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Let $S\subseteq \mathbb{R}$. Consider the statement 

“There exists a continuous function $f:S\rightarrow S$ such that $f(x) \neq x$ for all $x \in S.$ ”

This statement is false if $S$ equals

  1. $[2,3]$
  2. $(2,3]$
  3. $[-3,-2] \cup [2,3]$
  4. $(-\infty,\infty)$
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2 Answers

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http://mathworld.wolfram.com/FixedPointTheorem.html

2<=x<=3 value represented only rationals.

so answer is (A)

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correct answer is option A, the complete explanation is provided in given link below , where the use of intermediate value theorem and fixed point property is used.

(A) is false, i.e. [2,3]-> [2,3] has the Fixed Point Property. Consider a continuous function

 

f:[2,3]→[2,3]

and define

 

g:[2,3]→R

g(x)=f(x)−x

It follows that g(2)≥0 and g(3)≤0. By the intermediate value theorem there is x0∈[2,3] such that g(x0)=0 . Thus f(x0)−x0=0 and so f(x0)=x0.

https://math.stackexchange.com/questions/2552345/is-there-exists-a-continuous-function-f-s-%E2%86%92-s-such-that-fx-%E2%89%A0-x-for-all-x-%E2%88%88-s

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