Cardinals In Boolean-valued models

Question

I had two questions on Cardinals in Boolean-valued models.

First question: What is the difference between $\hat{\aleph_{\alpha}}$ and $\aleph_{\hat{\alpha}}$ ? (If I understand correctly, the first one is a "standard cardinal" in our Boolean-valued model and the second one is a cardinal in our Boolean-valued models (but not necessarily a standard one).

Second question: Could you help me to complete the following proof:

(Claim): $\mathbf{V}^{\mathbb{B}} \models \hat{\aleph_0} = \aleph_0$.

Proof. Since $x = \aleph_0$ is a $\Delta_0$-formula we have for some $u \in \mathbf{V}^{\mathbb{B}}$ we have$\mathbf{V}^{\mathbb{B}} \models \hat{u} = \hat{\aleph_0} $.

Answer 1

It's interesting, cause there are actually three variations that might show up in a statement in the forcing language:

1. $\aleph_{\check \alpha}$
2. $\check \aleph_\alpha$
3. $\aleph_\alpha$ (only makes sense if $\alpha$ is a ZFC-defined ordinal, e.g. $0,$ $\omega$, $\omega_1,$ $\omega_{\omega},$ the least $\aleph$ fixed point, etc, not an arbitrary ordinal)

I'll defer discussion of the first option till later. The second option means what it says literally. Maybe it will demystify it if we write $\aleph_\alpha=\gamma,$ since after all, $\aleph_\alpha$ is just some ordinal. Then we can say:

• $\check\aleph_\alpha$ just means $\check \gamma$... that is all.

The third choice comes up explicitly in your second question, where there's no check/hat on the $0$ in the RHS $\aleph_0.$ What is meant here is that the formula $"x=\aleph_\alpha"$ is an abbreviation for formula defining $\aleph_\alpha.$ So, $"x=\aleph_0"$ is just the formula saying $\mbox{"$x$ is the least limit ordinal"},$ or whatever of the many equivalent-over-$\sf ZFC$ definitions of $\aleph_0$ we might choose.

So, when they say $"x = \aleph_0"$ is a $\Delta_0$ formula, they mean $\mbox{"$x$ is the least limit ordinal"}$ is a $\Delta_0$ formula. Then you use the absoluteness theorem

If $\varphi(x)$ is a $\Delta_0$ formula and $a$ is any set, then $\varphi(a)$ holds if and only if $V^B\models \varphi( \check a )$

with $\varphi(x):=\mbox{"$x$ is the least limit ordinal"}$ and $a = \aleph_0.$ Since $\mbox{"$\aleph_0$ is the least limit ordinal"}$ holds, the absoluteness theorem tells us $V^B\models\mbox{"$\check \aleph_0$ is the least limit ordinal"},$ or, re-introducing the abbreviation, $V^B\models \check \aleph_0=\aleph_0.$

Now, what about $\aleph_{\check \alpha}?$ Well, any function you might define in ZFC has an analogue in the Boolean-valued model, and here the relevant function is the class function $\alpha\mapsto \aleph_\alpha.$ More formally, in ZFC we can prove some theorem that looks like

For any ordinal $\alpha,$ there is a unique cardinal $\aleph_\alpha$ such that $\varphi( \alpha,\aleph_\alpha)$ holds

where $\varphi(x,y)$ is the formula defining the class-function. So if we abbreviate that theorem as $\psi,$ we have $V^B\models \psi$ since $V^B$ satisfies ZFC. Since $"\mbox{$x$ is an ordinal}"$ is $\Delta_0,$ for any ordinal $\alpha,$ we have $V^B\models \mbox{"$\check \alpha$ is an ordinal"},$ so combining that with $\psi,$ it makes sense to use $\aleph_{\check \alpha}$ as an abbreviation inside sentences in the forcing language, and, e.g., we have $V^B\models \mbox{"$\aleph_{\check \alpha}$ is a cardinal"}.$

(Side remark: Note that in general, not every ordinal in the boolean-valued model is of the form $\check \alpha.$ Any mixture of $\check \alpha$'s is also an ordinal in $V^B$, and can be fed into $V^B$'s version of the function $\alpha \mapsto \aleph_\alpha$ to get a cardinal.)

As to the difference between $\check \aleph_\alpha$ and $\aleph_{\check \alpha},$ note that we've just shown the latter is always a cardinal in the BVM, whereas there is no reason to expect the former to be, since $\mbox{"$x$ is a cardinal"}$ is $\Pi_1,$ not $\Sigma_1$ or $\Delta_0.$ On the other hand, note that because it is $\Pi_1,$ we can say that if $V^B\models \mbox{"$\check \kappa$ is a cardinal"},$ then $\kappa$ is a cardinal.

If you've studied the standard approach to forcing, this should sound familiar. Every cardinal in the forcing extension is a cardinal in the ground model, but the converse is not necessarily the case: cardinals in the ground model can collapse in the forcing extension. This is not a coincidence. In fact we have the following theorem:

In general, for any ordinal $\alpha,$ we have $V^B \models \check \aleph_\alpha \le \aleph_{\check \alpha}.$ If $B$ satisfies the ccc, then for any ordinal $\alpha,$ we have $V^B \models \check \aleph_\alpha = \aleph_{\check \alpha}.$

So the difference between $\check\aleph_{\alpha}$ and $\aleph_{\check \alpha}$ (or lack thereof) is a reflection of $B$ collapsing (or preserving) cardinals.