Continuous mapping $f: [0,1]\rightarrow (0,1)$ CSIR December $2013$

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.

Solution 1:

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!

Solution 2:

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.