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By Cantor's intersection theorem every decreasing nested sequence of nonempty compact sets has a common point.

A superficially similar result holds that every decreasing nested sequence of nonempty internal sets in an ultrapower model of a hyperreal field ${}^\ast\mathbb{R}$ has a common point, a property known as countable saturation.

Is such a resemblance more than superficial?

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    $\begingroup$ It's probably worth mentioning also that even the nonstandard rationals $\mathbb{Q}^*$ and the nonstandard integers $\mathbb{Z}^*$ can exhibit countable saturation (and it is possible for all these to be much more saturated than just countably saturated). Meanwhile, saturation is not a consequence of the transfer principle, since one can easily construct nonstandard models lacking saturation. In light of this kind of variation, it's not really correct to speak of the hyperreals. $\endgroup$
    – Joel David Hamkins
    Nov 11, 2016 at 13:26
  • $\begingroup$ @JoelDavid, a nested sequence of internal sets in $\mathbb Z^\ast$ would in particular be a nested sequence of internal sets in $\mathbb R^\ast$ (and any common point would necessarily be a hyperinteger). $\endgroup$
    – Mikhail Katz
    Nov 13, 2016 at 12:34
  • $\begingroup$ Yes, of course. My point was that since $\mathbb{Q}^*$ can be saturated, but intervals in $\mathbb{Q}$ are not compact, it may somewhat undermine an expectation of a connection here. (Although your way of stating the principle for nested compact sets is fine for $\mathbb{Q}$.) $\endgroup$
    – Joel David Hamkins
    Nov 13, 2016 at 12:51
  • $\begingroup$ @JoelDavid, obviously this can't be interpreted naively at the level of subsets of $F^\ast$ (for whatever $F$). I was wondering whether perhaps there is an overarching framework where both nesting phenomena would arise as particular manifestations of a common principle. It's really more a question about Los' lemma than nonstandard analysis perhaps, but we don't have a tag for that. $\endgroup$
    – Mikhail Katz
    Nov 13, 2016 at 12:56
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    $\begingroup$ Of course compactness theorems in logic are connected to topological compactness. To give a simple example: the main content of compactness for propositional logic is that any Boolean algebra $B$ (e.g., a Lindenbaum algebra of a propositional theory) that is nontrivial (i.e., $0 \neq 1$) admits a Boolean algebra map $\phi: B \to 2$; one can construct a model from this datum $\phi$. This can be inferred from compactness of the Cantor space $2^X$ where $X$ is a set of generators for $B$, and is a starting point of Stone duality. See Prop. 3.1 here: ncatlab.org/nlab/show/Tychonoff+theorem $\endgroup$
    – Todd Trimble ♦
    Nov 13, 2016 at 13:44

2 Answers 2

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As Todd Trimble pointed out in the comments, the use of the term "compactness" in the Compactness Theorem does refer to the topological notion: the Stone space of the Lindenbaum algebra of a theory is compact. For saturation, the relevant notion are the spaces of complete $n$-types. These are naturally compact spaces. If $A_0 \supseteq A_1 \supseteq A_2 \supseteq \cdots$ are nonempty internal sets, then $\{x \in A_0,x \in A_1,x \in A_2,\ldots\}$ extends to a complete $1$-type $p(x)$ by compactness. A countably saturated model is one which realizes every complete type over a countable parameter set (e.g., $A_0,A_1,A_2,\ldots$). Therefore, a countably saturated model of non-standard analysis must realize the type $p(x)$. Because models of non-standard analysis satisfy comprehension, one can do without the syntactic notion of types and simply work with sets. This way, countable saturation for such models is equivalent to saying that the intersection of a countable family of internal sets is nonempty.

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    $\begingroup$ This does not answer the question concerning the relation between Cantor's lemma for nested compact sets on the one hand, and an analogous property for internal sets. $\endgroup$
    – Mikhail Katz
    Dec 1, 2016 at 12:01
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It turns out that there is a direct relationship between Cantor's lemma for nested compact sets $K_n$, on the one hand, and nested sequences of internal sets, on the other. Namely, the latter can be viewed as a special case of the former as follows.

The decreasing nested sequence of internal sets, $\langle{}^\ast\!K_n\colon n\in\mathbb N\rangle$, has a common point $x$ by saturation. But for a compact set $K_n$, every point of ${}^\ast\!K_n$ is nearstandard, i.e., infinitely close to a point of $K_n$ (in other words, every point is in the halo of a standard point of $K_n$). In particular, $st(x)\in K_n$ for all $n$, as required.

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