What's the order type of the least good pair with that additional qualification?

Question

In this posting I'll add a further condition on good pairs which were defined in a prior posting, I'll re-iterate the definition with the added condition:

A good pair $\langle A,B \rangle$ is an ordered pair fulfilling the following conditions:

• $A,B$ are disjoint nonempty well ordered sets.
• Let $\varphi^A,\varphi^B$ be the respective initial segments (i.e., subsets of $A,B$ closed under the respective well order) that are definable in the language of set theory (+ primitive constants $A,B$) by the formula $\varphi$ relativized to the union of the respective set with its power; then: $$\varphi^A \cong \varphi^B \iff \varphi^A \neq A$$
• $|A|,|B|$ are regular.

The idea is to blow up these pairs to the inaccessible realm.

Now adding primitive constants $A,B$ to the language of ZFC, and working in ZFC + a good pair exists:

What would be the least possible order type of set $A$?

Answer 1

My previous answer was incorrect (and the comments below are no longer relevant). In fact, it's easy to see that now you do reach all the way up to inaccessibles.

To avoid confusion with previous questions, I'll call your modified notion "very good pairs."

For $\alpha$ a regular cardinal, let $reg(\alpha)$ be the ordertype of the set of regular cardinals $<\alpha$. Clearly $reg$ is injective on the regular cardinals: if $\alpha<\beta$ are regular then $reg(\alpha)<reg(\beta)$. And letting $Reg(\alpha)$ be the initial segment of $\alpha$ with ordertype $reg(\alpha)$, it's not hard to show that $Reg(\alpha)$ is a second-order-definable subset of $\alpha$.

Now suppose $(\alpha,\beta)$ is a very good pair. Then since $reg$ is injective we have $Reg(\alpha)\not\cong Reg(\beta)$. Hence we must have $Reg(\alpha)=\alpha$ - which is to say, $\alpha$ must be inaccessible.

• Note that we're getting all of this strength from the new direction of the bi-implication in the definition: $(\alpha,\beta)$ is a merely good pair if (not iff) $\alpha$ isn't second-order definable in $\beta$, and that's a phenomenon which happens pretty low down by a simple pigeonhole argument.

In fact, though, we can go much further. Let $inac, Inac$ be the analogous "inaccessible-counting" functions. Suppose $(\alpha,\beta)$ is a very good pair. We know that $\alpha$ is inaccessible, so $inac(\alpha)<ina(\alpha)+1\le inac(\beta)$ and consequently we have $Inac(\alpha)\not\cong Inac(\beta)$. This means we must have $inac(\alpha)=\alpha$. So $\alpha$ is $2$-inaccessible. And we keep going up the inaccessibility hierarchy for as long as we can keep everything nicely second-order definable!

What about an upper bound? Well, there's a second-order sentence $\pi$ such that $\gamma\models\pi$ iff $\gamma$ is "sufficiently greater than" the least very good pair (specifically: $\gamma$ needs to be bigger than the powerset of the right coordinate of some very good pair). So we trivially get - assuming that very good pairs exist in the first place - that the following ordinal is an upper bound for the least very good pair:

Let $\theta$ be the smallest ordinal such that for every second-order sentence $\pi$, if all sufficiently large ordinals satisfy $\pi$ then $\theta$ satisfies $\pi$.

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Of course this leaves open the big question:

Is the existence of very good pairs in fact consistent with ZFC at all?

I suspect the answer is yes, but at the moment I don't see it.