0.1 Upcrossings
Let \((y_{n})\) be a sequence of real numbers and fix a finite interval \([a,b]\). We define \[ T_{1}=\min\{ n:y_{n}\leq a \}\quad \& \quad T_{2}=\min\{ n > T_{1}:y_{n}\geq b \}. \] The interval \([T_{1}, T_{2}]\) is called an upcrossing of \([a,b]\).
For \(k\geq 2\) we define \[ T_{2k-1}=\min\{ n>T_{2k-2}:y_{n}\leq a \}\quad\&\quad T_{2k}=\min\{ n>T_{2k-1}:y_{n}\geq b \}. \]
The number of upcrossings of \([a,b]\) be the finite sequence \((y_{0}, \dots, y_{n})\) is denoted \(U_{n}(a,b;y)\). The total number of upcrossings of \([a,b]\) or the infinite sequence \(y_n\) is defined as \[ U(a,b;y):=\lim_{ n \to \infty } U_{n}(a,b;y). \] Clearly one can define downcrossings similarly by reversing the inequalities and flipping \(a\) and \(b\) \[ T_{1}=\min\{ n:y_{n}\geq b \}\quad \& \quad T_{2}=\min\{ n > T_{1}:y_{n}\leq a \}. \]
If a sequence of real numbers \((y_{n})\) diverges, then there exists \(a<b\in \mathbb{Q}\) s.t. \(U(a,b;y)=\infty\).
\begin{proof} If a sequence \(\{ y_{n} \}\) diverges, then there exist rational numbers \(a<b\) and subsequences \(\{ y_{n_{k}} \}\) and \(\{ y_{m_{l}} \}\) such that \(\lim_{ k \to \infty }y_{n_{k}}<a<b<\lim_{ l \to \infty }y_{m_{l}}\). Then \(y_{n}\leq a\) for infinitely many indices \(n\) and \(y_{n}\geq b\) for infinitely many indices \(n\), that is \((y_{n})\) will cross the interval \([a,b]\) infinitely many times \(U(a,b;y)=\infty\).\end{proof}