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Let $f: [0,10) \to [0,10] $ be a continous map then

(a) $f$ need not have any fixed point

(b) $f$ has atleast $10$ fixed point

(c) $f$ has atleast $9$ fixed point

(d) $f$ has atleast one fixed point

Taking counterexample $f(x) = 1$; I can easily eliminate option (b) and (c) but I have no idea about first and last options.

Any hints will be helpful

Thank you!

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2 Answers 2

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Let $f(x)=10$ for $x\in [0, 10)$.

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Take $f(x)=5+x/2$. This will eliminate (d). So…

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  • $\begingroup$ Thanks for the answer! . Just one question : How did you came up with this counter example so quickly ? It will be helpful for me to know the thought process behind it so that I can use it solve further problems. $\endgroup$
    – user435638
    May 17, 2019 at 12:09
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    $\begingroup$ I drew (in my head) the graph of $x\mapsto x$ (with $x\in[0,10]$) and I thought about what should be the graph of $f(x)$ so that the two graps don't intersect. What I got was a straight line going from $(0,5)$ to $(10,10)$. $\endgroup$ May 17, 2019 at 12:33

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