Let $f$ be a continuous mapping from closed interval $[0,1]$ to itself. I need to prove that $f(x)=x$ for at least one $x \in [0,1]$. I want to do it by contradiction i.e. assuming that $f(x) \neq x \forall x \in [0,1]$ can we find an $x \in [0,1]$ such that there exist a sequence $x_n -> x$ but $f(x_n)$ does not tend to $f(x)$. I have no clue how to proceed. Some hint will be useful.
2 Answers
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Since $f$ is continuous and maps $[0,1]$ to $[0,1]$, you can apply the intermediate value theorem to the function $g$ defined by : $g(x) = f(x) - x$ for all $x \in [0,1]$. $g$ is continuous and $g(0)g(1) \leq 0$ by hypothesis.
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$\begingroup$ yes, this solution is nice. $\endgroup$– user21982Oct 4, 2013 at 13:34
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Hint: Using your approach, $f(0)\neq0$, thus $f(0)>0$. Similarly, $f(1)\neq1$, so $f(1)<1$. In particular, $f(0)-0>0$ while $f(1)-1<0$. Can you make your conclusion from here?
