# proof of Poincaré recurrence theorem 2

Let $\{{U}_{n}:n\in \mathbb{N}\}$ be a basis of open sets for $X$, and for each $n$ define

$${U}_{n}^{\prime}=\{x\in {U}_{n}:\forall n\ge 1,{f}^{n}(x)\notin {U}_{n}\}.$$ |

From theorem 1 we know that $\mu ({U}_{n}^{\prime})=0$. Let $N={\bigcup}_{n\in \mathbb{N}}{U}_{n}^{\prime}.$
Then $\mu (N)=0$. We assert that if $x\in X-N$ then $x$ is recurrent. In fact,
given a neighborhood^{} $U$ of $x$, there is a basic neighborhood ${U}_{n}$ such that $x\subset {U}_{n}\subset U$, and since $x\notin N$ we have that $x\in {U}_{n}-{U}_{n}^{\prime}$ which by definition of ${U}_{n}^{\prime}$ means that there exists $n\ge 1$ such that ${f}^{n}(x)\in {U}_{n}\subset U$; thus $x$ is recurrent. $\mathrm{\square}$

Title | proof of Poincaré recurrence theorem 2 |
---|---|

Canonical name | ProofOfPoincareRecurrenceTheorem2 |

Date of creation | 2013-03-22 14:29:58 |

Last modified on | 2013-03-22 14:29:58 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 5 |

Author | Koro (127) |

Entry type | Proof |

Classification | msc 37A05 |

Classification | msc 37B20 |