# Cardinalities of Collections of Models

Let $T$ be a complete theory in a countable language (with only infinite models). Recall the spectrum function: $I(\aleph_\alpha,T)=$ the number of non-isomorphic models of $T$ of cardinality $\aleph_\alpha$. We also recall that $I^*(\aleph_\alpha,T)=|\{M\models T:|M|\leq\aleph_\alpha\}|$ (clearly, we mod-out by isomorphism).

Edit:

Note that $I^*(\aleph_\alpha,T)=|\cup_{\beta \leq \alpha}I(\aleph_\beta,T)|$

Example: Note that if $T$ is dense linear ordering without endpoints or the random graph, then $I^*(\aleph_1,T)=2^{\aleph_1}$

Furthermore, we note that under GCH, we have that if $T$ is uncountably categorical, $I^*(\aleph_{\omega_2},T)=2^{\aleph_1}$

My question is as follows: does there exist some $\kappa$ such that for all $T$ we have $I^*(\xi_T,T)=\kappa$ for some $\xi_T$? What if we assume GCH?

Thanks!

• I can't quite parse the question. – Asaf Karagila Jun 24 '14 at 17:29
• @AsafKaragila; Hope my edit helps. – Kyle Gannon Jun 24 '14 at 17:42
• Correction: $I^*(\aleph_1,DLO)=2^{\aleph_1}$, and the same for the random graph. Every countable unstable theory (and more generally, all countable theories on the wrong side of Shelah's Main Gap) have $2^\kappa$ models in all uncountable $\kappa$. – Alex Kruckman Jun 24 '14 at 20:02
• @AlexKruckman; You're right. I'll fix that. – Kyle Gannon Jun 24 '14 at 20:11
• I should also mention that I cheated a little and tweaked the definition of $I^*$ to be $I^*(\aleph_\alpha,T)=|\{M\models T: |M|<\aleph_\alpha\}|$. – Kyle Gannon Jun 26 '14 at 21:58

I finally made an easy proof of this:

From the Main Gap Theorem, we have that for any complete theory $T$ in a countable language, one of the following holds:

$$\forall \alpha>0, I(T,\aleph_\alpha)=2^{\aleph_\alpha}$$

or

$$\forall \alpha>0, I(T,\aleph_\alpha)<\beth_{\omega_1}(\alpha)$$

Let $\kappa=\beth_{\omega_1}$

Case 1: $\forall \alpha>0, I(T,\aleph_\alpha)=2^{\aleph_\alpha}$

Note that $I^-(T,\beth_\omega)=\bigcup_{\aleph_\beta < \beth_{\omega_1}}I(T,\aleph_\beta)=\bigcup_{\aleph_\beta < \beth_{\omega_1}}2^{\aleph_\beta}=\beth_\omega$

Case 2: $\forall \alpha>0, I(T,\aleph_\alpha)<\beth_{\omega_1}(\alpha)$

Note that $I^-(T,\aleph_{\beth_{\omega_1}})=\bigcup_{\beta<\omega_1}I(T,\aleph_\beta)$

Now, we have that $\bigcup_{\beta<\omega_1}\{1\}\leq \bigcup_{\beta<\omega_1}I(T,\aleph_\beta)\leq \bigcup_{\beta<\omega_1}\beth_{\omega_1}(\beta)$

$\implies \beth_{\omega_1}\leq \bigcup_{\beta<\omega_1}I(T,\aleph_\beta)\leq \bigcup_{\beta<\omega_1}\beth_{\omega_1}$

$\implies\beth_{\omega_1}\leq \bigcup_{\beta<\omega_1}I(T,\aleph_\beta)\leq \beth_{\omega_1}$

$\implies \beth_{\omega_1}=\bigcup_{\beta<\omega_1}I(T,\aleph_\beta)= I^-(T,\aleph_{\beth_{\omega_1}})$