Does there exist a non-continuous clopen function $g: \mathbb R \to \mathbb R$? What about $\mathbb R^n\to \mathbb R^m$?

Question

Inspired by Can an open and closed function be neither injective or surjective., but focusing on the case where $X,Y=\mathbb R$. First off, because the only nonempty clopen set in $\mathbb R$ is $\mathbb R$ itself, any clopen $g:\mathbb R\to \mathbb R$ must be surjective. Injectivity is a much more interesting question.

Imagining $g$ to be continuous and drawing some pictures one may be brought to the conclusion that such $g$ must be increasing/decreasing (i.e. no wiggles/peaks/valleys), or more rigorously:

1. Lemma 1: open maps $f: \mathbb R \to \mathbb R$ can not attain extreme values on open sets, like interiors of sets $S\subseteq \mathbb R$ (c.f. the maximum principle for harmonic functions).

   Proof: if the maximum value say $M$ is attained at $x\in S$, then the image $f(S)$ does not contain an open neighborhood of $M=f(x)$, contradicting that $o$ should be an open map.

2. Lemma 2: if an open map $f:\mathbb R \to \mathbb R$ attains its extreme values on every bounded closed interval $[a,c]\subseteq \mathbb R$ (or in fact $\subseteq I$ for any closed interval $I\subseteq \mathbb R$), then $f$ must be purely (weakly) increasing or purely (weakly) decreasing on $I$.

   Proof: otherwise, there are $a<b<c$ s.t. $f(a)<f(b)>f(c)$ (or instead of a peak, a valley), which implies that maximum of $f$ on $[a,c]$ is attained in the interior $(a,c)$ of $[a,b]$, contradicting Lemma 1 above.

In fact, we can upgrade to strictly increasing/decreasing:

3. Lemma 3: a open map $f:\mathbb R \to \mathbb R$ that is weakly monotone on a closed interval $I \subseteq \mathbb R$ is in fact strongly monotone on $I$.

   Proof: otherwise, supposing w.l.o.g. $f$ is increasing, we have $a<b<c$ s.t. $f(a)<f(b)=f(c)$ (or $f(a)=f(b)<f(c)$ or something similar), so we see that the maximum value $f(c)$ of $f$ on $[a,c]$ is attained at $b\in (a,c)$, the interior of $[a,c]$, again contradicting Lemma 1.

Finally, I argue that

4. Lemma 4: an open map $U:\mathbb R \to \mathbb R$ that is monotone on an open interval $U \subseteq \mathbb R$ must be continuous on $U$.

   Proof: suppose w.l.o.g. $f$ is increasing. If $f$ is not continuous at a point $x_0$, then we have $$\lim_{x\nearrow x_0} f(x) = \sup_{x < x_0} =:S < I:= \inf_{x>x_0} f(x) = \lim_{x \searrow x_0} f(x).$$ Then a neighborhood $(x_0-\delta, x_0 + \delta)$ of $x_0$ gets mapped to a set containing points $\leq S$ arbitrarily close to $S$, then nothing larger until $f(x_0)$, and nothing larger again until points $\geq I$ (arbitrarily close to $I$). And this set, no matter how nice things may look $\leq S$ and $\geq I$, and no matter where $f(x_0)$ lies in $[S,I]$, does not contain an open interval around $f(x_0)$; contradiction to openness of $f$.

So, since continuous maps attain extreme values on all compact subsets, we have that

• Corollary: if $U$ is an open interval, $f: U \to \mathbb R$ is an open map, then we have the following implications: [$f$ continuous on $U$] $\implies$ [$f$ attains its extreme values on every bounded closed interval $\subseteq U$] $\implies$ [$f$ is (weakly) monotone] $\implies$ [$f$ is strictly monotone (in particular injective)] $\implies$ [$f$ is continuous on $U$], so in fact all of those things are equivalent (in the context of $f$ being an open map on the open interval $U$)

Also, we can conclude that a focusing on continuous $g: U \to \mathbb R$ ($U$ an open interval), $g$ clopen implies $g$ is strictly monotone and surjective; and the converse is also true: a strictly monotone surjective continuous function $g:U\to \mathbb R$ is also a clopen map (this direction uses the fact that a bijective open map is also a closed map). Thus we have the following Characterization: the continuous clopen maps $g: \mathbb R \to \mathbb R$ are exactly the continuous strictly increasing surjective functions.

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Ok, now what about non-continuous clopen maps $g: \mathbb R \to \mathbb R$?

I know there are non-continuous open maps: this post Open maps which are not continuous gives the example of an "everywhere locally surjective" function (e.g. the Conway Base-13 function), where the image of every non-empty open set is the entirety of $\mathbb R$, which is indeed open.

We can slightly enlarge this class of open maps to functions that are everywhere locally surjective only on an open interval $U=(a,b) \subseteq \mathbb R$, and say a continuous strictly monotone surjective function on $(-\infty, a)$ and $(b, \infty)$. This class essentially contains the Cantor set based example in the comments of the above linked MSE thread.

Unfortunately, everywhere locally surjective maps $s:U \to \mathbb R$ are not closed: this is because I can choose $x\in U$, and then pick a point $x_n \in (x-\frac \delta n, x-\frac \delta{n+1})$ for all $n\in \mathbb N^+$ s.t. $s(x_n)=s(x)+1+\frac 1n$. Then $\{x_n: n\in \mathbb N^+ \} \cup \{x\}$ is a closed subset (infinite collection of points with one accumulation point $x$), but its image is the set $s(x)+(\{1+\frac 1n:n\in \mathbb N^+\}\cup\{0\})$ which is not closed.

So none of those examples help us.

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Here's what I've been able to prove so far. If clopen $g: \mathbb R \to \mathbb R$ is not continuous, then by the above Corollary, there must be some bounded closed interval $[a,b]$ on which $g$ does not attain its extreme values. Say w.l.o.g. doesn't attain its supremum $S$ on $[a,b]$. If $S<\infty$, then the image $g([a,b])$ has limit point $S$ but does not contain $S$, which contradicts closedness of $g$.

So, $g$ must be unbounded on $[a,b]$. So there exist $x_n\in [a,b]$ s.t. $g(x_n)\geq n$. We want to try to exploit this unboundedness, so let us "zoom in" on where it's happening: Bolzano-Weierstrass tells us there is a convergent subsequence of the $\{x_n\}$, converging to say to $x\in [a,b]$. We abuse notation and identify this subsequence as $\{x_n\}$. $%\text{Say w.l.o.g. that (a subsequence of) $\{x_n\}$ is approaching $x$ from the left.}$

$%\text{We know that $g$ is an open map, so it maps every open neighborhood $\subseteq \mathbb R$ of $x$ to an open set containing $g(x)$. So for every $\epsilon>0$, we know for every $\delta>0$ that there is $y_\delta^\epsilon \in (x-\delta, x+\delta)$ s.t. $|g(y_\delta^\epsilon)-g(x)|<\epsilon$.}$ We know every open neighborhood $(x-\delta,x+\delta)$ of $x$ maps to an open set $O_\delta:=g((x-\delta,x+\delta))$ containing the point $g(x)$, and is also unbounded (in the positive direction). We know what open subsets of $\mathbb R$ look like: $O_\delta$ is a union of countably many disjoint open intervals, $\bigsqcup_{n=1}^\infty (c_n^\delta, d_n^\delta)$. We know that $ g([x-\delta,x+\delta])$ is a closed set containing $O_\delta$, so in particular $g([x-\delta,x+\delta])$ is a closed set containing $\bigcup_{n=1}^\infty [c_n^\delta, d_n^\delta]$.

The fact that $O_\delta$ is a disjoint union of the $(c_n^\delta, d_n^\delta)$ implies that it does not contain any of the points $c_n^\delta, d_n^\delta$ for any $n\in \mathbb N$ (ignoring those that $=\pm\infty$). But the image $g([x-\delta,x+\delta])$ has at most 2 more points than $g((x-\delta, x+\delta))$! The only way this can happen is if $|\{c_n^\delta, d_n^\delta:n\in \mathbb N\}|\leq 2$ (again ignoring those that $=\pm\infty$), or in other words $O_\delta$ must be something like

• $(-\infty, \alpha_\delta) \cup (\beta_\delta, \infty)$ for $\alpha_\delta <\beta_\delta$ where either $g(x)<\alpha_\delta$ or $g(x)>\beta_\delta$; $\qquad \leadsto \overline{O_\delta} = (-\infty, \alpha_\delta] \cup [\beta_\delta, \infty)$, which respectively either contains $(-\infty, g(x)]$ or $[g(x),\infty)$.
• $(\ell_\delta, \alpha_\delta) \cup (\alpha_\delta, \infty)$ for $-\infty \leq \ell_\delta < \alpha_\delta$ with $\ell_\delta<g(x) \neq \alpha_\delta$; $\qquad \leadsto \overline{O_\delta} = [\ell_\delta, \infty) \supseteq [g(x),\infty)$
• or $(\ell_\delta,\infty)$ for $-\infty\leq \ell_\delta<g(x)$. $\qquad \leadsto \overline{O_\delta} = [\ell_\delta, \infty)\supseteq [g(x),\infty)$

In any case, we can define inductively an increasing sequence $z_n\nearrow x$ s.t. $g(z_n)= g(x)+1+\frac 1n$ or $g(z_n) = g(x)-1-\frac 1n$, so then $\{z_n:n\in \mathbb N^+\}\cup \{x\}$ is a closed subset of $\mathbb R$ that does not get mapped to a closed subset of $\mathbb R$; contradiction to closedness of $g$. Thus, there are NO non-continuous clopen functions $g:\mathbb R\to \mathbb R$. In other words, for maps $\mathbb R\to \mathbb R$, clopen $\iff$ homeomorphism.

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Is this proof correct? Is there a cleaner proof, such as one that does not utilize so heavily properties of $\mathbb R$, so as to generalize to higher dimension clopen maps $g:\mathbb R^n \to \mathbb R^m$? Or are there in fact non-continuous clopen functions in higher dimensions?