Proving some statements only by the definition of Real numbers. Let $f \colon \ [0.1] \to \mathbb R$ is monotonically increasing function and $f(0)>0$ and $f(x)\neq x $ for all $x\in [0,1]$. $$A=\{x\in [0,1] : f(x)>x \}$$ 

We know: every non-empty subset of $\mathbb R$ bounded above has a least upper bound. 

A is non-empty subset of $\mathbb R$ because $0\in A$ and also $A$ is bounded so $\text{sup} A$ exists. Now I want to show that:


*

*$\text{sup}A\in A$ and consequently $\text{sup}A=\text{max}A$.

*$f(1)>1$


From 1 and 2 conclude that $\text{sup}A=\text{max}A=1$
I am suspicious to own solution.  I need the proof in details.
thanks
 A: Let $\sup(A)=a$ , since $A\subseteq [0,1] $ so $\sup(A)\in [0,1]$ so $a\in [0,1]$ now since f is increasing function and $f(0)>0$ so $0<f(0)\le f(a)\le 1 $. if $a=\sup(A)$ does not belong to $A$ then $f(a)\le a$ but $f(a)\neq a$ so $f(a)< a$ now hence: $$0<f(a)<a\le 1 $$
since $\sup(A)=a$ ,  definition of supermum says: there must exist $b\in A$ such that $b<a$ and $f(a)<b<a $ . because of  $b\in A$ we can say: $$ b<f(b)$$ 
in the other hand since $b<a$ so $f(b)\le f(a)$ so $ b<f(a) $ but it is contradiction with $f(a)<b<a $ and therefore :  $$\sup(A)\in A$$
for the second part, suppose $f(1)<1$. at the previous part we showed if $\sup(A)=a$ then $f(a)\in [0,1]$ so $f(f(a))$ is meaningful and since $1>f(1)\ge f(a)>a$ we have $$1>f(1)\ge f(f(a))>f(a)$$ so $f(a)\in A$ . So we conclude $\sup(A)=a<f(a)$ and $f(a)\in A$ but it is contradiction with $\sup(A)=a$ so we must have: $$f(1)>1$$ 
Now by first and second part we conclude that $1\in A$ and $\sup(A)=a\le 1$ so $$\sup(A)=a=1$$
A: This is the proof of first theorem:
Let's $\sup (A)=a$. If $A$ is finite then the proof is trivial. If $A$ is infinite we can choose the monotonically increasing sequence $\{x_n\}$ such that $x_n\in A$ and $x_n\rightarrow a$. And we have:
$$
f(a)\ge\lim_{n\rightarrow\infty}(f(x_n))\ge \lim_{n\rightarrow\infty}x_n=a
$$,
But $f(a)\ne a$, so $f(a)\gt a$. 
