alt Hacker NewsNot a mathematician, so this question may be a bit thick. I see the problem with the set of reals > 0, but is it perhaps that in this case > 0 is the problem and for sets specified as >= 0 it's fine because 0 is a nameable real and the smallest element. Obviously you can't just exclude certain expressions arbitrarily though, so I don't know how you could justify that mathematically.
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We just used the standard ordering < to define the set, it has nothing to do with the candidate well-ordering. If that's confusing, consider the set { 10^-x | x \in N } instead. It also has no minimum element in the standard ordering.
A well ordering on a set is a total order such that all non empty subsets have a minimum element with respect to this order. The standard ordering of the reals is not a well ordering, but the axiom of choice is equivalent to the statement that all sets possess a well-ordering. A well-order of the reals would probably look pretty chaotic though.
If there is any non-empty subset that has no smallest element, then the ordering in question is not a well-ordering. You can of course define some subsets of the reals that do have a smallest element in the standard ordering, for example all of the reals that are greater than or equal to 0. But there are also subsets that do not have a smallest element, and that is enough to show that the standard ordering on the reals cannot be a well-ordering.