TITLE: infinite permutations
QUESTION [16 upvotes]: This question is related to this one: Continued fractions using all natural integers. Suppose we have the set of natural numbers $N$ with order and we perform permutation on it. So we obtain the same elements with different order. Suppose we describe such permutations by usual notation when (1,2,3,4,5...) means identity permutation. Then lets say that permutation denoted by (1,3,2,4,5,6...) ( from the 4th place there is list of natural numbers in usual order) is finite because it only mixes numbers 1,2,3 -> 1,3,2 and for remaining elements it is identity permutation. As I find here there is definition of such objects, namely a few possibilities as states the answer of Qiaochu Yuan. 
Questions:

Are  infinite permutation decomposable into cycles? Transpositions?
Is possible to find such permutation of natural numbers that it cannot be a limit of finite permutations?

REPLY [7 votes]: Every permutation decomposes $\mathbb N$ into orbits. You can arrange these into a cycle decomposition if you allow infinite cycles.
I'm not comfortable with all infinite products, particularly ones which do not define permutations at all times. 
One notion of an infinite product of transpositions is a limit under pointwise convergence. That is, say $\pi$ is an infinite product of transpositions if there is a sequence of permutations $e=\pi_0, \pi_1, \pi_2...$ which converges pointwise to $\pi$ so that $\pi_{n+1} \pi_n^{-1}$ is a transposition.  
Every permutation is an infinite product of transpositions: Define $\pi_n = \bigg(\pi(n) ~~ \pi_{n-1}(n)\bigg)\pi_{n-1}.$ Then  $\pi_n$ agrees with $\pi$ on $\{1,...,n\}$, so $\pi_0, \pi_1, \pi_2, ... \to \pi$. 
Therefore, every permutation is a pointwise limit of what you called finite permutations.