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I search for an example of a function defined on $[0,1]$, strictly increasing on $A=\{\{n\sqrt{2}\}|~n\in \mathbb{Z}\}$, which is not monotone on any interval $I\subset [0,1]$

Maybe there is a simple example, but I cannot see it yet...

By a density result (of the set $A$ in $[0,1]$) it follows that such a function cannot be continuous on $[0,1]$.

JohnnyC
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    Here's a really dumb example that answers both your titular question and your question in the body (which shouldn't be different, so you should change one of them): $$f(x)=\begin{cases}1-x&x\in\mathbb Q\x&x\notin\mathbb Q\end{cases}$$ – Rushabh Mehta Mar 04 '21 at 17:33
  • Why is continuity a bad thing? And $A$ is not dense... – Rushabh Mehta Mar 04 '21 at 17:48
  • For any irrational $a$ the set ${{na}}$ is dense in $[0,1)$. – JohnnyC Mar 04 '21 at 18:29
  • That is false. As a counterexample, note that for any irrational $a$, $|a|>0$, so let $\varepsilon=\frac{|a|}5$. Thus a ball of radius $\varepsilon$ around each element of ${{na}}$ doesn't contain any other element. – Rushabh Mehta Mar 04 '21 at 18:57
  • https://math.stackexchange.com/questions/272545/multiples-of-an-irrational-number-forming-a-dense-subset – JohnnyC Mar 04 '21 at 19:20
  • Ah, you mean the fractional component. I see. – Rushabh Mehta Mar 04 '21 at 19:41

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