Let $f:[0,10)\rightarrow [0,10]$ be a continuous mapping.Then

$(A)f$ need not have any fixed point.

$(B)f$ has atleast 10 fixed points.

$(C)f$ has atleast 9 fixed points.

$(D)f$ has atleast one fixed points.

WHAT I THINK- Since $[0,10)$ under usual metric is not complete.So,by Banach fixed point theorem $f$ need not have any fixed point i.e.,(A) is true.

Please give your advice/suggestions/opinions about my choice.

If there is any other method by which we can deal with this problem,please share...

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    $\begingroup$ Just to say that Banach fixed point theorem isn't really relevant here because we don't know that $f$ is a contraction map. $\endgroup$ – Teddy38 Sep 25 '17 at 8:02
  • $\begingroup$ @EdMo38:Gotcha!! $\endgroup$ – P.Styles Sep 25 '17 at 8:03
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    $\begingroup$ 10 mapping to 10 could be the fixed point but it doesnt exist in this mapping $\endgroup$ – user29418 Sep 25 '17 at 8:04
  • $\begingroup$ @user29418:thanks for giving words to my thought... $\endgroup$ – P.Styles Sep 25 '17 at 8:06

Just take $f(x)=10$ for all $x\in [0,10)$. As we drop $10$, there are no fixed points. This counterexample immediately proves that statement $A$ is true.

EDIT: Suppose the question was whether a continuous map $f:[0,10]\rightarrow [0,10]$ has fixed points, how could we solve it?

If $f(0)=0$ or $f(10)=10$ there are fixed points, so assume this is not the case. Consider the function $g(x)=f(x)-x$. Clearly $f(0)>0$ and $f(10)<0$. By the intermediate value theorem there must exist an $c\in (0,10)$ such that $g(c)=0$. Equivalently, $f(c)=c$.

So in this case you still don't need the machinery of Banach's fixed point theorem to solve the question.

  • $\begingroup$ :Can Banach fixed point theorem be applied here? $\endgroup$ – P.Styles Sep 25 '17 at 8:02
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    $\begingroup$ No, the metric space $[0,10)$ is not complete and we don't know whether $f$ is contracting. Also, as I gave a counterexample, it should be clear that the Banach fixed point theorem should fail here. $\endgroup$ – Mathematician 42 Sep 25 '17 at 8:07
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    $\begingroup$ @PKStyles Banach fixed point theorem cannot be applied because its aplication (if applicable) would tell us that $f4 has a fixed point, which it does not have $\endgroup$ – Hagen von Eitzen Sep 25 '17 at 8:08
  • $\begingroup$ I edited my answer to give an elegant solution to a related exercise. This also shows that you don't always need advanced machinery to solve problems. $\endgroup$ – Mathematician 42 Sep 25 '17 at 8:44

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