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What would be a sequence of partitions of [0,1] to show that the function

$f(x)$ = \begin{cases} 0, & \text{if $x$ $\in$[0,1]$\cap$$Q^c$} \\ -x, & \text{if $x$$\in$[0,1]$\cap$$Q$} \end{cases}

is not of bounded variation on [0,1]?

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    $\begingroup$ Alternate between rational and irrational points. $\endgroup$ – Daniel Fischer Nov 26 '16 at 20:47
  • $\begingroup$ It is a variation at least $1/2$ between rational and irrational points for $x\ge 1/2$, and there are infintely many such changes in this part of the interval. $\endgroup$ – A.Γ. Nov 26 '16 at 20:49
  • $\begingroup$ @DanielFischer thank you. It works. $\endgroup$ – Janitha357 Nov 26 '16 at 21:19
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Try partition

$x_n = \frac{1}{n}$ , if $n$ is odd

$=\frac{1}{n+n^\frac{1}{n}}$ , if n is even.

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  • $\begingroup$ The site supports latex, write $5\cdot5$ and you get $5\cdot 5$. $\endgroup$ – peterh May 28 '18 at 13:49

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