0
$\begingroup$

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}.

  • 0
    I don't understand what the second part means, please edit the question to make it clear.2012-10-11
  • 0
    I edited the question, but which part isn't clear me too I can't prove that.2012-10-11
  • 0
    The set $\mathbb{N}^*$ should be the positive integers?2012-10-11
  • 0
    I think so. n>0 / n \in \mathbb {N}2012-10-11
  • 0
    Then you can let $a_n=1$ for all $n>1$. Since $1^n=1$ for all $n$, this works.2012-10-11

1 Answers 1