Are the Hyperreals complete?

Question

Since $^*\mathbb{R}$ does not form a metric space then it can not satisfy the Cauchy conditions for completeness. However, my intuition is telling me that it would satisfy conditions of completeness in the order theory sense that it has no "holes". Where am I going wrong here?

Answer 1

Depending on what definition of completeness you use, you will not have uniqueness.

For example, you have one notion of completeness saying "An ordered field $K$ is complete if and only if every Cauchy sequence converges to a unique limit in $K$." The condition "unique" can be dropped if $K$ also forms a metric space because then every limit is automatically unique. In $^*\mathbb{R}$ the limit of a Cauchy sequence is not unique. It converges to every point which is infinitessimally close to any other limit point.

Or if you use the notion of supremum completeness "Every subset of $K$ which is bounded above has a least upper bound in $K$", you will see that this does not hold in $^*\mathbb{R}$. Every subset which is bounded above has an upper bound in $\mathbb{R}$ but there is no least one, because you can always find one which is infinitessimally smaller.

Completeness can be rather a counterintuitive if you interpret it as "without gaps". It also includes something like "every gap is only filled once".

Answer 2

As far as "order theory sense" is concerned, there actually are "holes": consider for example the "hole" between all finite numbers and positive infinite hyperreals.

I am not sure why you say the hyperreal field does not form a metric space. Every field is a 1-dimensional vector space over itself, and the hyperreals are no exception, and in particular the hyperreals form a metric space with distances in the field.