Question: Let I = $[0,1]$. Suppose $f$ is a continuous mapping of I into I. Prove that $f(x) = x$ for at least one $x\in I$.
Note: Define first($[A,B]$) = $A$ and second($[A,B]$) = $B$, where $[A,B]$ is an interval in R.
Proof: let M = $sup\ f(I)$ and m = $inf\ f(I)$. Assume that M $\neq$ m and that $f(x) \neq x$ for all $x\in I$.Then $f$ is strictly monotonic and so let $L_1$ = $[f(m),f(M)]$ and define $L_n = [f(first(L_{n-1}), f(second(L_{n-1}))]$ ($n\ge 2$). Then V=$ \bigcap L_n$ is non empty. Now let $y\in V$. Then there exists $x_1\in V$ such that $f(x_1) = y$. Then if $x_1 \neq y$, we have $f(x_1) \neq y$, which is a contradiction and so $x_1 = y$. But this is contradicting our assumption. For the case where M = m it is trivial as the function would be a constant one.
Is it possible to generalize this to $R^k$ or perhaps to metric spaces?
Consider $g: x \in [0,1] \mapsto f(x) - x$, $ \ g(0)≥ 0 \ ; g(1) ≤0$. $g$ is continuous so...