37th Iranian Mathematical Olympiad

Let $S_1=\{1\}$ and for every $i\ge 1$, $$S_{i+1}=\{x\mid g(x)\in S_i\}-\{1\}.$$ Also, $$\bigcup _{i=1}^{\infty }{S}_i=A=\{x\mid \exists n\ge 0:g^n(x)=1\}.$$ Clearly $A$ is non-empty. If $[0,1]-A=B$ is also non-empty, partitions $A,B$ of $[0,1]$ will lead to a contradiction: If $x\in A$: $$\exists i:x\in S_i⟹g(x)\in S_{i-1}\subseteq A.$$ And if $g(x)\in A$: $$\exists i:g(x)\in S_i⟹x\in S_{i+1}\subseteq A.$$ Therefore, $B$ is non-empty. Now if $S_2$ has a finite number of elements, it has an element that is largest of all the elements. Let us denote it by $z$. Since $g(x)>x$, $$\max (x\mid x\in S_{i+1})\le \max (x\mid x\in S_i).$$ Therefore, all the elements of $A$ except $1$ are less than or equal to $z$. So, the non-empty interval $(z,1)$ doesn't exist in $A$, which is a contradiction. ■