# Every continuous open mapping of R into R is monotonic

I’m willing to bet that most students who have used Rudin’s Principles of Mathematical Analysis have encountered this problem:

15. Call a mapping of $$X$$ into $$Y$$ open if $$f(V)$$ is an open set in $$Y$$ whenever $$V$$ is an open set in $$X$$.

Prove that every continuous open mapping of $$\mathbb{R}$$ into $$\mathbb{R}$$ is monotonic.

## Obvious?

Here is one “solution” that is fairly intuitive. It relies on finding a minimum or maximum and considering the image of a small neighborhood around that min/max:

Lemma 1. Let $$f:\mathbb{R}\to\mathbb{R}$$ be a continuous open map. Then there cannot be three points $$p_1 < p_2 < p_3$$ such that $$f(p_1),f(p_3) < f(p_2)$$ ($$\wedge$$ shape) or $$f(p_1),f(p_3) > f(p_2)$$ ($$\vee$$ shape).

Proof. Assume that the first case holds and let $$M=\sup f([p_1,p_3])$$; since $$f([p_1,p_3])$$ is compact, there is a point $$x\in[p_1,p_3]$$ such that $$f(x)=M$$. Since $$f(p_1),f(p_3) < f(p_2)$$, we must have $$x \in (p_1,p_3)$$. But then $$f((p_1,p_3))$$ is not open, which is a contradiction. $$\square$$

Now we can prove the main theorem.

“Proof” of main theorem. If $$f$$ is not monotonic, then there are three points as in Lemma 1. $$\square$$

The problem with this “proof” is the assumption: surely we must have a $$\wedge$$ shape or a $$\vee$$ shape, but how do we really prove its existence? Negating the definition of “monotonic” merely gives us the existence of four points $$a,b,c,d \in \mathbb{R}$$ satisfying the following: \begin{align}
a < b &\quad\mathrm{and}\quad f(a) < f(b), \\ c < d &\quad\mathrm{and}\quad f(c) > f(d).
\end{align}

Then we use a purely combinatorial argument by a math.SE user:

If $$a=c$$, use $$adb$$ to form $$\vee$$ or $$abd$$ to form $$\wedge$$.

If $$a=d$$, use $$cdb$$ to form $$\vee$$.

If $$b=c$$, use $$abd$$ to form $$\wedge$$.

If $$b=d$$, use $$acd$$ to form $$\wedge$$ or $$cab$$ to form $$\vee$$.

Now we are left with the case when $$a,b,c,d$$ are pairwise distinct.

If $$a < b < c < d$$, use $$abd$$ to form $$\wedge$$ or $$acd$$ to form $$\wedge$$. If $$a < c < b < d$$, use $$abd$$ to from $$\wedge$$ or $$acd$$ to form $$\wedge$$. If $$a < c < d < b$$, use $$acd$$ to form $$\wedge$$ or $$cdb$$ to form $$\vee$$. If $$c < a < d < b$$, use $$cdb$$ to form $$\vee$$ or $$cab$$ to form $$\vee$$. If $$c < d < a < b$$, use $$cdb$$ to form $$\vee$$ or $$dab$$ to form $$\vee$$.

## A different approach

Such a case-by-case analysis is not particularly enlightening, so is there a different approach? This is the best I could come up with when I was attacking the problem:

We first start with a lemma. Call a function $$f:\mathbb{R}\to\mathbb{R}$$ locally monotonically increasing/decreasing if for every $$x \in \mathbb{R}$$, there is a neighborhood of $$x$$ on which $$f$$ is monotonically increasing/decreasing.

Lemma 2. A continuous open map $$f:\mathbb{R}\to\mathbb{R}$$ is monotonically increasing if it is locally monotonically increasing. The same applies if “increasing” is replaced with “decreasing”.

Proof. If $$f$$ is not monotonically increasing then we can find points $$x < y$$ such that $$f(x) > f(y)$$. Choose a neighborhood $$U \subseteq (-\infty,y)$$ of $$x$$ on which $$f$$ is monotonically increasing. $$f$$ cannot be constant on $$U$$, for otherwise $$f(U)$$ would not be open. So there is a point $$x’ \in U$$ such that $$f(x’) \ne f(x)$$. If $$x < x'$$ then $$f(x) < f(x')$$ and we can form a $$\wedge$$ shape, and if $$x' < x$$ then $$f(x') < f(x)$$ and we can again form a $$\wedge$$ shape. Both cases contradict Lemma 1. $$\square$$

Using this, we have:

Proof of main theorem. Let $$f:\mathbb{R}\to\mathbb{R}$$ be a continuous open map. Let $$A$$ be the set of all points $$x$$ such that $$f$$ is monotonically increasing in a neighborhood of $$x$$. Define $$B$$ similarly, but with “monotonically decreasing” instead. These two sets are clearly open. They are disjoint, for if $$x \in A \cap B$$ then there is a neighborhood $$U$$ of $$x$$ on which $$f$$ is both monotonically increasing and decreasing, i.e. constant; this is a contradiction since $$f(U)$$ is not open. If we can show that $$A \cup B = \mathbb{R}$$ then by the connectedness of $$\mathbb{R}$$ either $$A=\mathbb{R}$$ and $$B=\emptyset$$ or $$A=\emptyset$$ and $$B=\mathbb{R}$$, in which case we can apply Lemma 2 to conclude that $$f$$ is monotonically increasing or decreasing. Suppose $$x \in \mathbb{R} \setminus (A \cup B)$$. If there exists a point $$y > x$$ with $$f(y) > f(x)$$ then since $$f$$ is continuous, there is a neighborhood $$U$$ of $$x$$ such that $$f(U) \subseteq (-\infty,f(y))$$. But $$f$$ fails to be monotonically increasing on $$U$$, so we have a $$\vee$$ shape that contradicts Lemma 1. By a similar argument, there cannot be a point $$y > x$$ with $$f(y) < f(x)$$. Therefore $$f$$ is constant on $$(x,\infty)$$, which contradicts the fact that $$f$$ is open. $$\square$$

## An exercise

Prove Lemma 2 without assuming that $$f$$ is continuous or open. Hint: if $$x < y$$ then $$[x,y]$$ is compact.

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