If $f $ is real valued; Let $A =\{x : f(x)>x\}.$ Prove that $\sup A \in A.$ Let $f$ be a real-valued function defined on $[0,1]$ such that $f(0)>0, f(x) \ne x ~\forall x,$ and $f(x) \leq f(y)$ whenever $x \le y.$ Let $A =\{x : f(x)>x\}.$ Prove that $\sup A \in A.$
Attempt: Let us suppose that $\sup A = a \notin A.$ Then $f(a) \leq a$. But, since, $f(x) \ne x \implies f(a) < a~~~~ ..........(1)$
Now, by the definition of supremum, every open ball $B(a,r) $ contains at least one element $y$ of $A$ . Then, $y \in B(a,r)$ .
Since, $y \in A \implies f(y) > y ~~......(2)$
$ y < a$ and $f$ is increasing $ \implies f(y) \leq f(a)~~~~~~........(3)$

From $(1),(2),(3) : y <f(y) \leq f(a)<a$

How do I bring some contradiction now?
Thank you for your help..
 A: Direct proof: for any $x\in A$, we have
$$
f(\sup A)\geq f(x)>x.
$$
The first inequality uses monotonicity and the second uses $x\in A$. This means that $f(\sup A)$ is an upperbound for $A$. In particular, $f(\sup A)\geq\sup A$. But $f(\sup A)\neq \sup A$, so we must have $f(\sup A)>\sup A$. That is, $\sup A\in A$.
A: From your work, you get $y < f(a) < a$ for all $y \in A$ sufficiently close to $a$ (i.e. in some ball $B(a,r)$). Taking supremums over $A$, we see that 
$$
a = \sup A = \sup_{y \in A} y \le \sup_{y \in A} f(a) \le \sup_{y \in A} a = a.
$$
Therefore $a = f(a)$, a contradiction by (1).
Hope that helps,
A: You want to use that there is such a $y$ for each $r > 0$, and that then $a - r < y_r$. Together with what you got you end up with:
$\forall r > 0 \ \left ( a - r < f(a) < a \right )$
and this yields the contradiction.
Edit: To see the latter claim, pick $r := a - f(a)$. As $f(a) < a$, this satifies $r > 0$. Then $a - r < f(a)$ evaluates to $f(a) < f(a)$, which is a contradiction.
