Proof of Breakpoint Reversal Sorting Approximation Algorithm
Denote the strips in your permutation as $\langle S_1, S_2, \cdots, S_n \rangle$. Suppose $k-1$ is in segment $S_i$, and $k$ is in segment $S_j$, for some $i, j \in [n], i \neq j$.

Recall by definition of $k$ that $k-1$ must be the maximum and therefore rightmost element in an increasing strip, and $k$ the minimum and therefore also rightmost in a decreasing strip. Consider the two cases:


$i &lt; j$. The increasing strip comes before the decreasing. $k-1$ is on the 'right' end of $S_i$, but $k$ is on the 'right' end of $S_j$ also. In order for $k-1$ and $k$ to be adjacent, reverse the segments $S_{i+1}, \cdots, S_j$.
$j &lt; i$. The decreasing strip comes before the increasing. $k$ is on the 'right' end of $S_j$, but $k-1$ is on the 'right' end of $S_i$ also. Again, we reverse $S_{i+1}, \cdots, S_j$.


So, "reversing the segment between $k$ and $k-1$" means to reverse all the strips in-between the two strips containing $k$ and $k-1$, excluding the 'leftmost' strip, but including the 'rightmost' strip.