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If $f\colon (a,b)\rightarrow \mathbb{R}$ is continuous, then is it always possible to find a countable collection $\{J_i\}$ of sub-intervals of $(a,b)$ (which could be closed-open, open-closed, closed or open) such that

  • $J_i'$s cover $(a,b)$ except countably many points;

  • $f$ is monotonic on each $J_i$?

(I came to this question by considering the $\sin(\frac{1}{x})$ function on $(0,1)$.)

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    $\begingroup$ No. The Weierstrass nowhere differentiable function is monotonic on no interval, e.g.. $\endgroup$ – David Mitra Nov 19 '14 at 9:18
  • $\begingroup$ @DavidMitra I think what you wrote is acrually a full answer to the question, not just a comment. Perhaps post it as such? $\endgroup$ – 5xum Nov 19 '14 at 9:19
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No. The Weierstrass nowhere differentiable function is monotonic on no interval, e.g..

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  • $\begingroup$ Oh! Thanks. This example is important in other parts of analysis also. Thanks for the nice solution. $\endgroup$ – Groups Nov 19 '14 at 11:24

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