Is there a function $f\colon \mathbb{R} \to\mathbb{R} $ such that $ f(f(x)) = -x$ ?

  • 2
    $\begingroup$ Answer to a generalization of this question here. $\endgroup$ – Did Jan 28 '12 at 16:12
  • 2
    $\begingroup$ An answer on MO demonstrating that there is no such continuous $f$. $\endgroup$ – Zev Chonoles Jan 28 '12 at 16:13

There is.

Let $\{A,B\}$ be a partition of the positive reals such that $|A|=|B|=|\mathbb{R}|$, and let $\varphi:A\to B$ be a bijection. Define $f:\mathbb{R}\to\mathbb{R}$ as follows:

$$f(x)=\begin{cases} 0,&\text{if }x=0\\ \varphi(x),&\text{if }x\in A\\ -\varphi^{-1}(x),&\text{if }x\in B\\ -\varphi(-x),&\text{if }-x\in A\\ \varphi^{-1}(-x),&\text{if }-x\in B\;. \end{cases}$$


$$\begin{align*} f(f(x))&=\begin{cases} 0,&\text{if }x=0\\ f(\varphi(x)),&\text{if }x\in A\\ f(-\varphi^{-1}(x)),&\text{if }x\in B\\ f(-\varphi(-x)),&\text{if }-x\in A\\ f(\varphi^{-1}(-x)),&\text{if }-x\in B\;. \end{cases}\\\\ &=\begin{cases} 0,&\text{if }x=0\\ -\varphi^{-1}(\varphi(x)),&\text{if }x\in A\\ -\varphi(\varphi^{-1}(x)),&\text{if }x\in B\\ \varphi^{-1}(\varphi(-x)),&\text{if }-x\in A\\ \varphi(\varphi^{-1}(-x)),&\text{if }-x\in B \end{cases}\\\\ &=-x\;. \end{align*}$$

The idea is simply that $f$ permutes the sets $A,B,-A$, and $-B$ in the order

$$A\stackrel{f}\longrightarrow B\stackrel{f}\longrightarrow -A\stackrel{f}\longrightarrow -B\stackrel{f}\longrightarrow A$$

while leaving $0$ fixed. (Here $-A= \{-a:a\in A\}$, and similarly for $-B$.)

Added: It is possible, though a bit messy, to define $A,B$ and $\varphi$ explicitly. We may, for example, set $A=(0,1]$ and $B=(1,\to)$ and define $\varphi$ as follows. First, for $n\in\omega$ let $\varphi(2^{-n})=2^{n+1}$, so that $\varphi(1)=2,\varphi(1/2)=4,\varphi(1/4)=8$, and so on. Then let $\varphi$ map the interval $(2^{-(n+1)},2^{-n})$ to the interval $(2^n,2^{n+1})$ in the obvious way, taking $x$ to $1/x$.

| cite | improve this answer | |
  • $\begingroup$ Your addendum looks very much like what is done in the post I mentioned. $\endgroup$ – Did Jan 28 '12 at 17:11
  • $\begingroup$ @Didier: I’ve not yet had a chance to look at that one. I was just trying to find a simple, explicit bijection between a half-closed and an open interval. $\endgroup$ – Brian M. Scott Jan 28 '12 at 17:30
  • $\begingroup$ More accurately, the idea on the page I referred you to (and here) is to exchange infinitely many disjoint intervals in a bijective way. $\endgroup$ – Did Jan 28 '12 at 18:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.