4 ms·

The reals can be ordered, just use x < y. I think you mean that if ZFC is true, we could enumerate unnameable reals (choose one with the axiom of choice, remove

The reals can be ordered, just use x < y. I think you mean that if ZFC is true, we could enumerate unnameable reals (choose one with the axiom of choice, remove it, choose another one, etc.), but you could not enumerate them all. But it is true that you could get a "first" unnameable real.

pdonis 1mo ago
> The reals can be ordered, just use x < y. That ordering is not a well-ordering, which is what the GP specified. A well ordering requires that every non-empty subset has a smallest element. That's not true for the reals ordered by x < y: for example, the set of all reals > 0 has no smallest element. No one has explicitly shown that the reals can be well ordered, but it's a consequence of the axiom of choice that every set can be well-ordered. So in ZFC there must be a well ordering of the reals, even though no one has found one. Issues like this are why not all mathematicians accept the axiom of choice.
- simonh 1mo ago
  Not 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.
  - GPerson 1mo ago
    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.
  - luc4 1mo ago
    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.
  - pdonis 1mo ago
    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.
- Smaug123 1mo ago
  > no one has found one More than that: there is no way to build one (assuming the word "build" means some concrete construction), because it's consistent with ZF that the reals admit no well-ordering. Indeed, you can use forcing to construct a model of R in which there is an infinite but Dedekind-finite subset of R; and you can't well-order such a set, because a well-ordering would turn it into an ordinal, and any Dedekind-finite ordinal is finite. You must use some sort of choice principle to construct a well-ordering. (Of course, it's consistent that they can be well-ordered, too, as you say; or e.g. under the hypothesis V=L, where there's even a canonical well-ordering given by the lexicographic well-ordering L comes with.)

pdonis 1mo ago

> The reals can be ordered, just use x < y. That ordering is not a well-ordering, which is what the GP specified. A well ordering requires that every non-empty subset has a smallest element. That's not true for the reals ordered by x < y: for example, the set of all reals > 0 has no smallest element. No one has explicitly shown that the reals can be well ordered, but it's a consequence of the axiom of choice that every set can be well-ordered. So in ZFC there must be a well ordering of the reals, even though no one has found one. Issues like this are why not all mathematicians accept the axiom of choice.

simonh 1mo ago
Not 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.
- GPerson 1mo ago
  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.
- luc4 1mo ago
  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.
- pdonis 1mo ago
  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.
Smaug123 1mo ago
> no one has found one More than that: there is no way to build one (assuming the word "build" means some concrete construction), because it's consistent with ZF that the reals admit no well-ordering. Indeed, you can use forcing to construct a model of R in which there is an infinite but Dedekind-finite subset of R; and you can't well-order such a set, because a well-ordering would turn it into an ordinal, and any Dedekind-finite ordinal is finite. You must use some sort of choice principle to construct a well-ordering. (Of course, it's consistent that they can be well-ordered, too, as you say; or e.g. under the hypothesis V=L, where there's even a canonical well-ordering given by the lexicographic well-ordering L comes with.)