continuous monotonic function 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. 
 A: 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.
A: Hopefully my answer is simpler than the ones already there (in particular I only use the mean value theorem, not Bolzano-Weierstrass).
Let $X = (a,b)$ be an open interval. Let $f(x) \in f(X)$,  for $a < x < b$. Choose $u,v$ such that $a < u < x < v < b$ (note that $u,v \in X$). then by monotonicity, $f(u) < f(x) < f(v)$. By the MVT, the whole of $(f(u), f(v))$ is included in $f(X)$, and $f(x) \in (f(u), f(v))$, and therefore $f(x)$ is an interior point of $f(X)$.
A: Consider the open interval $(a,b)$. First note that $[a,b]$ is compact and connected, so $f[ [a,b]]$ is also compact and connected, and so must be a closed bounded interval (closed and bounded from compact and interval from connectedness), say $f[[a,b]] = [c,d]$. Now monotonicity implies that in fact $f[(a,b)] = (c,d)$ as well (the minimum goes to the minimum and the same for the maximum). 
A: As you have noticed $f(X)$ is an interval. To prove that it is open notice that given any $x\in X$ there exist $\varepsilon>0$ such that $[x-\varepsilon,x+\varepsilon]\subset X$. Since $f$ is continuous $f(X)$ contains $[f(x-\varepsilon),f(x+\varepsilon)]$ and since $f$ is strictly monotone this is a neighbourhood of $f(x)$ (e.g. if $f$ is strictly increasing $f(x-\varepsilon) < f(x) < f(x+\varepsilon)$).
