Can the hyper hyper real numbers be constructed?

Question

The hyperreal numbers can be constructed as $\mathbb{R}^{\mathbb{N}}/U$ given some ultra filter and this allows first-order statements to be transferred over to $\mathbb{R}^*$. Can this be done again to get $\mathbb{R}^{**}\equiv (\mathbb{R}^*)^\mathbb{N}/U$ to get the hyper hyper reals and expect this new thing to still be an ordered field? Is anything lost in this expansion, does it even work to begin with, can this be repeated indefinitely, will the cardinal it’s of the resulting set ever exceed $2^{\aleph_0}$, is the thing still complete, and can it be used for anything?

Edit: After posting, I though of $(\mathbb{R}^*)^{\mathbb{N}^*}/U$ where $\mathbb{N}^*$ is the set of hyper natural numbers. Does this do anything to the repeated construction mentioned above or anything else interesting like that?

Answer 1

Nicholas has given a good answer with some nice references (+1), but let me make a more general point that the object you're looking at is just another ultrapower of $\mathbb{R}$! Let me first fix some notation to make sure we're all on the same page. Given a set $I$, an ultrafilter $U$ on $I$, and a first-order structure $M$, we say two functions $f,g:I\to M$ are "$U$-equivalent" if $\{i\in I:f(i)=g(i)\}\in U$. We write the $U$-equivalence class of $f:I\to M$ as $[f]_U$. Thus the elements of $M^I/U$ are precisely the $U$-equivalence classes.

Okay, now here's a useful construction. Let $U,V$ be ultrafilters on sets $I,J$ respectively; we define the "tensor product" $U\otimes V$ as the set of all subsets $X\subseteq I\times J$ such that $\{i\in I:\{j\in J:(i,j)\in X\}\in V\}\in U$. I'll leave it to you to check that $U\otimes V$ is an ultrafilter on $I\times J$.

Now we have the following: suppose $M$ is any first-order structure in a first-order language $L$. Then $M^{I\times J}/U\otimes V$ is in fact naturally isomorphic to $(M^I/U)^J/V$! Why? Essentially via currying. Given a map $f:I\times J\to M$ and an element $j\in J$, let $f_j:I\to M$ be the map given by $f_j(i)=f(i,j)$, and let $\tilde{f}:J\to M^I/U$ be the map given by $\tilde{f}(j)=[f_j]_U$. Now define $\alpha:M^{I\times J}/U\otimes V\to(M^I/U)^J/V$ by taking $\alpha([f]_{U\otimes V})=[\tilde{f}]_V$. It's a good (albeit slightly tedious exercise) to show that this gives a well-defined isomorphism of $L$-structures.

So, in the case when $M=(\mathbb{R},+,\cdot)$ and $I=J=\mathbb{N}$, you get that the "hyper-hyper-real field" $(\mathbb{R}^{\mathbb{N}}/U)^\mathbb{N}/V$ is just isomorphic to $\mathbb{R}^{\mathbb{N}\times\mathbb{N}}/U\otimes V$! In particular, since $\mathbb{N}\times\mathbb{N}$ is the same cardinality as $\mathbb{N}$, this means that there is an ultrafilter $W$ on $\mathbb{N}$ such that $(\mathbb{R}^{\mathbb{N}}/U)^\mathbb{N}/V$ is isomorphic to $\mathbb{R}^\mathbb{N}/W$, so any "hyper-hyper-real field" is actually just a "hyper-real field" to begin with.

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In particular, you ask if a hyper-hyper-real field can be complete; the answer is no, for the exact same reason that a hyper-real field can never be complete.

Answer 2

Yes, we can form an ultrapower $S^*$ of any set $S$, and that will have the properties of a nonstandard extension; see Theorem 2.28 in [1]. Since we can do this for any set, it can be repeated indefinitely. The proof of this theorem uses an arbitrary infinite index set, so I don't think using $\mathbb N^*$ will do anything interesting, but I don't know for sure. As I understand, we usually only make use of the transfer principle or other such properties which are generic to nonstandard extensions, so the specific ultrapower construction is irrelevant.

As for cardinality, I believe the cardinality of a nonstandard extension depends highly on the specific ultrapower you choose. In the case of a nonstandard extension of superstructure, [1] has Proposition 7.3 which states that, for a $\kappa$-saturated nonstandard extension, every infinite internal set has external cardinality $\geq\kappa$, and Theorem 7.13 proves the existence of such nonstandard extensions for every uncountable $\kappa$. What this means is that, at least in the superstructure setting, you can make $\mathbb R^*$ larger than any cardinality you choose.

If you're interested specifically in the case of iterating ultrapowers with index set $\mathbb N$, then I don't know what happens with the cardinalities.

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Constructions like $\mathbb R^{**}$ (in fact higher iterations) are indeed useful, or at least interesting, though I am not familiar with them in that particular form. What I am familiar with is an axiomatic approach to nonstandard analysis which is *relativized". Karel Hrbacek constructs a model of such an axiomatic system[2] (part of the series [2,3,4]) with model theory that goes way over my head, but it does involve some gnarly iterated ultrapowers in the spirit of $\mathbb R^{**}$. He also references other such systems which inspire his.

Relativized nonstandard analysis posits the existence of an elementary (in the same way as $\in$) predicate $a \sqsubseteq b$ on arbitrary objects, to be read "$a$ is observable relative to $b$" (which may be though of as corresponding to being standard in regular nonstandard analysis). Objects like $0, 1, 2, 3, \mathbb N, \pi, \mathbb R, \{\}$ are observable relative to all other objects, and all such objects could be considered the "regular" objects of mathematics (but there is an alternative interpretation I prefer but won't get into here). However, every infinite set (and certain finite sets) contains unobservable elements; in fact, for $S$ an infinite set and any $a$, there is $x \in S$ such that $x \not\sqsubseteq a$.

As an example, for any $a$ we can define an infinitesimal relative to $a$, or an $a$-infinitesimal, as an $x \in \mathbb R$ such that $x < y$ for every $y \in \mathbb R$ where $y \sqsubseteq a$. Such an $a$-infinitesimal necessarily has $x \not\sqsubseteq a$, and we can also form $x$-infinitesimals which are infinitesimal relative to $x$. We can do this ad infinitum.

The downside to such a system is that set formation can be tricky; for example $\{n \in \mathbb N \;:\; n \sqsubseteq a\}$ for any $a$ cannot be a set. This is essentially because sets in this relativized approach correspond to internal sets in the superstructure approach. To my knowledge, there is no satisfactory relativized system which also handles external sets; Hrbacek discusses this somewhat in [3].

If you're interested in this sort of relative approach, the book [5] by Hrbacek, Lessman, and O'Donovan is excellent and very approachable (with the claim it was used successfully in a high school class).

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[1] L.O. Arkeryd, N.J. Cutland, C.W. Henson, Nonstandard Analysis: Theory and Applications, Springer Netherlands, Dordrecht, 1997.

[2] K. Hrbacek, Relative set theory: Internal view, Journal of Logic and Analysis. 1 (2009).

[3] K. Hrbacek, Relative set theory: Some external issues, Journal of Logic and Analysis. 2 (2010).

[4] K. Hrbacek, Relative set theory: Strong stability, Journal of Logic and Analysis. 4 (2012).

[5] K. Hrbacek, O. Lessmann, R. O’Donovan, Analysis with ultrasmall numbers, CRC Press, Boca Raton, FL, 2015.