# Does every Cauchy net of hyperreals converge?

This came up in a discussion with Pete L. Clark on this question on complete ordered fields. I argued that every Cauchy sequence in the hyperreal field is eventually constant, hence convergent; he asked whether the same is true for arbitrary Cauchy nets in $\mathbb{R}^*$. I'm not sure how to deduce this either from the transfer principle ("every Cauchy net converges" is a very second-order statement) or from the ultraproduct condition of $\mathbb{R}^*$. Does anyone know the answer?

(I agree that if $f: \mathbb{N}^* \to \mathbb{R}^*$ is an internal Cauchy net, then $f$ has a limit.)

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If $(x_\lambda)$ is a Cauchy net in a compact space $X$ then it converges to a cluster point $x$. So I guess not every Cauchy net of hyperreals converge since it seems to depend on compactness. – PEV Mar 22 '11 at 4:15

Hints (general ordered field, not just "the" hyperreal field.)

(a) Can you show that a convergent net is Cauchy?

(b) Are there convergent nets not eventually constant?

(c) Conclude that there are non-constant Cauchy nets.

OF COURSE you need to define "Cauchy net" before you can even ask the question...

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okay, I can take the hint: consider the directed set $I$ of all open neighborhoods of $0$. If for each $U \in I$ we take $x_{\mathcal{U}}$ to be a nonzero element of $U$, then we get a net which converges to $0$ and is not eventually constant. (Here we are using, at most, that $0$ is a nonisolated point of a $T_1$-topological space.) But my question was not this exactly but rather: is there (a version of) the hyperreals in which every Cauchy net converges? – Pete L. Clark May 12 '11 at 18:38
There is a nonstandard model of $\mathbb R$ in which not every Cauchy sequence converges. For the general nonstanderd model of $\mathbb R$ and the net case (or equivalently, every Cauchy filter converges) probably you need to get a real model theorist. Usually in non-archimedean ordered fields they talk not about completeness but pseudo-completeness. I don't know if that is because completeness never holds or for some other reason. – GEdgar May 13 '11 at 15:53
hmm, the plot thickens. After I wrote my last comment, it seems to me that one could at least try to take the "full Cauchy completion" (i.e., the construction using equivalence classes of Cauchy nets) of any non-Archimedean ordered field. I see two potential problems: (i) it may not be completely for free that this gives an ordered field (this cannot be the case for Dedekind completion!). I'm guessing it does though. And (ii): does what one gets deserve to be called a hyperreal field? I don't know anything about this... – Pete L. Clark May 13 '11 at 16:49