If $f:[a,b]\to[a,b]$ is increasing, continuous, and $f(a)=a$, how to prove $f(E)=E$ where $E=\{x:a\le x\le b,f(x)\ge x\}$? Let $f:[a,b]\to[a,b]$ satisfy:


*

*$f$ is   increasing


*$f$ is  continuous


*$f(a)=a$

If $E=\{x:a\le x\le b,f(x)\ge x\}$, then how can we prove that $f(E)=E$?
 A: Suppose that $x\in E$. Then $x\le f(x)$, and since $f$ is increasing, $f(x)\le f\big(f(x)\big)$ and hence $f(x)\in E$; this shows that $f[E]\subseteq E$. 
Now suppose that $x\in E\setminus f[E]$; clearly $f(x)>x$. The set $\{y\in[a,x]:f(y)=y\}$ is closed, so it has a largest element $u$. Then $f(u)=u<x<f(x)$, and $f$ is continuous, so there must be a point $y\in(u,x)$ such that $f(y)=x$. This is a contradiction, so $E\setminus f[E]=\varnothing$, and hence $f[E]=E$.
A: Consider the set $U=\{x\;;\; f(x)>x\}$. If it is empty, the statement is trivial since $E$ boils down to the fixed points of $f$. 
This is were the assumptions $f(a)=a$ and $f(b)\leq b$ come into play. Since $f$ is continuous $U$ is open in $[a,b]$, ie there is $V$ open in $\mathbb{R}$ such that
$$
U=V\cap [a,b].
$$
But since $a,b$ do not belong to $U$, we have 
$$
U=V\cap(a,b)
$$
open in $\mathbb{R}$.
So assume it is not empty and write it as a countable disjoint union of open intervals
$$
U=\{x\;;\; f(x)>x\}=\bigcup_{n\geq 1}(a_n,b_n).
$$
Now observe that $f(a_n+1/k)>a_n+1/k$ for all $k$  sufficiently large. So $f(a_n)\geq a_n$. And of course we do not have $f(a_n)>a_n$ so 
$$
f(a_n)=a_n\qquad\mbox{and}\qquad f(b_n)=b_n
$$
likewise.
So
$$
E=\bigcup_{n\geq 1}[a_n,b_n]\cup\{x\;;\;f(x)=x\}.
$$
Note that all interval bounds of the lhs set belong to the rhs set, so the union is not disjoint. But this is not a problem.
Since $f$ is increasing and continuous, the intermediate value theorem shows
$$
f([a_n,b_n])=[f(a_n),f(b_n)]=[a_n,b_n].
$$
Now 
$$
f(E)=f\left(\bigcup_{n\geq 1}[a_n,b_n] \cup\{x\;;\;f(x)=x\}\right)=\bigcup_{n\geq 1}f([a_n,b_n])\cup f(\{x\;;\;f(x)=x\})
$$
$$
=\bigcup_{n\geq 1}[a_n,b_n]\cup\{x\;;\;f(x)=x\}=E.
$$
A: For such a function, $f(E)$ may not be the same as $E$. For example, take $f(x) = x^2$ for all $x\in[1, 2]$. 


*

*$f$ is increasing on $[1, 2]$

*$f$ is certainly continuous

*$f(1) = 1$


Also note that $f(x) \geq x$ for all $x\in[1, 2]$, so $E = [1, 2]$. However, $f(E) = [1, 4]$. 
