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This question is crossposted at MSE; however, it has not gotten any responses there despite a bounty, so I'm asking it here.

Say that a theory $T$ in the language of ordered fields + constants is $\mathbb{R}$-satisfiable if it has a model whose ordered field part is $\mathbb{R}$, with the usual ordered field structure. I'm interested in what compactness-like properties $\mathbb{R}$-satisfiability has.

To begin with, we can easily knock off compactness itself: since $\mathbb{R}$ is Archimedean, $\mathbb{R}$-satisfiability is not compact. But things get murkier when we consider “compactness at higher cardinalities.”

For cardinals $\kappa<\lambda$, say that $\mathbb{R}$-satisfiability is $(\kappa, \lambda)$-compact if whenever $\Gamma$ is a set of sentences of cardinality $<\lambda$, and every subset of cardinality $<\kappa$ is $\mathbb{R}$-satisfiable, then $\Gamma$ is $\mathbb{R}$-satisfiable. (So usual compactness is $(\omega, \infty)$-compactness, and countable compactness is $(\omega,\omega_1$)-compactness.)

It’s easy to show that $\mathbb{R}$-satisfiability is not $(\omega_1,\omega_2)$-compact: any countable linear order embeds into $\mathbb{R}$, but $\omega_1$ does not. And it follows from Easton’s theorem that for every cardinal $\kappa$ with $cf(\kappa)>\omega$, it is consistent with ZFC that $\mathbb{R}$-satisfiability is not $(\kappa^+, \kappa^{++})$-compact (look at a theory asserting the existence of $\kappa^+$-many distinct reals assuming $\mathfrak{c}=\kappa$).

This suggests two natural questions:

  • Can ZFC prove that $\mathbb{R}$-satisfiability is $(\omega_{\omega+1}, \omega_{\omega+2})$-compact?

  • Is it consistent with ZFC that $\mathbb{R}$-satisfiability is $(\omega_2, \omega_3)$-compact?

I believe the answer to the second question should be a relatively easy "yes", while the first question should be "no" but might require some work. However, I don't immediately see how to resolve either piece.

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  • $\begingroup$ In real-closed fields, every formula is equivalent to an existentially quantified identity. It follows easily that in order to test $(\kappa,\lambda)$-compactness of $\mathbb R$-satisfiability, it is enough to consider theories $T$ axiomatized by integer polynomial equations in the constants. $\endgroup$ – Emil Jeřábek Oct 18 '16 at 19:15
  • $\begingroup$ (You may also severely restrict the form of the polynomials, if desired. For example, it is enough to have only equations of the form $c_i+1=0$, $c_i+c_j+c_k=0$, and $c_ic_j+c_k=0$.) $\endgroup$ – Emil Jeřábek Oct 18 '16 at 19:38
  • $\begingroup$ Is it consistent with ZFC at all that there exist $\kappa<\lambda$ such that $\mathbb R$-satisfiability is $(\kappa,\lambda)$-compact? $\endgroup$ – Emil Jeřábek Oct 19 '16 at 12:00
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    $\begingroup$ @EmilJeřábek I believe that if $\kappa$ is measurable, then $\mathbb{R}$-satisfiability is $(\kappa, \kappa^+)$-compact as follows. Let $L$ be the language of arithmetic augmented by $\kappa$-many constants $c_\eta$, and let $T$ be an $L$-theory. Write $T=\bigcup_{\eta<\kappa}T_\eta$, where each $T_\eta$ only uses constant symbols $c_\alpha$ with $\alpha<\eta$. Suppose every size-$<\kappa$ subset of $T$ is $\mathbb{R}$-satisfiable. Let $F_\eta:\eta\rightarrow\mathbb{R}$ yield an interpretation $\mathbb{R}$-satisfying $T_\eta$. (cont'd) $\endgroup$ – Noah Schweber Oct 20 '16 at 21:59
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    $\begingroup$ We now take the average of the $F_\eta$s. Fix a measure $U$ on $\kappa$, and let $F: \kappa\rightarrow\mathbb{R}$ be given by $F(\alpha)=r$ iff $\{\eta: F_\eta(\alpha)=r\}\in U$; note that $F$ is defined everywhere since $\vert\mathbb{R}\vert<\kappa$. It's now not hard to show that $F$ yields an interpretation $\mathbb{R}$-satisfying $T$. $\endgroup$ – Noah Schweber Oct 20 '16 at 22:01
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The answer to the first question is no, by an easy argument.

What I claim is that ZFC proves that $\newcommand\R{\mathbb{R}}\R$-satisfiability is not $(\mathfrak{c},\mathfrak{c}^+)$-compact. (This improves upon your observation about $(\kappa^+,\kappa^{++})$.) To see this, let $T$ be the elementary diagram of $\R$, with a constant for every real, plus another constant, asserted to be none of them. Every size-less-than-continuum subtheory of this theory is $\R$-satisfiable. But for the whole theory, the only way to interpret the $\R$-constants in an $\R$-structure is as themselves, since the rational constants will be interpreted as themselves and the other constants are then constrained by the type of their cuts in the rationals. So there is nobody left for the one extra constant.

In particular, if $2^\omega=\aleph_{\omega+1}$, then this shows $\R$-satisfiability needn't be $(\omega_{\omega+1},\omega_{\omega+2})$-compact.

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  • $\begingroup$ Ouch, that's really obvious in retrospect - +1. Do you see how to attack the other question? $\endgroup$ – Noah Schweber Oct 18 '16 at 13:53
  • $\begingroup$ Unfortunately, I don't yet have any attacks on the second question. Do you know anything assuming CH? $\endgroup$ – Joel David Hamkins Oct 18 '16 at 14:02
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    $\begingroup$ If CH holds, then $\R$ is not $(\omega_2,\omega_3)$-compact, by the argument in your post: consider the theory of $\omega_2$-many distinct constants. And the case $2^\omega=\omega_2$ is handled by my argument. So we have only the case $2^\omega\geq\omega_3$ remaining. $\endgroup$ – Joel David Hamkins Oct 18 '16 at 19:48

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