Prove that $f$ is continuous in $\mathbb{R}$

Question

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.

Answer 1

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 $|f(x)-f(x_0)| < \epsilon' \le \epsilon$.

Answer 2

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.