How to prove that strictly monotonic continuous function carries open intervals to open intervals?
if $f:\mathbb{R}\to \mathbb{R}$ continuous and monotonic, we need to to Prove that If $X \subset$ R is an open interval then $f(X)$ is an open interval.
Clearly $f(X)$ is an interval(as continuous functions takes connected sets to connected sets and connected sets in $\mathbb{R}$ is an interval ) we have to prove that it is open.
i am not able to use the monotonicity to prove an arbitrary point is an interior point.
Best Answer
I give the proof for the maximum, and the minimum is likewise. suppose the domain is(a,b), and the range is $X$.
i.e. $$x\in(a,b)$$.
without loss of generality, we suppose the function is strictly monotonic increasing
Let M denote the supremum of the range of f(x), $$f(x)\le M$$ for $x\in (a,b)$.
Now, we prove M cannot be reached. we demonstrate a proof by contradiction.
Suppose not, then then there is a $x_0\in (a,b)$, such that $$ f(x_0)=M$$
since $x_0<b$, we can find a $x'$ such that, $$x_0<x'<b$$. The function is strictly increasing(as supposed above).
then $$M=f(x_0)<f(x')$$
so this means that $M$ is not the supremum of f(x), which is a contradiction.So the supremum M cannot be reached for $x\in (a,b)$. we finish our proof.