Is this true? For any hyperfinite $n$ that isn't finite, there is a hyperfinite set $A$ such that $\mathbb R \subset A$ and $|A|\le n$ (that's the crucial part, of course)? Intuitively it seems right, but I haven't found a reference and I am not very good at NSA.
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1$\begingroup$ The buzzword here is "enlargement". A nonstandard extension is an enlargement if each standard $A$ is a subset of a hyperfinite subset of ${}^\ast A$. $\endgroup$– Kevin O'BryantCommented Nov 8, 2012 at 17:52
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$\begingroup$ Kevin, do you really mean every $A$? That seems impossible without some restriction on the sizes of $A$ to be considered. For example, if $A$ has cardinality larger than $\mathbb{N}^\ast$, then it cannot be covered with any hyperfinite set. $\endgroup$– Joel David HamkinsCommented Nov 8, 2012 at 18:53
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$\begingroup$ I do mean every $A$, but I was just giving a definition, not a proof of existence! Technically, it goes like this: you fix a superstructure $V$ that contains all of the sets of your interest, say $V$ has cardinality $\kappa$. Then you take a nonstandard extension of $V$ that is $\gamma$-saturated, for some $\gamma>\kappa$, and that extension is an enlargement. This avoids your concern because $\gamma$-saturation forces ${}^\ast {\mathbb N}$ to be sufficiently enormous. $\endgroup$– Kevin O'BryantCommented Nov 9, 2012 at 2:57
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$\begingroup$ I see. By fixing the superstructure, you have in effect limited the cardinality of the $A$ that arise, as I suspected. (By the way, I notice that you always but the asterisk to the left, whereas I have also elsewhere seen it on the right---which is best or more correct?) $\endgroup$– Joel David HamkinsCommented Nov 9, 2012 at 4:16
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$\begingroup$ I learned to put the asterisk on the left, but it shouldn't go so far to the left as I put it. At least, that's how I see it in articles concerning NSA. It's certainly easier to TeX it to the right, though. $\endgroup$– Kevin O'BryantCommented Nov 9, 2012 at 4:58
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2 Answers
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This depends on how strong your axioms for non-standard analysis are. Certainly any sufficiently saturated model will have an $A$ of the sort you ask about. [Proof: The collection of formulas consisting of "$x$ is a set", "$|x|\leq n$", and "$r\in x$" for all standard reals $r$ is finitely satisfiable. So by saturation there is an $A$ satisfying the whole collection.) But there are various weaker assumptions than $(2^{\aleph_0})^+$-saturation that are sometimes used in NSA and that might not suffice to provide the $A$ that you want.
answered Nov 8, 2012 at 14:12
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$\begingroup$ Andreas, do you know a context in which having less saturation is better/important/useful? I agree that in principle one could have less saturation, but I´m wondering if you really meant the "sometimes used in NSA". $\endgroup$ Commented Nov 8, 2012 at 15:18
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$\begingroup$ Ramiro, one answer to your question could be that there are advantages to building $\mathbb{R}^\ast$ as an ultrapower by an ultrafilter on $\mathbb{N}$, and these are not supersaturated. $\endgroup$ Commented Nov 8, 2012 at 16:04
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$\begingroup$ My recollection (perhaps faulty after all these years) is that some of the early work on nonstandard analysis used "richness" conditions that varied from one paper to another. I believe that some of these were essentially saturation weakened to allow only parameters from the standard model, so that I couldn't use the parameter $n$ in my answer. $\endgroup$ Commented Nov 8, 2012 at 17:51
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Andreas has pointed out that in any sufficiently saturated nonstandard model $\mathbb{R}^\ast$, the answer is yes.
Meanwhile, let me point out that if, as is commonly done, one builds one's hyperreals $\mathbb{R}^\ast$as the ultrapower of $\mathbb{R}$ by an ultrafilter on $\mathbb{N}$, then the answer is no. Indeed, in such a nonstandard $\mathbb{R}^\ast$, there is no hyperfinite cover of the standard reals at all. To see this, suppose that $x$ is a nonstandard hyperfinite set of reals. In the ultrapower, $x$ is represented by a function $f$ from $\mathbb{N}$ to the finite sets of reals, so that $f(k)$ is a finite set of reals. By the Los theorem, the reals $r$ with $r^\ast\in x$ must all have $r\in f(k)$ for almost all $k$. But this is a countable set, so $x$ contains at most countably many reals.
answered Nov 8, 2012 at 16:02