Prove that $f$ is continuous in $\mathbb{R}$ Let  $f:\mathbb{R} \to \mathbb{R}$ be a monotone function such that $f(V)$ is open for every open set $V\subset \mathbb{R}$. Prove that $f$ is continuous in $\mathbb{R}$. 
Any hint for proving this I will appreciate. 
 A: Without loss of generality we may assume that $f$ is non decreasing.
First, note that $f$ is injective. Suppose $f(x_1) = f(x_2)$, then if $x_1<x_2$,
the set $(x_1,x_2)$ would be open and the set $f((x_1,x_2)) = \{ f(x_1) \}$ is not open which is a contradiction. Hence $f$ is injective.
Now I claim that $f(\mathbb{R})$ is connected (that is, an interval). Suppose that there is some $y_0 \in (\inf f, \sup f) \setminus f(\mathbb{R})$ (note 
that $f(\mathbb{R}) \subset (\inf f, \sup f)$).
Then $I_-=f^{-1}((-\infty, y_0)), I_+f^{-1}((y_0, \infty))$ form a partition of $\mathbb{R}$ into two intervals.
There are two possibilities: (i) $I_- $ has the form $I_- = (-\infty, x_0]$, with
$f(x_0) < y_0$  and $y_0 < f(x) $ for all $x > x_0$. In particular,
$f(x_0) \in f(\mathbb{R})$ but $(f(x_0), y_0) \cap f(\mathbb{R}) = \emptyset$,
and hence $f(\mathbb{R})$ is not open.
(ii) $I_+ $ has the form $I_+ = [x_0, \infty)$, and a similar analysis
leads to the same contradiction.
Hence $f(\mathbb{R}) = (\inf f, \sup f)$,
in particular, it is an interval.
Now pick $x_0$, $\epsilon >0$. Choose $0< \epsilon' \le \epsilon$ such that
$\inf f < f(x_0) - \epsilon'$ and $f(x_0)+ \epsilon' < \sup f$. Now
choose $x_-, x_+ $ such that $f(x_-) = f(x_0) - \epsilon'$ and
$f(x_+) = f(x_0) + \epsilon'$.
Note that $x_- < x_0 < x_+$ and let $\delta = \min (x_0-x_-, x_+-x_0)$.
Then if $|x-x_0| < \delta$, we have $|f(x)-f(x_0)| < \epsilon' \le \epsilon$.
A: Perhaps the contrapositive:
If $f$ is monotonic and not continuous on $\mathbb{R}$, then $\exists$ an open $V \subset \mathbb{R}$ such that $f(V)$ is not open. 
EDIT:
proof:
Suppose $f: \mathbb{R} \to \mathbb{R}$ is monotonic and not continuous at $p \in \mathbb{R}$.  Let $A = (p-\delta, p+\delta)$ for some $\delta \gt 0$. A is open in $\mathbb{R}$.  The domain of $f$ is $\mathbb{R}$. Then $\forall x \in A, f(x) \in f(A)\subset\mathbb{R}$.
Let $s = sup \{f(x)\in f(A)|x \in (p-\delta, p)\}$ and $t = inf\{f(x)\in f(A)|x \in (p, p+\delta)\}$
Without loss of generality, suppose $f$ is non-decreasing.  if $s = t, f(p) \not= s = t$, otherwise $f$ would be continuous at p.  This would imply, for some  $x_1 \lt p \lt x_2$, $f(p) \gt f(x_1) \land f(p) \gt f(x_2)$ or $f(p) \lt f(x_1) \land f(p) \lt f(x_2)$. Then $f$ is not monotonic.  Hence $s \not = t$.
Then $s \lt t$ and $s \le f(p) \lt t \lor s \lt f(p) \le t$.
Assume without loss of generality that $s \lt f(p) \le t$.  Then $s \lt f(p)$, which implies that $\exists y \in \mathbb{R}$, such that $s \lt y \lt f(p)$.  But $\forall x \in (p-\delta,p), f(x) \le s$.  Thus $y \notin f((p-\delta,p])$. Therefore $\forall \alpha$ such that $0 \lt \alpha \le f(p) - y,$ $\exists y^{'} \in \mathbb{R}$ such that $y^{'} \notin f[(p-\alpha,p+\alpha)]\subset f(A)$.
Thus $f(p) \in f(A)$ cannot be an interior point of $f(A)\subset \mathbb{R}$, which implies that $f(A)$ is not open.  Because $p \in A$ and $A$ is open, $A$ is an open subset of $\mathbb{R}$ such that $f(A)$ is not open.
