Which of the following is false $?$
$A.$ Any continuous function from $[0,1]$ to $[0,1]$ has a fixed point.
$B.$ Any homeomorphism from $[0,1)$ to $[0,1)$ has a fixed point.
$C.$ Any bounded continuous function from $[0,\infty)$ to $[0,\infty)$ has a fixed point .
$D.$ Any continuous function from $(0,1)$ to $(0,1)$ has a fixed point.
Now , if we take $f(x)=x^2$ , then $f((),1))\subset (0,1)$ but it does not show any fixed point. So, option $D$ is our false statement .
And option $A$ is a well known result .
So, turns out that , the statements in option $B$ and $C$ are both correct. I need help to prove them.
For $B$ , my thought is that the said homeomorphism can be extended to $[0,1]$ and the fixed point theorem will apply but we will need to show that the fixed point is not the point $x=1$. Is that right $?$ But I don't know $C$ .
Thanks for any help.
Best Answer
For B consider that $0$ is the only point $p\in [0,1)$ such that $[0,1)\backslash \{p\}$ is a connected space, so a homeomorphism $f$ must have $f(0)=0.$ Another way is that $B=\{[0,d):0<d<1\}$ is a neighborhood base at $ 0$ such that $\forall b\in B\; (\bar b\backslash b$ has just one member.) No other point in $[0,1)$ has a nbhd base with this property, so $f(0)=0$.....For C if $f(0)=0$ we are done. If $f(0)>0$ the continuous function $g(x)=f(x)-x$ is positive at $x=0$ and negative when $x>\sup \{f(y) :y\geq 0\}$ so for some $x$ we have $ g(x)=0.$