$\textbf{The Problem:}$ Show that the function $F(x)=\cos(x^{-1})\mathbf1_{x>0}$ is not of bounded variation on $[0,1].$
$\textbf{My Thoughts:}$ I play the game of whipping up the right partition to make the sum blow up at infinity. I will use the partition given by the points where $F$ achieves its local maxima and minima, of which there are infinitely many in $[0,1].$ Mainly I will use the partition $$\mathcal{P}=\{t_0=1,t_1=\pi^{-1},t_2=(2\pi)^{-1},\dots,t_{N-2}=[(N-1)\pi]^{-1},t_{N-1}=(N\pi)^{-1},t_N=0\}.$$ Then the variation of $F$ over the partition $\mathcal P$ is given by \begin{align*}\sum^{N}_{k=1}\vert F(t_k)-F(t_{k-1})\vert&= \sum^{N}_{k=1}\vert \cos(t_k)-\cos(t_{k-1})\vert\\ &=1+\cos(1)+\sum^{N-1}_{k=2}2\\ &\overset{N\to\infty}{\longrightarrow}\infty \end{align*} so $F$ cannot be of bounded variation on the interval $[0,1].$
Could anyone please verify if my proof is correct? Any comments are welcomed, be it about the style, lack of details, and most important, the accuracy of the ideas.
Thank you for your time, and appreciate any feedback.