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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.

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    $\begingroup$ Yes, the homeomorphism can be extended to $[0,1]$, but that requires proof. For another way, what property distinguishes $0$ from the other points in $[0,1)$? (Thus every homeomorphism $[0,1) \to [0,1)$ must have $0$ as a fixed point. There may be others, but $0$ must be one.) For C, you could think about how the intermediate value theorem relates to the situation. $\endgroup$ – Daniel Fischer Dec 18 '15 at 10:43
  • $\begingroup$ A only is true if the function is surjective to me, maybe Im wrong. $\endgroup$ – Masacroso Dec 18 '15 at 11:14
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    $\begingroup$ Dear @Masacroso,A is always true. Consider $g(x)=f(x)-x$ and use Intermediate value theorem. $\endgroup$ – Dontknowanything Dec 18 '15 at 11:25
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    $\begingroup$ For part C, you can restrict the function from $[0,M]$ to $[0,M]$ where $M$ is an upper bound of the range. $\endgroup$ – Prince Kumar Dec 7 '16 at 11:20
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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.$

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HINT. The same proof based on the intermediate value theorem that you use to prove A works for D too. You don't even need $f$ to be bounded; any less-than-linear bound $f(x)\le Cx^\alpha,\ 0< \alpha < 1$ will suffice.

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