[proofplan] The equivalence between convergence in $\overline{\mathbb{R}}$ and finiteness of upcrossing counts is established by relating both conditions to the $\liminf$ and $\limsup$ of the [sequence](/page/Sequence). If the sequence converges, the $\liminf$ equals the $\limsup$, preventing infinitely many upcrossings of any interval $[a, b]$. Conversely, if the sequence does not converge, there exist rationals $a < b$ between the $\liminf$ and $\limsup$, and the sequence must cross $[a, b]$ infinitely often. [/proofplan] [step:Show convergence implies finite upcrossings for all rational $a < b$] Suppose $x_n \to L$ for some $L \in \overline{\mathbb{R}}$. Then $\liminf_{n} x_n = \limsup_{n} x_n = L$. If $N([a, b], x) = \infty$ for some rationals $a < b$, then the sequence drops below $a$ and rises above $b$ infinitely often. In particular, $\liminf_{n} x_n \leq a$ and $\limsup_{n} x_n \geq b$, giving $\liminf_{n} x_n \leq a < b \leq \limsup_{n} x_n$. This contradicts $\liminf = \limsup = L$. [/step] [step:Show finite upcrossings for all rational pairs implies convergence] Suppose $(x_n)$ does not converge in $\overline{\mathbb{R}}$. Then $\liminf_{n} x_n < \limsup_{n} x_n$ (the two may be in $\overline{\mathbb{R}}$, but they are distinct). Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exist rationals $a, b$ with $\liminf_{n} x_n < a < b < \limsup_{n} x_n$. The condition $\liminf_{n} x_n < a$ means $x_n < a$ for infinitely many $n$. The condition $\limsup_{n} x_n > b$ means $x_n > b$ for infinitely many $n$. Alternating between values below $a$ and above $b$ forces $N([a, b], x) = \infty$: after each time the sequence drops below $a$, it must eventually rise above $b$ (since $\limsup > b$), completing an upcrossing. This process repeats indefinitely. Taking the contrapositive: if $N([a, b], x) < \infty$ for all rationals $a < b$, then $(x_n)$ converges in $\overline{\mathbb{R}}$. [/step]