# “Bootstrapping” an unbounded class of inaccessible cardinals

The "richness principle" of set theory asserts roughly that "everything that happens once should happen an unbounded number of times".
An example would be the existence of an unbounded class of inaccessible cardinals.
Now there are many examples of large cardinals $$\kappa$$ whose existence guarantees an unbounded class of inaccessibles below $$\kappa$$.
So, my first question:
Is there any large cardinal such that "$$\kappa$$ exists" $$\implies$$ "There exists an unbounded class of inaccessible cardinals in V" ?

An even stronger example of this would be something like:
"There exists a large cardinal of Type A" $$\implies$$ "There exists an unbounded class of cardinals of Type A". Is there any large cardinal with this property?
I know that rank-into-rank cardinals satisfy an "upward reflection" property, but that only happens $$\omega$$-many times.
Would a Reinhardt cardinal (in ZF rather than ZFC say) reflect upwards unboundedly many times ?

• For your second question, if you consider first order expressibility in ZFC, then the answer is no: otherwise consider the second one of type $A$, and cut the universe at that level. – Mohammad Golshani Jul 22 '19 at 5:31
• If you go to ZF, then the existence of a regular cardinal ($\omega$) does not imply that there is a proper class of regular cardinals. – Asaf Karagila Jul 22 '19 at 6:11

$$\kappa$$ is superhuge if for any $$\gamma$$, there exists $$j: V\to M$$ such that $$crit(j)=\kappa,$$ $$\gamma and $${}^{j(\kappa)}M\subset M.$$ But $$j(\kappa)$$ (inaccessible in M) must be inaccessible in $$V$$ as well.

If $$\kappa$$ is extendible, then there exists a proper class of inaccessible cardinals (and even more).

• One question: Does this hold because $V_\kappa\prec_{\Sigma_3} V$? (So that "even more" could be any $\Pi_1$ notion) – Pedro Sánchez Terraf Jul 22 '19 at 12:39
• Could you elaborate a little more about this please ? Why is it true ? What do you mean by "even more" ? – Anindya Jul 23 '19 at 22:52

The maximality principle implies a sweeping uniform version of the idea you mention.

Namely, theorem 14 of that paper shows that if MP holds, then there is a proper class of inaccessible cardinals, if any; and similarly for Mahlo cardinals and indeed any large cardinal notion that is downward absolute to small-forcing grounds.

To explain, the maximality principle is the principle that any statement that is forceably necessary is already true. That is, if there is a forcing extension such that the statement is true in all further forcing extensions, then the statement is already true. This principle is expressible in modal terms by the scheme $$\Diamond\Box\varphi\to\varphi$$, and indeed, the modal logic of forcing was introduced in this paper specifically because of the connection with the maximality principle.

The point now is that if MP holds and there is an inaccessible cardinal, then there cannot be only one or a bounded number of them, because then we could collapse them all and thereby make it forcing necessarily true that there would be none; so under MP there would have to have been none to begin with. So if there is one, there must be a proper class of them.

There are three interesting types of upwards reflection. First off, reflecting upwards to an unbounded class of cardinals, or reflecting oneself upwards, or both. The other instance is when you can find a larger cardinal, but not an unbounded class. For instance, if $$\kappa$$ is $$\gamma$$-uplifting, then there is an inaccessible $$\gamma'\gt\gamma$$, and if $$\kappa$$ is superstrong, then there is some $$\gamma'\gt\kappa$$ such that $$V_\kappa\prec V_{\gamma'}$$.

If $$\kappa$$ is extendible or superhuge, and $$\lambda$$ is an ordinal then there is a proper class of $$\gamma$$ inaccessible, indescribable, measurable, Woodin, superstrong, and $$\lambda$$-supercompact. The reason for this is that, suppose $$\kappa$$ is inaccessible, and $$\alpha$$ is some ordinal. Then, if $$\eta\gt\alpha$$ and $$j: V_{\kappa+\eta+1}\prec V_{j(\kappa+\eta)+1}$$ with $$j(\kappa)\gt\eta$$, then $$V_{\kappa+\eta+1}\vDash\kappa\text{ is inaccessible}$$ and so $$V_{j(\kappa+\eta)+1}\vDash j(\kappa)\text{ is inaccessible}$$, and also $$V_{j(\kappa)+1}\in V_{j(\kappa+\eta)+1}$$, and inaccessibility is a $$\Pi_1^1$$ property of $$V_{j(\kappa)}$$.

Similarly Woodinness is a $$\Pi_1^1$$ property of $$V_{j(\kappa)}$$, and so if $$\kappa$$ is $$1$$-extendible then there is some Woodin cardinal above it. Measurability, on the other hand, is a $$\Sigma_1^2$$ property of $$V_{j(\kappa)}$$, and so if $$\kappa$$ is $$2$$-extendible, then there is a measurable cardinal above it, and full extendibility implies a proper class of measurable Woodin cardinals. Furthermore, $$\Pi_n^m$$ indescribability is a $$\Sigma_n^m$$ property of $$V_{j(\kappa)}$$, for $$n\gt 0$$ and $$m\gt 1$$, and so if $$\kappa$$ is $$n$$-extendible for $$n\gt 1$$, there is an $$n$$-indescribable, measurable, Woodin cardinal above $$\kappa$$.

Next, superstrongness is a bit trickier. Let $$\alpha$$ be some ordinal, $$E$$ an extender witnessing superstrongness, $$\eta$$ a limit $$\gt\alpha+rank(E)$$, and $$j: V_{\kappa+\eta+1}\prec V_{j(\kappa+\eta)+1}$$ with $$j(\kappa)\gt\eta$$. Then $$V_{\kappa+\eta+1}\vDash\kappa\text{ is inaccessible}$$ and so $$V_{j(\kappa+\eta)+1}\vDash j(\kappa)\text{ is superstrong}$$, and the property of being superstrong is absolute between $$V_\beta$$ for $$\beta$$ a limit.

Finally, if $$\kappa$$ is $$\lambda+1$$-extendible, as witnessed by $$j: V_{\kappa+\lambda+1}\prec V_{j(\kappa+\lambda)+1}$$, then $$j(\kappa)$$ is $$\lambda-$$supercompact, because $$P(P_\kappa(\lambda))\subseteq V_{j(\kappa+\lambda)+1}$$, and so $$D=\{X\subseteq P_\kappa(\lambda)|j"\lambda\in j(X)\}$$ witnesses the $$\lambda$$-supercompactness of $$j(\kappa)$$.

This same argument can be applied to get that if $$\kappa$$ is extendible and huge (E.g. Superhuge), then there is a proper class of huge cardinals. Here is another way to get this. Let $$P$$ be some large cardinal witnessed by some structure of limited rank, i.e. $$\Sigma_2$$. Assume $$P(\kappa)$$ and $$\kappa$$ is extendible. Let $$D$$ be some witness to $$P(\kappa)$$, and $$\eta$$ a limit $$\gt rank(D)$$. Now, as $$P(\kappa)\leftrightarrow\exists X(Q(\kappa,X))$$, where $$Q$$ is $$\Pi_1$$, we have $$V_{j(\kappa+\eta)}\vDash Q(\kappa,D)$$ and so there is a normal measure on $$\kappa$$ concentrating on those cardinals which have $$P(\lambda)$$.

This is a mirror of Reinhardt cardinals; if $$\kappa$$ is Reinhardt and $$P(\kappa)$$, then there is a normal measure concentrating on those cardinals which have $$P(\lambda)$$; if $$\kappa$$ is supercompact and $$P(\kappa)$$ is $$\Sigma_2$$, then there is a normal measure concentrating on those cardinals which have $$P(\lambda)$$. Furthermore, by if $$j: V\prec M$$ is a non-trivial elementary embedding with critical point $$\kappa$$, and $$M^\kappa\subseteq M$$, $$P(\kappa)$$ is $$\Pi_1$$, then there is a normal measure concentrating on those cardinals which have $$P(\lambda)$$ by downwards absoloutness of $$\Pi_1$$ formulas.

Now, if $$\kappa$$ is extendible, then it is $$\Sigma_3$$-reflecting. Then $$V_\kappa\vDash\text{There is a proper class of }\lambda\text{ such that }P(\lambda)$$. Therefore, as the previous assertion is $$\Pi_3$$, the same holds in $$V$$. Note that using extendibility and hugeness for a normal measure concentrating on the huge cardinals is a bit over shooting the goal; this follows from supercompactness and hugeness. Furthermore, if a cardinal is supercompact and rank-into-rank, there is a normal measure concentrating on those cardinals which are rank-into-rank, so that if $$\kappa$$ is extendible and rank-into-rank, then there is a proper class of rank-into-rank cardinals.

The other case is self-reflection. There are very few cases of self reflection. Typically, if a cardinal has some $$\Pi_n$$ property, the upwards reflection is derived from $$\Sigma_n$$-correction of $$V_\kappa$$ in $$V$$. This means that if $$\kappa'\gt\kappa$$ has that same property $$P$$, then $$V_{\kappa'}\vDash P(\kappa)$$. In fact, typically less is needed. For example, if $$\kappa$$ is extendible, then there is not necessarily a proper class of strongly unfoldable, strong, or supercompact cardinals (But there is a proper class of unfoldable cardinals, as every Ramsey cardinal is unfoldable).

However, in some instances this does not hold. Take rank-into-rank cardinals. If $$j: V_\lambda\prec V_\lambda$$ with critical point $$\kappa$$, then $$j(\kappa)$$ is also rank-into-rank, as witnessed by $$j^+(j)$$. In fact, each $$j^n(\kappa)$$ is rank-into-rank, as witnessed by $$j^{n+}(j)$$. Similar arguments work for $$j: V_{\lambda+1}\prec V_{\lambda+1}$$ and $$j: L(V_{\lambda+1})\prec L(V_{\lambda+1})$$, which are (Currently!) some of the strongest large cardinals. It is worth noting that the existence of some $$j: L(V_{\lambda+1})\prec L(V_{\lambda+1})$$ is in fact, $$\Sigma_2$$. It is equivalent to the existence of a normal non-principal $$L(V_{\lambda+1})$$-ultrafilter over $$V_{\lambda+1}$$, so that the least supercompact cardinals is larger than the least $$I0$$ cardinal, and if there is an extendible $$I0$$ cardinal, then there is a proper class of such cardinals.

Furthermore, if $$j: V\prec V$$ with critical point $$\kappa$$, then $$j(\kappa)$$ is also Reinhardt, as witnessed by $$j^+(j)$$. In fact, each $$j^n(\kappa)$$ is Reinhardt, as witnessed by $$j^{n+}(j)$$. However, it is not known if a Reinhardt cardinal need be supercompact or even $$\Sigma_2$$-reflecting, much less strongly extendible, though it must be weakly extendible. Furthermore, we can cut the universe off at the least inaccessible $$\lambda\gt\kappa_\omega(j)$$, to get $$V_\lambda$$ as a transitive model of $$ZF$$ with a Reinhardt cardinal.

Finally, we have large cardinals that self reflect upwards unboundedly many times. To my knowledge, only one such large cardinal exists. If $$\kappa$$ is super Reinhardt with target $$j(\kappa)$$ above $$\lambda$$, and super Reinhardt with target $$j'(\kappa)$$ above $$\alpha$$, then $$j(\kappa)$$ is super Reinhardt with target above $$\alpha$$, as witnessed by $$j^+(j')$$. In fact, each $$j^n(\kappa)$$ is super Reinhardt with target above $$\alpha$$, as witnessed by $$j^{n+}(j')$$. Super Reinhardt cardinals are some of largest large cardinals. Every super Reinhardt cardinals is reflecting, superhuge, and a stationary limit of such cardinals. The existence of a super Reinhardt cardinal is also significantly stronger than that of an $$I0$$ cardinal.