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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?

Dustin McPhate
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    The hyper-reals are very incomplete as linear orders - in fact, no countable increasing sequence in $^* \mathbb R$ has a supremum in $^* \mathbb R$. – Forever Mozart Dec 10 '14 at 13:25

2 Answers2

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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".

KDuck
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  • How do you define Cauchy sequences without a metric? – Tobias Kildetoft Dec 10 '14 at 10:22
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    An order relation induces something like a metric, only that on $^*\mathbb{R}$ the image of it will not be in $\mathbb{R}$, which is one of the conditions for a metric.

    We have a function $d: (^\mathbb{R})^2\to (^\mathbb{R}), (x,y)\mapsto |x-y|$. It has all the properties of a metric except that it does not map to $\mathbb{R}$.

    Convergence of $(a_n)$ to $a$ in $^*\mathbb{R}$ can now be defined as follows: $$\forall \varepsilon \in \mathbb{Q}_{>0} \exists N\in\mathbb{N} \forall n > N: d(a_n,a)<\varepsilon$$

     – KDuck Dec 10 '14 at 10:26
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    The natural thing to do here would be to use ${}^*\Bbb Q_{>0}$ for defining a Cauchy sequence, in which case the limits are unique again. – Asaf Karagila Dec 10 '14 at 10:32
  • @Asaf: Then sequences like $(1/n)$ would not be Cauchy any more? I think by defining Cauchy sequences like that even the field $^\mathbb{Q}$ would be complete: If $(a_n)$ is a Cauchy sequence in $^\mathbb{Q}$, then $$\forall \varepsilon\in (^*\mathbb{Q})_{>0} \exists N\in\mathbb{N}\forall n,m\geq N: |a_n-a_m|<\varepsilon.$$ So at some point the differences between the terms of the sequence $(a_n)$ become infinitessimally small and hence converge to a number $a$ with standard part in $\mathbb{Q}$ and infinitesimal non-standard part. – KDuck Dec 10 '14 at 10:52
  • Right. We need longer sequences too. – Asaf Karagila Dec 10 '14 at 10:59
  • Also, for a notion of completeness for general ordered fields we can only use terms we can define in every ordered field. So using $^*\mathbb{Q}_{>0}$ for Cauchy sequences would not be applicable to standard ordered fields. – KDuck Dec 10 '14 at 11:00
  • I'm not trying to argue that ${}^*\Bbb R$ is complete. Just what it would mean to talk about Cauchy completeness. – Asaf Karagila Dec 10 '14 at 11:19
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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.

Mikhail Katz
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