Let$ f : [0,1] \to [0,1] $ be a strictly increasing function. Show that it has a fixed point. Certainly for continuous function, it is true. What about this? We can narrow down it to the fact that $f(0)>0$ and $f(1)<1$. For some $x\in(0,1)$, $g(x) = f(x)-x$ has the property that $g(x)$ attains negative and positive value inside a small neighbourhood of $x$. 
 A: Assume that $f(0) > 0$ and $f(1) < 1$ (for otherwise we are done). Then
$$
x_0 := \sup\{x \in [0,1]:\ f(x) > x\}
$$
belongs to the open interval $(0,1)$.
Let us prove that $f(x_0) = x_0$.
Since, by definition of $x_0$, $f(x) \leq x$ for every $x\in (x_0, 1)$, by monotonicity it holds
$$
f(x_0) \leq f(x) \leq x, \qquad \forall x\in (x_0, 1),
$$
hence $f(x_0) \leq x_0$.
On the other hand, by definition of $\sup$, for every $\epsilon\in (0, x_0)$ there exists $x_\epsilon\in(x_0-\epsilon, x_0)$ such that $f(x_\epsilon) > x_\epsilon$, so that
$$
x_\epsilon < f(x_\epsilon) \leq f(x_0)
$$
and, finally, $x_0 \leq f(x_0)$.
A: Let $f:[0,1] \to [0,1]$ be increasing (but not necessarily continuous). Prove that there exists $x$ such that $f(x)=x$. 
Notice I do not assume strict increasing.
Proof:
Let 
$$A=\{t: f(t) > t \} \ .$$
If $A= \phi$, then we must have $f(0)=0$ and we are done. So, let $A \neq \phi$. Define
$$ x := \sup A \ .$$
Claim: $f(x)=x$.
1) $f(x) \geq x$: For any $a>0$ there exists a $t_n \in A$ such that
$$ x-a < t_n < x \ .$$
This forces
$$ x-a < t_n < f(t_n) \leq f(x) \ .$$
Since $a>0$ was arbitrary, we get
$$ x \leq f(x)  \ .$$
2) $f(x) \leq x $: If $x=1$ then this is obvious. Assume $x<1$. If we had $f(x) > x$ then for some $\delta > 0$, 
$$ f(x)=x+\delta \leq 1 \ .$$
Now, 
$$ f(x+\delta/2) \geq f(x) = x+ \delta > x+\delta/2 \ . $$
This shows $x+\delta/2 \in A$, which contradicts maximality of $x$. Therefore, we must have $f(x) \leq x$.
By 1) and 2) we get $f(x)=x$.
