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In one of the remarks for this highly upvoted unanswered question: Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not? , the author points out in the post that

a bijection that preserves connectedness on $\mathbf{R}$ must be monotone.

Why is this true?

I understand every single word in this statement, but I do not know how to prove it. To set up the notation,

Let $f:\mathbf{R}\to\mathbf{R}$ be a bijection such that for any connected subset $A$ in $\mathbf{R}$, the set $f(A)$ is also connected. How does one show that $f$ must be monotone?

To get a feeling for what could go wrong if $f$ is not monotone, I consider the simple case when $f(x)=x^2$. Obviously, $f(A)$ is connected for any connected set $A$ since $f$ is continuous; but it is not bijective. Other than this dumb example, I don't have any intuitions.

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    Not that it's not just $f(\mathbb{R})$ that must be connected, but $f$ applied to any connected subset of $\mathbb{R}$. – Chessanator Dec 02 '21 at 18:02
  • @Chessanator: ah, sure, thanks, I will edit the post. –  Dec 02 '21 at 18:04
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    The remark says: $f:\mathbb R\rightarrow \mathbb R$ bijective and connected $\implies$ $f$ monotone. It does not make any statement about $x^2$ because it is not bijective. – Andreas Tsevas Dec 02 '21 at 18:08
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    @Andreas: The contrapositive says something about functions that are not monotone. That's why I looked at that dumb example. –  Dec 02 '21 at 18:22

2 Answers2

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Suppose that $f$ is not monotonic. Then there are numbers $a,b,c\in\Bbb R$ such that $a<b<c$ and that $f(b)$ is greater than both $f(a)$ and $f(c)$ or that $f(b)$ is smaller than both $f(a)$ and $f(c)$. Suppose that we are in the first case. You have $f(a)>f(c)$ or $f(a)<f(c)$ or $f(a)=f(c)$. In this last case, $f$ is not injective, and we're done. If $f(a)>f(c)$, then, $f([b,c])$ is not an interval, since it contains $f(b)$ and $f(c)$, but not $f(a)$. And if $f(a)<f(c)$, then, $f([a,b])$ is not an interval, since it contains $f(a)$ and $f(b)$, but not $f(c)$.

The other case is similar.

José Carlos Santos
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  • It's obvious of course, but might we add that if $f$ is not continuous at a point $x$, say, then the image is not connected. – Robbie Dec 02 '21 at 18:10
  • @Robbie Why not? If, say,$$f(x)=\begin{cases}x\sin\left(\frac1x\right)&\text{ if }x\ne0\0&\text{ otherwise,}\end{cases}$$then $f$ is discontinuous at $0$, but, for any interval $I$ of $\Bbb R$, $f(I)$ is connected. – José Carlos Santos Dec 02 '21 at 18:12
  • I meant given $f$ is a bijection, sorry. We do need continuity of $f$ for your proof, that's all I meant. – Robbie Dec 02 '21 at 18:15
  • In order to apply the intermediate value theorem, do you need the extra assumption that $f$ is continuous? –  Dec 02 '21 at 18:26
  • @ripples I've edited my answer. What do you think now? – José Carlos Santos Dec 02 '21 at 18:41
  • @JoséCarlosSantos: In the first case where $f(b)>\max(f(a),f(c))$, and the first sub-case where $f(a)>f(c)$, one has $f(c)<f(a)<(b)$; how do you conclude that $f([b,c])$ does not contain $f(a)$? –  Dec 02 '21 at 20:31
  • Because $f$ is bijective; in particular, it's injective. So, since $a\notin[b,c]$, $f(a)\notin f([b,c])$. – José Carlos Santos Dec 02 '21 at 20:58
  • I don't see why being injective implies that. If $S$ is a finite set, I can see that $a\not\in S$ implies $f(a)\not\in f(S)$. But $[b,c]$ is an infinite set. I don't see why $a\not\in [b,c]$ implies $f(a)\not\in f([b,c])$. Could you elaborate? –  Dec 02 '21 at 21:05
  • I don't see what being finite has to do with this. If $f(a)\in f([b,c])$, then $f(a)=f(a')$ for some $a'\in[b,c]$. But $f(a)=f(a')\implies a=a'$, which is impossible, since $a\notin[b,c]$, whereas $a'\in[b,c]$. – José Carlos Santos Dec 02 '21 at 21:09
  • Okay. Thanks. I was thinking in terms of cardinality, which is not needed at all. +1 for your answer. I will wait and see if there will any inputs from others. –  Dec 02 '21 at 21:30
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Here is a slightly different presentation of the José's nice answer; which I find easier to digest for myself.

Let us prove by contradiction and assume that $f$ is not monotonic. We first prove the following lemma.

Lemma. If $f:\mathbf{R}\to\mathbf{R}$ is not monotonic, then there exists real numbers $a,b,c$ with $a<b<c$ and $$ (1)\quad f(b)> \max(f(a),f(c))\qquad \text{or}\qquad (2)\quad f(b)> \min(f(a),f(c)) $$

For case (1), there are three sub-cases:

(1.1) $f(a)>f(c)$. Then $f(c)<f(a)<f(b)$. Let $I=f([b,c])$. Since $f$ preserves connectedness, the set $I$ must be an interval and thus $[f(c),f(b)]\subset I$. It follows that $f(a)\in I$. But $f$ is bijective, so $a\in f^{-1}(I)=[b,c]$, which contradicts the assumption that $a<b<c$.

(1.2) $f(a)<f(c)$. Then $f(a)<f(c)<f(b)$. We can then argue similarly as in (1.1) and consider $I=f([a,b])$. We can then get the contradiction that $c\in[a,b]$.

(1.3) $f(a)=f(c)$. This is impossible since $f$ is bijective and particularly injective.

One can discuss (2) similarly.

Proof of the lemma. If $f$ is monotonic, then for any $a,b,c$ with $a<b<c$ one must have

  • either $f(b)\in[f(a),f(c)]$, if $f$ is increasing;
  • or $f(b)\in[f(c),f(a)]$, if $f$ is decreasing.

(1) and (2) are nothing but the negation of these two cases.