4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ You are asking about the possible values of what is usually denoted by $I(T,\kappa)$ for a countable theory $T$ and $\kappa$ one of $\aleph_1,\aleph_2,\aleph_{\omega_1}$. The possible values of this function for $\kappa$ uncountable were studied and essentially determined by Shelah. His analysis was later completed by Hart-Hrushovski-Laskowski, see here for the precise statement. Determining the possible values of $I(T,\aleph_0)$ is still open, and amounts to solving the famous Vaught's conjecture. $\endgroup$ – Andrés E. Caicedo Feb 5 '15 at 7:03
  • $\begingroup$ Clarification, by the way, since I want to edit my answer, is $T$ just saying that there are infinitely many infinite classes, or that every equivalence class is infinite, and there are infinitely many of them? Because in the former case it might be the case that there are additional finite classes. $\endgroup$ – Asaf Karagila Feb 5 '15 at 10:36
  • $\begingroup$ @AsafKaragila I believe it's the former case, there may be some finite classes. Does this change the number of models that may satisfy T? $\endgroup$ – chibro2 Feb 5 '15 at 14:18
1
$\begingroup$

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.


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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.