Real Analysis – Continuous Mapping f: [0,1] ? (0,1) CSIR December 2013

continuityreal-analysis

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.

Best Answer

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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