Suppose some theory T has countably many axioms, how many models of $T$ are there of cardinality $\aleph_1$,$\aleph_2$,$\aleph_{\omega_1}$?

Question

Setting

Let $\mathcal{L} = \{E\}$ where $E$ is a binary relation symbol. Let $T$ be the $\mathcal{L}$-theory of an equivalence relation with infinitely many infinite classes. So we see $T$ has countably many axioms. How many models of $T$ are there of cardinality $\aleph_1$,$\aleph_2$,$\aleph_{\omega_1}$?

Problem

First, I'd like to clarify that by "model $\mathcal{M}$ of cardinality k", it means the universe underlying $\mathcal{M}$ has cardinality $k$, correct?

My next confusion lies in in cardinal arithmetic more than anything else. I can see $T$ admits only one model of cardinality $\aleph_0$ up to isomorphism. But for $\aleph_1$ I do not know how to "count" the number of models. My argument is below.

so we have $$\text{cardinality of each model} \times \text{cardinality of number of models} = \aleph_1.$$ This relation maybe satisfied if \begin{align*} &(\text{cardinality of each model} = \aleph_1) ~\vee~ (\text{cardinality of number of models} = \aleph_1)\\ &~\vee~ (\text{cardinality of each model} = \aleph_1 \wedge \text{cardinality of number of models} = \aleph_1). \end{align*}

Hence we have three "classes" of models that are each isomorphic within themselves.

But this looks terribly wrong. For example, is there a notion of multiplication within and between cardinalities? Could someone teach me how to think about different cardinalities of infinities?

Edit

With regard to the aleph-omega-1 case. My problem is I am not familiar with how to count up to this size. I see we have a continuum of options for the number of equivalent classes, and a continuum of options for sizes of each class so that the result sums up to $\aleph_{\omega_1}$. But if I would like to say something more satisfying than just "the number of equivalent classes is uncountable", how would I proceed?

Could someone to explain to me how to think about sets when they are this large. All I know about $\aleph_{\omega_1}$ is that is comes after this sequence:

$$\aleph_0,\aleph_1,\ldots,\aleph_{\omega},\aleph_{\omega+1},\ldots,\aleph_{\omega_1}.$$

Hardly enough knowledge to reason with.

Answer 1

I'm assuming no finite equivalence classes are allowed.

A model $\mathcal{M}$ of this theory is basically just a family $\{E_\eta: \eta<\lambda\}$ of sets (the equivalence classes), where each $E_\eta$ has cardinality $\lambda_\eta\ge\aleph_0$.

EXERCISE: two models $\mathcal{M}=\{E_\eta: \eta<\lambda_0\}$ and $\mathcal{N}=\{F_\eta: \eta<\lambda_1\}$ are isomorphic iff for each $\theta$, $$\vert\{\eta: \vert E_\eta\vert=\theta\}\vert=\vert\{\eta: \vert F_\eta\vert=\theta\}\vert,$$ that is, if they have the same number of equivalence classes of each size. Note that each equivalence class in $\mathcal{M}$ has cardinality at most that of $\mathcal{M}$, and there are at most cardinality-of-$\mathcal{M}$-many equivalence classes in $\mathcal{M}$

So to count the models, it's enough to count the number of (numbers of equivalence classes of each side) there are. Let's look at $\aleph_1$ first.

There are two possible sizes for the equivalence classes: $\aleph_0$, and $\aleph_1$. Moreover, the number of equivalence classes of a given size must be a number in $\{0, 1, 2, . . . , \aleph_0, \aleph_1\}.$ So there are at most $\aleph_0^2=\aleph_0$ many possible models. Now, of course, we're overcounting - for instance, we can't have 17 equivalence classes of each infinite cardinality $\aleph_0$ and $\aleph_1$, since that would mean there were only finitely many equivalence classes total - but it does give us an upper bound.

EXERCISE: show that there are, in fact, $\aleph_0$-many models of $T$ of cardinality $\aleph_1$. HINT: how many models are there of cardinality $\aleph_1$ which have $\aleph_0$-many equivalence classes of size $\aleph_0$?

Okay, so let’s keep going. The upper bound we can abstract from the previous example is $$B(\kappa)=C(\kappa)^{I(\kappa)},$$ where $C(\kappa)$ is the number of cardinals less than or equal to $\kappa$, and $I(\kappa)$ is the number of infinite cardinals $\le\kappa$. Note that this is in general a poor upper bound - set $\kappa=\aleph_0$. But let’s look at it anyways:

• $B(\aleph_1)=\aleph_0^2=\aleph_0$.

• $B(\aleph_2)=\aleph_0^3=\aleph_0$.

• $B(\aleph_3)=\aleph_0^4=\aleph_0$.

• …

Since each $\aleph_n$ (for $n\in\omega$) has only countably many cardinals below it - namely, $0, 1, 2, . . . , \aleph_0, \aleph_1, . . . , \aleph_n$ - $B(\aleph_n)$ is always just $\aleph_0$. EXERCISE: this bound is sharp - that is, the number of models of $T$ of size $\aleph_n$ is exactly $\aleph_0$, for each $n\in\omega$.

What about something nastier, like $\aleph_{\omega_7+5}$? Well, $I(\aleph_{\omega_7+5})=\vert \omega_7+6\vert$ - in general, $I(\aleph_\alpha)=\vert \alpha+1\vert$ - so $I(\aleph_{\omega_7+5})=\omega_7$. Meanwhile, note that $C(\kappa)=I(\kappa)+\aleph_0$, so in this case we have $C(\aleph_{\omega_7+5})=I(\aleph_{\omega_7+5})=\omega_7$ - so we’re left with an upper bound of $B(\aleph_{\omega7+5})=\omega_7^{\omega_7}=2^{\omega_7}$.

EXERCISE: show that this is optimal.

I’ll leave the general pattern for you to figure out, but hopefully this will get you started.

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CHALLENGE: what about something "super-singular" like $\aleph_{\omega_{\omega_1}}$?