proof of Poincaré recurrence theorem 1


Let An=∪k=n∞f-k⁢E. Clearly, E⊂A0 and Ai⊂Aj when j≤i. Also, Ai=fj-i⁢Aj, so that μ⁢(Ai)=μ⁢(Aj) for all i,j≥0, by the f-invariance of μ. Now for any n>0 we have E-An⊂A0-An, so that

μ⁢(E-An)≤μ⁢(A0-An)=μ⁢(A0)-μ⁢(An)=0.

Hence μ⁢(E-An)=0 for all n>0, so that μ(E-∩n=1∞An)=μ(∪n=1∞E-An)=0. But E-∩n=1∞An is precisely the set of those x∈E such that for some n and for all k>n we have fk⁢(x)∉E. □

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