The first thing to notice is that infinite permutations may have infinite support, that is, they may move infinitely many elements. Therefore, we cannot expect to express them as finite compositions of permutations having only finite support. 

But if we allow (well-defined) infinite compositions, then the answer is that every permutation is a composition of transpositions. To see this, suppose that f is a permutation of &omega;. First, we may divide f into its disjoint orbits, where the orbit of n is defined as all the numbers of the form f<sup>k</sup>(n) for any integer k. The action of f on each of these orbits commute with each other, because the orbits are disjoint. The finite orbits are just finite cycles, which can be expressed as a product of transpositions in the usual way. An infinite orbit looks like a copy of the integers, since it will be infinite in both directions. All such cycles are isomorphic to the shift map on the integers, which can be represented in cycle notation as (... -2 -1 0 1 2 ...). This permutation is equal to the following product of transpositions:
 [(0 1)(0 2)(0 3)....][...(-3 0)(-2 0)(-1 0)]. First, let me explain what this product means. The first factor operates on the non-negative integers, and sends 0 to 1, 1 to 0 to 2, 2 to 0 to 3, etc., so it behaves properly on the non-negative integers, and the second factor fixes all positive integers. So the composition is well-defined with the right answer for non-negative naturals. Secondly, the first factor fixes all negative integers, and then the second factor sends them in finitely many moves to the right place. e.g. -3 goes to 0 and then to -2, etc. So altogether, the product is operating correctly. This same idea can be used to represent any infinite orbit, and so any permutation is a suitable well-defined product of transpositions, as desired.

Note: I just realized that I am using backwards composition notation! I'll fix this in an edit.