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Question is :

Suppose $f: [0,1]\rightarrow (0,1)$ is Continuous then which of the following is NOT true..

  • $F\subseteq[0,1]$ is closed set implies $f(F)$ is closed in $\mathbb{R}$
  • If $f(0)<f(1)$ then $f([0,1])$ must be equal to $[f(0),f(1)]$
  • There must exist $x\in(0,1)$ such that $f(x)=x$
  • $f([0,1])\neq (0,1)$

Continuous map need not map closed sets to closed sets..

So, first option is not true...

Continuous maps takes connected sets to connected sets ...

So $f([0,1])$ must be connected and it is equal to $[f(0),f(1)]$.. So, Second option is true..

Continuous maps takes compact sets to compact sets...

So, $f([0,1])\neq (0,1)$ and so fourth option is true...

I guess third option is also false though I can not think of any example..

Continuous map from compact set to itself has a fixed point.

But how do i conclude that this would imply third option is true/false.

Please help me to clear this...

Thank you.

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    $\begingroup$ In general continuous maps don't map closed sets to closed sets, but if $F \subseteq [0,1]$ it is compact, so $f(F)$ is compact and therefore closed. $\endgroup$ – Michael Albanese Dec 22 '13 at 11:00
  • $\begingroup$ When studying the third claim consider Bolzano and the function $g(x)=f(x)-x$. Together with Michael's comment this leaves option ... $\endgroup$ – Jyrki Lahtonen Dec 22 '13 at 11:02
  • $\begingroup$ The second option is not true, consider $f(x) = \frac14\bigl(1 + x(3/2-x)\bigr)$. $\endgroup$ – Daniel Fischer Dec 22 '13 at 11:02
  • $\begingroup$ @DanielFischer : I could not understand how did you come up with such an example very fast :O why would my justification is wrong :O Please explain a bit more.... $\endgroup$ – user87543 Dec 22 '13 at 11:04
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    $\begingroup$ @JyrkiLahtonen : Oh yes... $g(x)=f(x)-x$ implies $g(0)=f(0)>0$ and $g(1)=f(1)-1<0$ thus there exists a point in between $0$ and $1$ such that $g(a)=0$ i.e., $f(a)=a$ Done... Thank you... $\endgroup$ – user87543 Dec 22 '13 at 11:11
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Constructing the counterexample to (2) is easy. Draw a picture! Can you draw a squiggly line from $f(0)$ to $f(1)$ that goes below $f(0)$ say? That's your counterexample!

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  • $\begingroup$ I am sorry, I could not visualize your statement could you please explain a bit more... $\endgroup$ – user87543 Dec 22 '13 at 11:13
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    $\begingroup$ @PraphullaKoushik Do exactly as I have said. Take a pen out, draw the $x-y$ plane, draw $f(0)$ and $f(1)$ with $f(1) > f(0)$. Now draw a curve connecting the two points, and draw it so that some portion of the curve goes below the horizontal line $y = f(0)$. $\endgroup$ – user38268 Dec 22 '13 at 11:14
  • $\begingroup$ Oh... I got it :) Thank you.. That was wonderful... $\endgroup$ – user87543 Dec 22 '13 at 11:17
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To see that "There must exist $x\in(0,1)$ such that $f(x)=x$" observe that since $f$ is continuous and domain is a closed interval, $f$ is bounded.

Now consider the function

$g(x)=f(x)-x$. It is trivial to show that $g(x)$ is continuous. Now $g(0)=f(0)>0$ and $g(1)=f(1)-1 <0$. By intermediate value theorem there must exist a $c \in (0,1)$ such that $g(c)=0$ which gives the desired result.

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  • $\begingroup$ I have solved this in the comments.. Thank you anyways... $\endgroup$ – user87543 Dec 23 '13 at 3:19
  • $\begingroup$ may be you can like it $\endgroup$ – tattwamasi amrutam Dec 23 '13 at 3:20

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