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I came across this question when I tried to solve a math problem, and my solution was heavily based on this statement above. I find it very intuitive because you can't draw a continous function from one point of the plane to another (with different y's) without having to draw the line such that it is possible to find a very small interval on which the function is strictly increasing/decreasing. My question is the following: is this statement provable (and how the proof would unfold) or even true? I asked my math teacher whether there is a theorem about this topic, and he couldn't remember one.

Andrei Secuiu
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    Possible duplicate of Nowhere monotonic continuous function (googling for "continuous, nowhere monotone functions" gave me that as a result). – Arthur May 10 '19 at 10:45
  • At the opposite extreme there is a continuous surjection $f:[0,1]\to [0,1]$ such that (I): For any $m,n\in \Bbb N$ with $m\le 4^n$ there exists $p\in \Bbb N$ with $p\le 2^n$ such that $f$ maps $[(m-1)4^{-n},m4^{-n}]$ onto $[(p-1)2^{-n},p2^{-n}] $, and (II): For any $x\in [0,1]$ and any nbhd $U$ of $x,$ the set ${y\in U: f(y)=f(x)}$ is uncountable – DanielWainfleet May 10 '19 at 15:24
  • Perhaps a simpler example. We can represent the Koch Snowflake as the the image $f([0,1])$ of a continuous$ f:[0,1]\to \Bbb R^2$ with $f(0)=f(1)$ with $f$ not constant on any interval of positive length. Let $g(t)=x$ when $f(t)=(x,y). $ Then $g$ is continuous and nowhere monotonic. Extend $g$ to domain $\Bbb R$ by $g(t+n)=g(t)$ when $n\in \Bbb Z$ and $t\in [0,1].$... See Koch Snowflake in wikipedia. – DanielWainfleet May 10 '19 at 15:41

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