has a fixed point $p$. Therefore, $f^{-1}(g(f(p)))=p \implies g(f(p))=f(p)$ and we have that $f(p)$ is a fixed point of $g$.