5
$\begingroup$

Let $f: \mathbb{R}\to \mathbb{R}$ be strictly increasing function such that $f(f(f(x)))=x$ then by this $f(x)=x.$ Does the result fail if we drop the hypothesis that f is strictly increasing? I find such an example if I change the domain and codomain to $\mathbb{R}^2$ and $f=rotation \ by\ 120^o,$ but could not think of a map from $\mathbb{R}\to \mathbb{R}.$ Any help is appreciated.

$\endgroup$

4 Answers 4

10
$\begingroup$

Let $g:\mathbb{R^1}\to\mathbb{R^2}$ be a one-to-one mapping (which exists, because both sets have the same cardinality). Now set $f=g^{-1}\circ F\circ g:\mathbb{R}^1\to\mathbb{R}^1$, where $F$ is the rotation by $120^\circ$. So the result is not true: this is the desired counterexample.

$\endgroup$
2
  • 2
    $\begingroup$ Do you mean one-to-one and onto? Otherwise $g^{-1}$ doesn't make sense $\endgroup$
    – Exit path
    Commented Jan 30, 2019 at 20:27
  • $\begingroup$ @leibnewtz Yes, sure. $\endgroup$
    – Vladimir
    Commented Jan 30, 2019 at 20:30
4
$\begingroup$

Define $f(0)=1, f(1)=2, f(2)=0$ and $f(x)=x$ for $x \not\in \{0,1,2\}$.

$\endgroup$
3
$\begingroup$

You can make $f$ act as a 3-cyclic permutation

$ f(x) = \begin{cases} 0 & \text{if $x = -1$ } \\ 1 &\text{if $x = 0$} \\ -1 &\text{if $x = 1$} \\ x &\text{otherwise} \end{cases}$

I suppose it's also possible to make a continuous version of this.

$\endgroup$
3
  • 2
    $\begingroup$ Continuous is not possible, because then f would be increasing. $\endgroup$
    – user345777
    Commented Jan 30, 2019 at 20:29
  • $\begingroup$ Sorry, it was a stupid guess. Can I ask you why it would be increasing? $\endgroup$
    – CNS709
    Commented Jan 30, 2019 at 20:33
  • 1
    $\begingroup$ Agree with @user345777 $f$ is obviously a bijection. A continuous bijection from $\Bbb{R}$ to itself is either increasing or decreasing. The 3-fold iteration of a decreasing function is itself decreasing which is absurd. So it must increasing and that case has been handled. $\endgroup$ Commented Jan 30, 2019 at 20:34
2
$\begingroup$

If you allow yourself to consider the projective line (i.e., the reals with a point at infinity), then the function $$ f(x) = \frac{1}{1-x} $$ is a nice continuous solution to the problem (although you have to make sense of "continuous" for this extended line). Pierre Samuel's book on Projective Geometry has a nice exposition of this around page 58.

(By the way, this answer was inspired by @DonaldSplutterwit's deleted answer, although I might have stumbled on it myself, since I've been thinking about projective geometry and looking at Samuel's book today anyhow. Thanks, Donald!)

$\endgroup$
1
  • $\begingroup$ Then you could "cram $\infty$ into $\mathbb{R}$" using a variant of the infinite hotel paradox if you really wanted to, and then apply the same sort of construction as in Vladimir's solution. Of course, that won't result in a continuous function $\mathbb{R} \to \mathbb{R}$. $\endgroup$ Commented Jan 30, 2019 at 20:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .