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Suppose $f$ is a function as $f:[0,1]\to[0,1]$ and continuous on $[0,1]$. How can I prove that $\exists x_0 \in [0,1]$ such that $f(x_0)= x_0$. Also, how can I prove that $\forall n \in \mathbb {N} \ast \exists a_n \in [0,1] such that {a}{n} = {a}{n}^{n}.

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I don't understand what the second part means, please edit the question to make it clear. – Michael Greinecker Oct 11 '12 at 19:51
I edited the question, but which part isn't clear me too I can't prove that. – pourjour Oct 11 '12 at 20:04
The set $\mathbb{N}^*$ should be the positive integers? – Michael Greinecker Oct 11 '12 at 20:07
I think so. n>0 / n \in \mathbb {N} – pourjour Oct 11 '12 at 20:10
Then you can let $a_n=1$ for all $n>1$. Since $1^n=1$ for all $n$, this works. – Michael Greinecker Oct 11 '12 at 20:13

Show that the function $x\mapsto f(x)-x$ is continuous too, is nonpositve at one point, nonegative at one point and therefore zero at some point by the intermediate value theorem.

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