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I'm not convinced this is possible, as soon as you have $2$ distinct elements mapping to the same number, the function is no longer $1$-$1$ and therefore not a bijection.

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  • $\begingroup$ yes, it is possible, but it cannot be continuous. $\endgroup$
    – Masacroso
    Nov 27, 2017 at 17:15
  • $\begingroup$ Non monotone does not imply that two elements must map to the same target in general; you are using your intuition about continuous functions. A bijection here will not be continuous. $\endgroup$ Nov 27, 2017 at 17:15

3 Answers 3

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Let $f(x) = x$ for rational $x$ and $f(x) = 1-x$ for irrational $x$. This function is not only non-monotone, it is nowhere monotone (meaning that it is non-monotone on any subinterval of $[0,1]$).

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  • $\begingroup$ that's interesting, for some reason I was stuck on thinking it has to be continuous. thanks $\endgroup$
    – user480172
    Nov 27, 2017 at 17:21
  • $\begingroup$ Oh I see. If the function is continuous, then I believe it must be monotone as a consequence of the intermediate value theorem. $\endgroup$
    – User8128
    Nov 27, 2017 at 17:22
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How about the bijection $f:[0,1]\to [0,1]$ with $f(x)=\begin{cases} x, \text{if}~~ 0<x<1\\ 1,\text{if}~~ x=0\\ 0,\text{if}~~ x=1\end{cases}$

This is not strictly monoton.

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  • $\begingroup$ why is that not strictly monotone? looks like it to me $\endgroup$
    – user480172
    Nov 27, 2017 at 17:18
  • $\begingroup$ @user480172 Perhaps you should check carefully the meaning of "monotone", strictly or whatever. $\endgroup$
    – DonAntonio
    Nov 27, 2017 at 17:19
  • $\begingroup$ It is strictly monotone on $(0,1)$ but not on the whole intervall $[0,1]$, since $0<1$ but $f(0)>f(1)$. $\endgroup$
    – Cornman
    Nov 27, 2017 at 17:19
  • $\begingroup$ OH I get it nice thank you $\endgroup$
    – user480172
    Nov 27, 2017 at 17:20
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Remark that if a bijection is monotonous on $[0,1]$ you can build piecewise non-monotonous function based on the restriction of this bijection scaled to smaller intervals.

For instance $f:\begin{cases} x\in[0,\frac 14[ & 4x^2\\ x\in[\frac 14,\frac 12[ & \dfrac 34-x\\ x\in[\frac 12,\frac 34[ & x\\ x\in[\frac 34,1] & \dfrac{\cos(4\pi x-3\pi)+7}8\end{cases}$

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