Let $ A $ be a non-empty subset of $\mathbb R$ and $f : A \to \mathbb R$ be a continuous function on $A$ such that $f(V)$ is an open set for any open set $V$ , then how to prove that $f$ is monotone ?
Asked
Active
Viewed 299 times
1
-
I think you should assume that $A$ is connected. Consider $f(x) = x$ for $x ∈ (0, 1)$ and $f(x) = -x$ for $x ∈ (1, 2)$ otherwise. – user87690 May 30 '14 at 13:56
-
This will help http://wj32.org/wp/2013/01/15/every-continuous-open-mapping-of-r-into-r-is-monotonic/ – sas May 30 '14 at 13:56
2 Answers
3
One method:
Suppose $A$ is connected (otherwise, the statement isn't generally true).
If $f$ is continuous but not monotonic, there exists $a,b,c \in A$ with $[a,c] \subset A$ and $f(a) = f(c) < f(b)$ or $f(a) = f(c) > f(b)$ (verify that this must be the case). Show that $f((a,c))$ cannot be open (hint: why does $f$ attain a maximum/minimum on $(a,c)$?).
Thus, $f$ not monotonic $\implies$ $f(V)$ is not always open for open $V$. The conclusion follows.
Ben Grossmann
- 225,327
1
As you hace been said, we need to assume that $A$ is connected. Then, it is possible to assume WOLOG that $V$ is connected, hence, $f(V)$ is connected.
Suppose that $f(V)$ is not open. Then, there exists some $y\in f(V)$ (say, $f(x)=y$) such that for every $\epsilon>0$ there exists $y_\epsilon\in(y-\epsilon,y+\epsilon)$. Therefore, $\sup$ or $\inf f(V)=y$, that is, $f$ meets an extremus at $x$.
ajotatxe
- 65,084