# proof of convergence of a sequence with finite upcrossings

We show that a sequence $x_{1},x_{2},\ldots$ of real numbers converges to a limit in the extended real numbers if and only if the number of upcrossings $U[a,b]$ is finite for all $a.

Denoting the infimum limit and supremum limit by

 $l=\liminf_{n\rightarrow\infty}x_{n},\ u=\limsup_{n\rightarrow\infty}x_{n},$

then $l\leq u$ and the sequence converges to a limit if and only if $l=u$.

We first show that if the sequence converges then $U[a,b]$ is finite for $a. If $l>a$ then there is an $N$ such that $x_{n}>a$ for all $n\geq N$. So, all upcrossings of $[a,b]$ must start before time $N$, and we may conclude that $U[a,b]\leq N$ is finite. On the other hand, if $l\leq a$ then $u=l and we can infer that $x_{n} for all $n\geq N$ and some $N$. Again, this gives $U[a,b]\leq N$.

Conversely, suppose that the sequence does not converge, so that $u>l$. Then choose $a in the interval $(l,u)$. For any integer $n$, there is then an $m>n$ such that $x_{m}>b$ and an $m>n$ with $x_{m}. This allows us to define infinite sequences $s_{k},t_{k}$ by $t_{0}=0$ and

 $\displaystyle s_{k}=\inf\left\{m\geq t_{k-1}\colon X_{m} $\displaystyle t_{k}=\inf\left\{m\geq s_{k}\colon X_{m}>b\right\},$

for $k\geq 1$. Clearly, $s_{1} and $x_{s_{k}} for all $k\geq 1$, so $U[a,b]=\infty$.

Title proof of convergence of a sequence with finite upcrossings ProofOfConvergenceOfASequenceWithFiniteUpcrossings 2013-03-22 18:49:39 2013-03-22 18:49:39 gel (22282) gel (22282) 4 gel (22282) Proof msc 40A05 msc 60G17