Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $( \mathbb{R}^\mathbb{N}/\mathcal{U} )_{\mathcal{U}\in\beta\mathbb{N}}$ be the set of all the hyperreal number systems, does there exist a set $\mathbb{X}%$ and embeddings $i_{\mathcal{U}}: \mathbb{R}^\mathbb{N}/\mathcal{U} \hookrightarrow \mathbb{X}$ for all $\mathcal{U} \in \beta\mathbb{N}$ ? If so, does anybody know about the (topological, algebraic, analytic) structure of such a structure $\mathbb{X}$ ?

Edit: Chris Eagle suggested to look at $\bigcup_{\mathcal{U}\in\beta\mathbb{N}} \mathbb{R}^\mathbb{N}/\mathcal{U}$. This set obviously has the properties I am so bluntly asking for, but I want some more structure. Something like the construction of the maximal exotic $\mathbb{R}^4$, in which every exotic one embeds, allthough this isn't a union of all the exotic $\mathbb{R}^4$'s (not that I am aware of).

I am looking for a way, is essence, to "ditch" the ultrafilter in the definition, but I don't think this is so easy to get rid of, then.

share|improve this question
1  
Obviously given a set of sets, one can find a set into which each of them embeds: their union, for example. So what do you really want? –  Chris Eagle Jul 14 '12 at 15:44
    
In large cardinals there is a notion of extender, which is a direct limit of ultraproducts. If you can order the filters in such way that ultraproducts give an embedding order induced by the filters then you can define this extender-like notion. I'm not sure how you would do that, though. –  Asaf Karagila Jul 15 '12 at 23:33
    
In some sense, the ring of functions $\mathbb{N} \to \mathbb{R}$ itself is such a "maximal" object with a rich structure. e.g. if you grind through things, I'm pretty sure this gives Chris Eagle's set the structure of a bundle over $\beta \mathbb{N}$. –  Hurkyl Jul 16 '12 at 0:09

1 Answer 1

There is an elementary embedding $j:\mathbb{R}\to\mathbb{R}^*$ for which all the ultrapower embeddings $j_U$ arise naturally as factors $i_U$ as you request via a universal property, and this would seem to be what you want. In particular, this $\mathbb{R}^*$ is an ordered field satisfying exactly the same theory as $\mathbb{R}$, and looks very like the ultrapowers $\mathbb{R}^{\mathbb{N}}/U$, but is much larger. This map $j$ may be taken to be elementary with respect to any additional structure placed on $\mathbb{R}$, since actually we have a map $j$ defined on the entire set-theoretic universe.

One way to find a such model $\mathbb{R}^*$ is to realize that since the product $U\times V$ of ultrafilters on $\mathbb{N}$ is an ultrafilter on $\mathbb{N}\times\mathbb{N}$, we have a natural directed system of ultrapower embeddings here. Namely, let $I=\beta\mathbb{N}$ and consider the finite subsets of $I$ as product ultrafilters on $\mathbb{N}$, with respect to a fixed linear ordering. Since the finite subsets are closed under unions, we have a corresponding directed system of ultrapower embeddings. Specifically, all the ultrapowers fit naturally together into a giant directed system, and you may take $\mathbb{X}=\mathbb{R}^*$ as the limit of this system. This gives it a very rich structure indeed, a structure closely connected with the individual $\mathbb{R}^{\mathbb{N}}/U$, taken finitely many at a time with the product ultrafilters, and enjoying the universal property of the limit of those ultrapowers. (I would draw more of the commutative system here, but I'm not sure how much diagrams tex is possible here.)

One can specifically represent the elements of the limit $\mathbb{R}^*$ with functions $f:\mathbb{N}^{\beta\mathbb{N}}\to \mathbb{R}$ having finite support. That is, each $f$ depends only on the coordinates corresponding to finitely many ultrafilters, and the product of those ultrafilters are used when defining equivalence $f\equiv g$ and so on. Let $j:\mathbb{R}\to\mathbb{R}^*$ be the embedding corresponding to this, which is the limit of the directed system. Every ultrafilter $U\in\beta\mathbb{N}$ has its ultrapower $j_U:\mathbb{R}\to \mathbb{R}^{\mathbb{N}}/U$ as a factor of $j$. This idea also underlies the concept of extender embeddings mentioned by Asaf in the comments.

Another different but more concrete way to do it, with a well-ordered iteration, is first to enumerate all the ultrafilters $U_\alpha$ in a well-ordered sequence, and then form the corresponding iterated ultrapower system $$\mathbb{R}=\mathbb{R}_0\to \mathbb{R}_1\to\cdots\to\mathbb{R}_\alpha\to\mathbb{R}_{\alpha+1}\to\cdots,$$ where the embedding from $\mathbb{R}_\alpha$ to $\mathbb{R}_{\alpha+1}$ is the ultrapower by $U_\alpha$, and one takes limits at limit ordinals. The limit of this system of embeddings produces an elementary map $j_\infty:\mathbb{R}\to \mathbb{R}^*$, for which the separate ultrapowers $j_{U_\alpha}$ can be seen to arise as factors.

Finally, let me remark that I object to the idea that the term "hyperreals" is to be used only for models of the form $\mathbb{R}^{\mathbb{N}}/U$ for a (nonprincipal) ultrafilter $U$ on $\mathbb{N}$. Rather, this is merely a convenient way to produce particular models of hyperreals. But there are many more. To my way of thinking, any elementary extension of the real field $\langle \mathbb{R},{+},{\cdot},0,1,{\lt}\rangle$ counts as an instance of the hyperreals, and not all of these arise as ultrapowers by ultrafilters on $\mathbb{N}$. For example, one can show that the limit models I describe above are not ultrapowers, and it is also easy to build models via the Lowenheim-Skolem theorem that are simply too large to arise as ultrapowers. Furthermore, ultrapowers always exhibit a certain degree of saturation, such as having uncountable cofinality, that not all nonstandard extensions of $\mathbb{R}$ need exhibit.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.