Classification Theory

Results

$I(T,\kappa )$ denotes the number of isomorphism type of cardinality $\kappa$ of a theory $T$, this is also called the ‘spectrum’ of $T$. This is known as the spectrum of a theory.

The spectrum of a sentence $\phi$ of some logic $\mathcal{L}$, is the class of cardinals $\kappa$ such that $\kappa$ is the cardinality of a model of $\phi$. That is $Sp(\phi )=\{\kappa |\mathfrak{A}\models \phi \wedge |\mathfrak{A}|=\kappa \}$ An alternative definition of The Löwenheim Number (using the spectrum of a sentence). $\ell (\mathcal{L})=\text{Sup}\{\text{min(c)}|\text{c is a spectrum of }\mathcal{L}\}$.

• $I(T,\kappa )=1⟹$ $T$ has one unique model up to isomorphism, that is, $T$ is $\kappa$-categorical (all models of $T$ of cardinality $\kappa$ are isomorphic).
• By Morley’s Categoricity Theorem, if $I(T,\kappa )=1$ for $\kappa$ is an uncountable cardinal and $T$ is a theory in a countable language, then $I(T,\lambda )=1$ for all $\lambda >{\aleph }_0$.
• Shelah’s generalization, if $I(T,\kappa )=1$ for $\kappa >|\mathcal{L}|$, then $I(T,\lambda )=1$ for all $\lambda >|\mathcal{L}|$.
• The Löwenheim-Skolem Theorem $⟹$ that $I(T,\kappa )$, the number of isomorphism types of the first-order theory $T$ (in a countable language) of cardinality $\kappa$, is non-zero (i.e., $I(T,\kappa )\ge 1$) for an infinite cardinal, then it is nonzero for all infinite cardinals Note: This isn’t constant throughout cardinals, if $I(T,\kappa )=\mu$ and $I(T,\lambda )=\delta$ such that $\kappa$ and $\lambda$ are infinite cardinals and $\kappa \ne \lambda$, then you can possibly have $\mu \ne \delta$. That is, the number of isomorphism types can change depending on the cardinal. This is due to the fact that for a model $\mathcal{M}$ and two elementary extensions ${\mathcal{N}}_1⪰\mathcal{M}$ and ${\mathcal{N}}_2⪰\mathcal{M}$ such that $|{\mathcal{N}}_1|=|{\mathcal{N}}_2|$ $/⟹$ ${\mathcal{N}}_1\cong {\mathcal{N}}_2$ as the exact structure may not match, only the structure of ${\mathcal{N}}_1$ and ${\mathcal{N}}_2$ restricted to the domain $M$ of $\mathcal{M}$ will match, but not necessarily the same structure (although these structure will be elementary equivalent).
• If $I(T,\kappa )=0$, then $T$ is either inconsistent (and so has ‘void’ models, i.e., $T$ has no models) or $T$ only has finite models (categoricity and the notation $I(T,\kappa )$ is only considered for infinite cardinals, i.e., $\kappa \ge {\aleph }_0$).

Łoś–Vaught test

Łoś–Vaught test: If a first-order ${\mathcal{L}}_{\omega ,\omega }$ theory of vocabulary $\tau$ is $\kappa$-categorical for some $\kappa >|\tau |$ and has no finite models, then it is complete.

Working in the empty language (i.e., w/ just equality =), there is a single second-order sentence $\varphi$ asserting that “the universe is infinite and does not have cardinality exactly ${\aleph }_1$“. The theory $\{\varphi \}$ is categorical in every cardinality except ${\aleph }_1$ but not complete due to the existence of non-standard models from assertions further restricting the size of the universe (i.e., “the universe is countable”). $\to$ So you can find theories $T$ which are categorical in some infinite $\kappa <|\tau |$ and are not complete as it is not categorical in $\lambda \ge |\tau |$.

The Löwenheim-Skolem Theorem theorem is integral to the Łoś–Vaught test, thus, the logic must be ${\mathcal{L}}_{\omega ,\omega }$ in order to show that for models $\mathcal{M}\models T$ and $\mathcal{N}\models T$ such that they are elementary equivalent ($\mathcal{M}\equiv \mathcal{N}$), there are elementary extensions ${\mathcal{M}}^{\prime }$ and ${\mathcal{N}}^{\prime }$ such that ${\mathcal{M}}^{\prime }\cong {\mathcal{N}}^{\prime }$.

As stated above, a condition for the Łoś–Vaught test is that $T$ has no finite models as there are $\mathcal{L}$-sentence $\phi$ which states the size of the model. Furthermore, there is an $\mathcal{L}$-sentence $\varphi$ that completely characterizes the model. Consider the following cases: i) Models $\mathcal{M}\models T$ and $\mathcal{N}\models T$ such that $M<N$; ii) $|M|<|N|$ where $\mathcal{M}$ is a finite model and $\mathcal{N}$ is a infinite model For case i), then $\mathcal{M}\cong \mathcal{N}$. For case ii) there is a contradiction as $\mathcal{N}$ would have to satisfy the formula $\varphi$ characterizing $\mathcal{M}$ but it is also true that $|M|<|N|$, so they cannot be isomorphic.

Proof (Łoś–Vaught test): We proceed by contradiction, assume $T$ is a theory which is not complete and has no finite models and is $\kappa$-categorical for $\kappa >|\tau |$.

Since $T$ is not complete, there must exist some sentence $\phi$ such that $T/\models \phi$ and $T/\models \neg \phi$ (i.e., $\phi$ is an independent statement).

$⟹$ That the theories $T\cup \{\phi \}$ and $T\cup \{\neg \phi \}$ have models.

Since $T$ has no finite models, then $T\cup \{\phi \}$ and $T\cup \{\neg \phi \}$ have only infinite models.

Since these theories have infinite models, and the language is of cardinality $\kappa$, then the (Upwards) Löwenheim-Skolem Theorem can be applied.

From the Upward Löwenheim-Skolem Theorem one can prove that there are models ${\mathcal{M}}_{\phi }$ and ${\mathcal{M}}_{\neg \phi }$ of $T\cup \{\phi \}$ and $T\cup \{\neg \phi \}$, respectively, of cardinality $\kappa$.

It can be observed that the models ${\mathcal{M}}_{\phi }$ and ${\mathcal{M}}_{\neg \phi }$ disagree about the truth value of the sentence $\phi$. $⟹$ That ${\mathcal{M}}_{\phi }\ncong {\mathcal{M}}_{\neg \phi }$ as isomorphisms preserve the truth value of sentences.

Thus, $T$ is not $\kappa$-categorical. A contradiction, so $T$ must be complete by necessity.

More Results

Never Two Theorem: $I(T,{\aleph }_0)\ne 2$ i.e., Given $T$ is a first-order theory $T$ in a countable language then it cannot have exactly two isomorphism types for countably infinite models (i.e., there cannot be exactly two non-isomorphic models up to isomorphism).

Morley: If $I(T,{\aleph }_0)$ is infinite, then it must equal ${\aleph }_0$ or ${\aleph }_1$ or $2^{{\aleph }_0}$. If the continuum hypothesis is false, it is not know if $I(T,{\aleph }_0)={\aleph }_1$. The Vaught conjecture asserts that ${\aleph }_0<I(T,{\aleph }_0)<2^{{\aleph }_0}$

Morley’s Problem: For $T$ is complete first-order theory, $I(T,\kappa )$ is non-decreasing in $\kappa$ for uncountable $\kappa$ (i.e., if $\kappa$ and $\lambda$ are uncountable cardinals such that $\kappa <\lambda$, then $I(T,\kappa )\le I(T,\lambda )$).

Shelah’s (nigh) Complete Solution: For a complete first-order theory $T$ w/in a countable language, then either $I(T,2^{\kappa })$ for all uncountable cardinals $\kappa$, or $I(T,{\aleph }_{\xi })<{\beth }_{{\omega }_1}(|\xi |+{\aleph }_0)$ for all ordinals $\xi$.

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Vaughtian Pairs: For any two cardinals $\lambda >\mu >\omega$, a two cardinal $(\lambda ,\mu )$-model of a theory $T$ is a model $\mathcal{M}\models T$ such that $|M|=\lambda$ and w/ a definable subset $\varphi (\mathcal{M})$ of size $\mu$.

Two Cardinal Theorem: If a theory has a $(\lambda ,\mu )$-model, then it has a $({\omega }_1,\omega )$-model

Talk about stability in Classification Theory

References

Embeddings, Elementary Embeddings, and Isomorphisms Categoricity The Löwenheim-Skolem Theorem [Łoś–Vaught test](Łoś-Vaught Test - ProofWiki) [Spectrum of a Theory](Spectrum of a theory - Wikipedia)