Does the concept of permutation make sense for a set indexed by the real numbers? I know that the concept of permutation makes sense for sequences, which are sets indexed by the natural numbers (if the sequence is infinite) or indexed by the first $n$ natural numbers (if the sequence is of length $n$). A permutation then is simply a reordering of the elements of such a sequence.
My question now is whether permutations also make sense for sets indexed by the real numbers, instead of the natural numbers.
My thinking is that it does because the real numbers are a totally ordered set, just like the natural numbers. But I am not sure.
Is the existence of a total order on the indexing set sufficient for a permutation of an indexed set to make sense?
 A: If you define a permutation as a bijection of the set into itself, the concept makes sense with any indexing set (not just totally ordered ones).
A: The main distinction is whether the index set is finite or infinite, not whether it is totally (or well) ordered.
You can speak of permutations of an infinite set just fine -- they are still the same as the bijections from the set to itself.
However, beware that several "standard" results from permutation theory actually only apply for permutations of a finite set. In particular, for an infinite set the permutation group is not generated by the transpositions, and there is no well-defined sign of a permutation.
A: There is no problem in considering permutations over any set $X$: a permutation is just a bijection $X\to X$. The concept of order is not relevant at all for defining permutations. 
You may be misled by the fact that an order on the (finite) base set can be used for computing the signature of a permutation, but the signature is independent of the chosen order (it can be defined independently of any order on the base set, if it is finite, of course).
