# proof of convergence of a sequence with finite upcrossings

We show that a sequence^{} ${x}_{1},{x}_{2},\mathrm{\dots}$ 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 $$.

Denoting the infimum limit and supremum limit^{} by

$$l=\underset{n\to \mathrm{\infty}}{lim\; inf}{x}_{n},u=\underset{n\to \mathrm{\infty}}{lim\; sup}{x}_{n},$$ |

then $l\le u$ 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 $$. If $l>a$ then there is an $N$ such that ${x}_{n}>a$ for all $n\ge N$. So, all upcrossings of $[a,b]$ must start before time $N$, and we may conclude that $U[a,b]\le N$ is finite. On the other hand, if $l\le a$ then $$ and we can infer that $$ for all $n\ge N$ and some $N$. Again, this gives $U[a,b]\le N$.

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

$$ | ||

${t}_{k}=inf\{m\ge {s}_{k}:{X}_{m}>b\},$ |

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

Title | proof of convergence of a sequence with finite upcrossings |
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Canonical name | ProofOfConvergenceOfASequenceWithFiniteUpcrossings |

Date of creation | 2013-03-22 18:49:39 |

Last modified on | 2013-03-22 18:49:39 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 4 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 40A05 |

Classification | msc 60G17 |