Models of the theory of real closed fields with extra constants

Question

Let $L$ be the language of real closed fields $\{0,1,+,-,\times\}$, and $T$ the theory of real closed fields, i.e. the $L$-sentences that are true in the standard $\mathbb{R}$ model. Let $L'$ be $L$ extended with a countable set of constants.

If $\Gamma$ is a satisfiable countable set of $L'$ sentences that contains $T$, must it have a model that can be obtained from the standard $\mathbb{R}$ model of $L$ by assigning a real to each of the added constants?

If yes, is the same true when there are $|\mathbb{R}|$-many added constants? When there are $|\mathbb{R}|$-many added constants and when $\Gamma$ has larger cardinality?

If you create a new theory of a model by augmenting the language with constants, does the new theory contain sentences that says anything "original"? is a related question.

Answer 1

No, add one new constant $c$ and add the axioms for your new theory, $$0<c<\frac{1}{n}.$$ This is consistent, since any finite number of the axioms are satisfiable.

Answer 2

The answer is no, because the reals are Archimedean.

So for instance, if you take just one constant $c$ and take for $\Gamma$ sentences $0<c$, and for all natural numbers, $n$, $c<1/n$, then $RCF+\Gamma$ is clearly satisfiable (say, by the real closure of ${\bf R}[x]$ with your favourite ordering), but you can't name such a $c$ in ${\bf R}$.

Note that what you ask is really if any type on countably many/ continuum many variables over the empty set can be realised in ${\bf R}$, which is very close (though slightly weaker in general) to asking if ${\bf R}$ is saturated in some cardinality.

In fact, this is equivalent to asking if every countable real closed field (or of size up to continuum) can be embedded into reals (by which I mean elementarily embedded, but for real closed fields this is the same as simply algebraically embedded, thanks to model completeness), i.e. about whether or not reals are an universal model (for the appropriate cardinality). I think it is not a hard exercise to see that there is a model $R$ of $RCF$ of size continuum which has your property for countable $\Gamma$ (take simply the direct limit of all countable models (up to isomorphism)). I also think it is true by a stability-theoretic argument (though I am not sure of it at the moment; I think it should also be doable in mostly order-theoretic manner), that such a model (of size continuum) does not exist for $\Gamma$ of size continuum.

Answer 3

The other two answers are absolutely correct; let me give an answer in a slightly different direction. Say that a theory $T$ in the language of ordered fields + constants is $\mathbb{R}$-satisfiable if it has a model which is an expansion of $\mathbb{R}$ with the usual ordered field structure; I'm interested in the model-theoretic properties of $\mathbb{R}$-satisfiability.

One natural question to ask is, "Is there a compactness theorem for $\mathbb{R}$-satisfiability?" The answer is no (and this provides another counterexample to the question you ask). Consider the theory $T$ consisting of the axioms of real closed fields, together with

• $c_i+1<c_{i+1}$ for $i>0$, and

• $c_0>c_i$ for $i>0$.

Then every finite subset of $T$ is $\mathbb{R}$-satisfiable, but $T$ itself is not, since $c_0$ would have to be infinite.

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What about "compactness in other cardinalities"? 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.)

Is $\mathbb{R}$-satisfiability $(\omega_1, \omega_2)$-satisfiable? No! Given $\omega_1$-many constants $c_\eta$ ($\eta<\omega_1$), consider the theory $S=\{c_\alpha<c_\beta: \alpha<\beta\}$. Every countable subset of $S$ is $\mathbb{R}$-satisfiable, since any countable linear order embeds into $\mathbb{R}$, but $S$ itself is not $\mathbb{R}$-satisfiable: if $c<d$ are reals, then there is a rational $c<q<d$, so the $\mathbb{R}$-satisfiability of $S$ would imply that the rationals are uncountable.

Is $\mathbb{R}$-satisfiability $(\omega_2,\omega_3)$-satisfiable? Not provably in ZFC! Suppose CH holds - that is, $\vert\mathbb{R}\vert=\aleph_1$ - and we have constant symbols $c_\alpha$ for $\alpha<\omega_2$. Then $\{c_\alpha\not=c_\beta: \alpha\not=\beta\}$ is not $\mathbb{R}$-satisfiable, but every subset of size $<\omega_2$ is $\mathbb{R}$-satisfiable. (Similarly, for any cardinal $\kappa$ of uncountable cofinality, it is consistent with ZFC that $\mathbb{R}$-satisfiability is not $(\kappa^+,\kappa^{++})$-compact.

This leaves two interesting questions:

• Can we prove in ZFC 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?

Off the top of my head I don't know the answer to either question.

If you're interested in this sort of question, you might enjoy looking at abstract model theory.

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Note: the general question of when an equational theory has a model compatible with a particular topological space has been extensively analyzed in universal algebra, particularly by Walter Taylor, extending the classic theorem of algebraic topology that $S^2$ does not admit a group structure.