# proof of Poincaré recurrence theorem 1

Let ${A}_{n}={\cup}_{k=n}^{\mathrm{\infty}}{f}^{-k}E$. Clearly, $E\subset {A}_{0}$ and ${A}_{i}\subset {A}_{j}$ when $j\le i$. Also, ${A}_{i}={f}^{j-i}{A}_{j}$, so that $\mu ({A}_{i})=\mu ({A}_{j})$ for all $i,j\ge 0$, by the $f$-invariance of $\mu $. Now for any $n>0$ we have $E-{A}_{n}\subset {A}_{0}-{A}_{n}$, so that

$$\mu (E-{A}_{n})\le \mu ({A}_{0}-{A}_{n})=\mu ({A}_{0})-\mu ({A}_{n})=0.$$ |

Hence $\mu (E-{A}_{n})=0$ for all $n>0$, so that $\mu (E-{\cap}_{n=1}^{\mathrm{\infty}}{A}_{n})=\mu ({\cup}_{n=1}^{\mathrm{\infty}}E-{A}_{n})=0$. But $E-{\cap}_{n=1}^{\mathrm{\infty}}{A}_{n}$ is precisely the set of those $x\in E$ such that for some $n$ and for all $k>n$ we have ${f}^{k}(x)\notin E$. $\mathrm{\square}$

Title | proof of Poincaré recurrence theorem 1 |
---|---|

Canonical name | ProofOfPoincareRecurrenceTheorem1 |

Date of creation | 2013-03-22 14:29:56 |

Last modified on | 2013-03-22 14:29:56 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 5 |

Author | Koro (127) |

Entry type | Proof |

Classification | msc 37A05 |

Classification | msc 37B20 |