# To show that a function is not of bounded variation on [0,1]

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]?

• Alternate between rational and irrational points. – Daniel Fischer Nov 26 '16 at 20:47
• 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. – A.Γ. Nov 26 '16 at 20:49
• @DanielFischer thank you. It works. – Janitha357 Nov 26 '16 at 21:19

$x_n = \frac{1}{n}$ , if $n$ is odd
$=\frac{1}{n+n^\frac{1}{n}}$ , if n is even.
• The site supports latex, write $5\cdot5$ and you get $5\cdot 5$. – peterh May 28 '18 at 13:49