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]?

  • 1
    $\begingroup$ 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

Try partition

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.