# Weak Indestructibility and Reflection

James Holland

(2022-04-12)

###### Abstract

We establish an equiconsistency between (1) weak indestructibility for all
$\kappa+2$-degrees of strength for cardinals $\kappa$ in the presence of a
proper class of strong cardinals, and (2) a proper class of cardinals that are
strong reflecting strongs. We in fact get weak indestructibility for degrees
of strength far beyond $\kappa+2$, well beyond the next inaccessible limit of
measurables (of the ground model). One direction is proven using forcing and
the other using core model techniques from inner model theory. Additionally,
connections between weak indestructibility and the reflection properties
associated with Woodin cardinals are discussed. This work is a part of my
upcoming thesis [thesis].

###### Contents

  1. 1 Introduction
    1. 1.A Background
    2. 1.B Limiting results
  2. 2 Strongs Reflecting Strongs in the Core Model
  3. 3 The Forcing Direction
    1. 3.A Forcing UWISS
    2. 3.B A proper class of strongs
  4. 4 Small Side Results

##  1 Introduction

Theorems like compactness for first-order logic tell us there is a great
degree of ambiguity for mathematical and set theoretic concepts. But much of
this disappears if we restrict ourselves to only considering models that, in
some sense, properly interpret membership (i.e. that are transitive). But this
does not get rid of all ambiguity. The advent of _forcing_ as a method of set
theory has revealed that even with this restriction, there’s a great amount of
ambiguity not about the membership relation, but instead about what sets
exist.

One response to this is to come up with axioms that help sharpen our
conception of the universe so as to become immune to methods of forcing in
some sense, and say more definitively what exists. For example, the axiom of
“$\mathrm{V}=\mathrm{L}$” yields models that are immune to forcing. Under
certain assumptions, the core model $\mathrm{K}$ has this property too.
Forcing axioms can be considered in a similar way, essentially stating that
we’ve already forced as much as we can.

The consistency of most of these axioms is unfortunately not something we can
know, as they often carry with them _large cardinal_ hypotheses. Such
hypotheses are used as a standard measure of the strength of statements: if we
want to know how strong a statement is, we show it is equiconsistent with some
large cardinal axiom. The question then becomes to what extent can these large
cardinals be immune to forcing?

Generally speaking, we cannot ensure large cardinal properties are immune to
forcing— _indestructible_ —because we may simply add via forcing a bijection
between the cardinal $\kappa$ and $\aleph_{0}$. But if we restrict our forcing
to be, say, smaller than $\kappa$, we can get preservation of some properties,
especially those involving elementary embeddings since measures in the ground
model generate measures in the generic extension. Large forcings can still
pose a problem, but the answer isn’t as clear if we restrict our attention to
large posets that don’t affect anything “small”. These are the notions we will
investigate here, and mostly we will consider $<\kappa$-strategically closed,
$\leq\kappa$-distributive posets.

This topic starts with Laver’s preparation for making a supercompact cardinal
$\kappa$ indestructible (by $<\kappa$-directed closed forcings) in
[laverprep]. Since then, there has been a great deal of literature about the
limits of this for other cardinals and varying degrees of strength.111By a
_degree of strength_ of a caridnal $\kappa$, I mean an ordinal $\rho$ such
that $\kappa$ is $\rho$-strong, defined with strongdef. Ignoring for the
moment what exactly a lot of this terminology means, a short selection of
results in this area is the following.

  * •

[apterhamkins] explores making all degrees of supercompactness and strength
indestructible while there is a supercompact.

  * •

[hamkinssuper] explores the ways in which indestructible cardinals can be made
destructible and subsequently resurrected.

  * •

[aptersargsyan] explores a weaker version of indestructibility for strength
while there is a strong cardinal, and gets an equiconsistency result:
universal weak indestructibility (while there is a strong cardinal) is
equiconsistent with a _hyperstrong_ cardinal.

  * •

[apterapplications] further explores this weaker version of indestructibility
for strength to get (weak) indestructibility for lots of (strong) supercompact
cardinals, and again establishes an equiconsistency for this indestructibility
for many strongs from a hyperstrong cardinal (a proper class if we cut off the
universe at the hyperstrong).

This work further explores [aptersargsyan] and [apterapplications]. Generally
speaking, these works focus on exploring indestructibility for large degrees
of strength and supercompactness in the presence of many strong or
supercompact cardinals. Ostensibly, this is a harder task than showing weak
indestructibility for smaller degrees of strength, simply because a strong
cardinal’s degrees of strength are arbitrarily large. But instead, this work
shows that to get universal weak indestructibility for small degrees of
strength and supercompactness, we actually need a large _increase_ in
consistency strength, much more than a proper class of hyperstrong or
hypercompact cardinals, just one of which is used in [aptersargsyan] and
[apterapplications] to get (many) weakly indestructible strongs and
supercompacts.

There is a balance to the amount of indestructibility one can have, and
generally speaking, indestructibility for large degrees of strength will
conflict with indestructibility for small degrees of strength. The largest
sense of “small” degrees is degrees of strength below the next cardinal
$\kappa$ that is $\kappa+2$-strong, and the smallest sense of “large” degrees
is degrees slightly above such a cardinal. So the universal indestructibility
results of [aptersargsyan] and [apterhamkins] critically proceed by making
sure there is no measurable above the resulting strong or supercompact
cardinal, thus being incompatible with the existence of multiple strong or
supercompact cardinals. The work in [apterapplications] considers multiple
strong and supercompact cardinals with indestructibility, but as a
consequence, ignores indestructibility for smaller degrees of strength.

The following theorem was the main goal of [aptersargsyan].

######  Theorem 1 • 1 (Apter–Sargsyan).

The following are equiconsistent with ZFC:

  1. 1.

There is a hyperstrong cardinal.

  2. 2.

There is a strong cardinal and universal weak indestructibility for all
degrees of strength holds.

Hence a hyperstrong cardinal is able to yield indestructibility for all small
degrees of strength, and produce a strong cardinal (with also indestructible
strength). Considering the balance of indestructible degrees of strength, a
natural strengthening of (2) is having a proper class of strong cardinals
while still ensuring universal weak indestructibility for small degrees of
strength. A natural guess at the consistency strength of this is the existence
of many hyperstrong cardinals. But this is insufficient, and indeed the
consistency strength of this is strictly larger than a proper class of
hyperstrong cardinals. The following is one of the main results of the
document that ensures this increase in strength, assuming for the moment that
a single strong reflecting strongs cardinal (with a strong above it) gives the
consistency of a proper class of hyperstrongs.

######  Theorem 1 • 2 (Main Goal).

The following are equiconsistent with ZFC:

  1. 1.

There is a proper class of strongs reflecting strongs.

  2. 2.

There is a proper class of strong cardinals and weak indestructibility for any
cardinal $\kappa$’s $\kappa+2$-strength.

  3. 3.

There is a proper class of strong cardinals and weak indestructibility for any
cardinal $\kappa$’s $\lambda$-strength where $\lambda$ is below the next
measurable limit of measurables above $\kappa$.

In my thesis [thesis], we also establish a similar result for
supercompactness, and the following just by continuing the preparation from
maingoal through to a Woodin cardinal, and a separate preparation where we
ignore small degrees of strength.

######  Corollary 1 • 3 (Side Result).

The following are equiconsistent with ZFC:

  1. 1.

There is a Woodin cardinal.

  2. 2.

There is a Woodin cardinal $\delta$ such that weak indestructibility for any
$\kappa$’s $\kappa+2$-strength holds below $\delta$.

  3. 3.

There is a Woodin cardinal $\delta$ such that every $<\delta$-strong cardinal
has weakly indestructible $<\delta$-strength.

This gives the consistency of a large number of weakly indestructible strong
cardinals from a Woodin cardinal. Moreover, it tells us that weak
indestructibility for large degrees of strength and for small degrees of
strength below a Woodin cardinal are equiconsistent. So while weak
indestructibility for small degrees (with, say, two strong cardinals) is
strictly stronger than for a proper class of weakly indestructible strong
cardinals, this difference levels out as we approach a Woodin.

###  1.A Background

Throughout we will use the following notation.

######  Definition 1.A • 1.

  * •

For $\delta\in\mathrm{Ord}$, write $\delta^{+\sharp}$ for the least measurable
cardinal $\kappa>\delta$.

  * •

Write $\delta^{+\sharp\sharp}$ for the least cardinal $\kappa>\delta$ that is
$\kappa+2$-strong.

  * •

Write $\delta^{+\P}$ for the least strong cardinal $\kappa>\delta$.

  * •

For a transitive model of set theory, $\mathrm{M}$, and
$\lambda\in\mathrm{Ord}^{\mathrm{M}}$, we write $\mathrm{M}\mid\lambda$ for
$\mathrm{V}_{\lambda}^{\mathrm{M}}$, the $\lambda$-th stage of the cumulative
hierarchy in $\mathrm{M}$.

  * •

For a poset $\mathbb{R}\in\mathrm{V}$, we write $\mathrm{V}^{\mathbb{R}}$ for
an arbitrary generic extension of $\mathrm{V}$ by $\mathbb{R}$, occasionally
working below some particular condition in $\mathbb{R}$.

######  Definition 1.A • 2.

Let $\kappa$ be a cardinal, $\varphi$ a property of cardinals, and $\psi$ a
property of posets.

  * •

$\kappa$ has $\varphi$ as _indestructible_ by posets with $\psi$ iff

$\forall\mathbb{P}\text{ a poset}\
(\psi(\mathbb{P})\rightarrow\mathbb{P}\Vdash\text{``}\varphi(\check{\kappa})\text{''})\text{.}$

  * •

$\kappa$ has $\varphi$ as _weakly indestructibile_ by posets with $\psi$ iff
it’s indestructible by $\leq\kappa$-distributive posets with $\psi$.

So the usual Laver indestructibility refers to indestructibility for all
degrees of supercompactness of a single (supercompact) cardinal $\kappa$ by
$<\kappa$-directed closed posets. For strength, we usually consider weak
indestructibility by $<\kappa$-strategically closed posets, and thus
indestructibility $<\kappa$-strategically closed, $\leq\kappa$-distributive
posets. $<\kappa$-strategic closure is a significant weakening of
$<\kappa$-directed closure and is defined as follows.

######  Definition 1.A • 3.

Let $\mathbb{P}$ be a poset and $\kappa,\lambda$ ordinals. The game
$\mathcal{G}_{\mathbb{P}}^{\lambda}$ is the two person game of length
$\leq\lambda$

${}\qquad\begin{array}[]{r *{20}{c}}\text{{{\boldmath
I}}:}&p_{0}=\textbf{1}^{\mathbb{P}}&&p_{2}&&\cdots&&p_{\omega}&&p_{\omega+2}&&\cdots\\\
\text{{{\boldmath
II}}:}&&p_{1}&&p_{3}&&\cdots&&p_{\omega+1}&&\cdots\text{,}\end{array}$

where $p_{\alpha}\leq^{\mathbb{P}}p_{\beta}$ for $\alpha\geq\beta$, and
assuming such a $p_{\alpha}$ exists, I plays $p_{\alpha}$ for even
$\alpha<\lambda$ (including limit $\alpha$) and II plays
$p_{\alpha}\in\mathbb{P}$ for odd $\alpha<\lambda$. I wins iff players play on
each turn: the resulting sequence of $p_{\alpha}$s has length $\lambda$.

  * •

$\mathbb{P}$ is _$\leq\kappa$ -strategically closed_ iff I has a winning
strategy in $\mathcal{G}_{\mathbb{P}}^{\kappa+1}$.

  * •

$\mathbb{P}$ is _$\kappa$ -strategically closed_ iff I has a winning strategy
in $\mathcal{G}_{\mathbb{P}}^{\kappa}$.

  * •

$\mathbb{P}$ is _$ <\kappa$-strategically closed_ iff $\mathbb{P}$ is
$\leq\alpha$-strategically closed for all ordinals $\alpha<\kappa$.

Basically whereas $<\kappa$-closure ensures we can always extend a
$\leqslant^{\mathbb{P}}$-decreasing sequence $\langle
p_{\alpha}:\alpha<\lambda\rangle$ whenever $\lambda<\kappa$,
$\kappa$-strategic closure only gives control over half of the sequence,
relying on I’s strategy to extend at limits.

Despite their differences, the similarities between $<\kappa$-closed and
$\kappa$-strategically closed preorders are enough to show to allow some
arguments about $<\kappa$-closure to go through about $\kappa$-strategic
closure. For example, $\kappa$-strategically closed preorders are
$\leq\kappa$-distributive by basically the same proof as with full closure.
There are many other common results in iterated forcing where using strategic
closure vs. full closure makes no difference [cummings].

######  Definition 1.A • 4.

Let $\kappa$ be a cardinal. A preorder $\mathbb{P}$ is _$
<\kappa$-distributive_ iff for every collection $\mathcal{D}$ of open, dense
sets of $\mathbb{P}$, if $|\mathcal{D}|<\kappa$ then
$\bigcap\mathcal{D}\neq\emptyset$ is open, dense. We may similarly define
$\leq\kappa$-distributive.

######  Corollary 1.A • 5.

For any infinite cardinal $\kappa$, if $\mathbb{P}$ is $<\kappa$-strategically
closed, then $\mathbb{P}$ is $<\kappa$-distributive.

It’s not hard to see that if $\kappa$ is measurable, its $\kappa+1$-strength
is weakly indestructible. So easy models of universal indestructibility for
small degrees of strength are those with no large cardinals, or those with
only measurables. Obviously if a poset _isn’t_ $<\kappa$-distributive, it can
collapse the strength of $\kappa$, e.g. through a trivial collapse
$\mathrm{Col}(\omega,\kappa)$.

######  Corollary 1.A • 6.

Let $\kappa$ be measurable. Therefore $\kappa$’s measurability (i.e. its
$\kappa+1$-strength) is weakly indestructible.

###### Proof.

If $U\in\mathrm{V}$ is a measure and $\mathbb{P}\in\mathrm{V}$ is a
$\leq\kappa$-distributive poset, then
$U\in\mathrm{V}\subseteq\mathrm{V}^{\mathbb{P}}$ is still a measure in
$\mathrm{V}^{\mathbb{P}}$.∎

The rest of this document will assume familiarity with strong cardinals and
their fundamental properties.

######  Definition 1.A • 7.

  * •

A cardinal $\kappa$ is _$\lambda$ -strong_ iff there is an elementary
embedding $j:\mathrm{V}\rightarrow\mathrm{M}$ (as a class of $\mathrm{V}$)
with $\operatorname{cp}(j)=\kappa$, $j(\kappa)>\lambda$, and
$\mathrm{V}\mid\lambda=\mathrm{M}\mid\lambda$.

  * •

A cardinal is _strong_ iff it is $\lambda$-strong for every
$\lambda\in\mathrm{Ord}$.

  * •

We call embeddings (extenders) witnessing $\lambda$-strength _$\lambda$
-strong embeddings (extenders)_.

Beyond their definition, strong cardinals are useful for the following
reflection principle.

######  Lemma 1.A • 8 ($\Sigma_{2}$-reflection).

Let $\kappa$ be a strong cardinal. Suppose
$\mathrm{V}\mid\delta\vDash\text{``}\varphi(\vec{x})\text{''}$ for some
formula $\varphi$ and parameters $\vec{x}\in\mathrm{V}\mid\kappa$. Therefore
there are unboundedly many $\alpha<\kappa$ such that
$\mathrm{V}\mid\alpha\vDash\text{``}\varphi(\vec{x})\text{''}$.

###### Proof.

Let $j:\mathrm{V}\rightarrow\mathrm{M}$ be a $\delta$-strong embedding so that
$j(\vec{x})=\vec{x}$, $\delta<j(\kappa)$, and
$\mathrm{M}\mid\delta=\mathrm{V}\mid\delta\vDash\text{``}\varphi(\vec{x})\text{''}$.
In $\mathrm{M}$, there is then some level of the cumulative hierarchy below
$j(\kappa)$ that satisfies $\varphi(j(\vec{x}))$ and so by elementarity, in
$\mathrm{V}$ there’s a level $\mathrm{V}\mid\alpha$ of the cumulative
hierarchy with $\operatorname{rank}(\vec{x})<\alpha<\kappa$ satisfying
$\varphi(\vec{x})$. Because we could have thrown in any fixed ordinal
$\beta<\kappa$ as a useless parameter into $\vec{x}$, the idea above gives an
$\alpha>\beta$ with this property. ∎

We now establish some definitions related to [aptersargsyan], explaining
aptsargmain.

######  Definition 1.A • 9.

Let $\kappa$ and $\alpha>0$ be ordinals. We define by transfinite recursion
what it means for $\kappa$ to be _$\alpha$ -hyperstrong_.

  * •

$\kappa$ is _$0$ -hyperstrong_ iff $\kappa$ is strong.

  * •

$\kappa$ is _$ <\alpha$-hyperstrong_ iff $\kappa$ is $\xi$-hyperstrong for
every $\xi<\alpha$.

  * •

$\kappa$ is _$\alpha$ -hyperstrong_ iff for any $\lambda>\kappa$, there is an
extender $E$ giving an ultrapower embedding
$j_{E}:\mathrm{V}\rightarrow\mathrm{M}$ with
$\operatorname{cp}(j_{E})=\kappa$, $j_{E}(\kappa)>|\mathrm{V}\mid\lambda|$,
$\mathrm{V}\mid\lambda\subseteq\mathrm{M}$, and
$\mathrm{M}\vDash\text{``}\kappa\text{ is
}<\alpha\text{-hyperstrong}\text{''}$.

  * •

$\kappa$ is _hyperstrong_ iff $\kappa$ is $\xi$-hyperstrong for every ordinal
$\xi$.

We will work with hyperstrongs mostly to show that they are insufficient in
establishing the main result of maingoal. We will also make use of strongs
reflecting strongs. We will discuss the interaction between these two
definitions in the next subsection.

######  Definition 1.A • 10.

Let $\kappa$ be a cardinal and $\lambda$ an ordinal.

  * •

We say $\kappa$ is _$\lambda$ -strong reflecting strongs ($\lambda$-_srs_)_
iff there’s a $\lambda$-strong $(\kappa,\lambda)$-extender $E$ giving an
elementary
$j_{E}:\mathrm{V}\rightarrow\mathrm{M}=\operatorname{Ult}_{E}(\mathrm{V})$
with

$\\{\xi<\lambda:\xi\text{ is
strong}\\}=\\{\xi<\lambda:\mathrm{M}\vDash\text{``}\xi\text{ is
strong}\text{''}\\}\text{.}$

  * •

We call such an embedding (extender) a _$\lambda$ -srs embedding
(_extender_)_.

  * •

We say $\kappa$ is _srs_ iff $\kappa$ is $\lambda$-srs for every
$\lambda>\kappa$; and we say $\kappa$ is $<\lambda$-srs if $\kappa$ is
$\alpha$-srs for each $\alpha<\lambda$.

For the most part, we only care about strongs reflecting strongs when they
have strongs above them since if there is no strong above $\kappa$, $\kappa$
is srs iff $\kappa$ is $1$-hyperstrong. But under the not-uncommon assumption
of a strong above, this is a strengthening of hyperstrong cardinals both in
the sense of being hyperstrong and in having a (much) higher consistency
strength. The (2) implies (1) direction of maingoal then tells us that a
proper class of hyperstrong cardinals is insufficient to ensure universal weak
indestructibility for very small degrees of strength with a proper class of
strongs.

For now, we end this subsection with the following definition.

######  Definition 1.A • 11.

_UWISS_ _(Universal weak indestructibility for small degrees of strength)_ is
the proposition that for all $\kappa$, if $\kappa$ is $\kappa+2$-strong, then
$\kappa$’s $\kappa+2$-strength is weakly indestructible by
$<\kappa$-strategically closed posets.

This will be equiconsistent with every $\rho$-strong $\kappa$ having weakly
indestructible $\rho$-strength for $\rho$ below the next measurable limit of
measurables above $\kappa$.

###  1.B Limiting results

The limit imposed on universal indestructibility was explored in
[apterhamkins] for full indestructibility, but we may consider weak
indestructibility in a similar way. The main limiting result is the following
with the proof adapted from [apterhamkins] and the Superdestruction Theorem
III from [hamkinssuper].

######  Result 1.B • 1.

Assume GCH.222Note that the assumption of GCH here isn’t much of a problem,
since in the context of indestructibility by a large class of posets, one can
just force the appropriate instances with Cohen forcing. More precisely, for
the theorem to go through, we merely need $\lambda=\kappa^{+\sharp\sharp}$ to
have $2^{2^{\lambda}}=\lambda^{++}$ or else we must consider a
$\lambda>\kappa$ that is $\gamma$-strong where
$2^{2^{\lambda}}=\lambda^{+\gamma}$. Suppose $\kappa$ is strong and this is
not destroyed by $\mathrm{Add}(\kappa^{+},1)$. Suppose
$\kappa^{+\sharp\sharp}$ exists. Therefore, there are arbitrarily large
$\delta<\kappa$ whose $\delta+2$-strength is destroyed by
$\mathrm{Add}(\delta^{+},1)$.

###### Proof.

Let $A$ be $\mathrm{Add}(\kappa^{+},1)$-generic over $\mathrm{V}$. By
hypothesis, $\kappa$ is still strong in $\mathrm{V}[A]$ and so we get a
sufficiently strong embedding $j:\mathrm{V}[A]\rightarrow\mathrm{M}[j(A)]$
with $\operatorname{cp}(j)=\kappa$ so that $\lambda=\kappa^{+\sharp\sharp}$ is
still $\lambda+2$-strong in $\mathrm{M}[j(A)]$.

######  Claim 1.

$\lambda$’s $\lambda+2$-strength is destroyed by $\mathrm{Add}(\lambda^{+},1)$
in $\mathrm{V}[A]$.

###### Proof.

Suppose not. Let $G$ be $\mathrm{Add}(\lambda^{+},1)$-generic over
$\mathrm{V}[A]$ and let $i:\mathrm{V}[A*G]\rightarrow\mathrm{N}[i(A*G)]$ be
$\lambda+2$-strong with $\operatorname{cp}(i)=\lambda$. Since
$A\subseteq\kappa<\operatorname{cp}(i)$, $i(A*G)=A*i(G)$. Without loss of
generality, $i$ is generated by extenders such that $\mathrm{N}[A*i(G)]$ is
closed under $\lambda$-sequences of $\mathrm{V}[A*G]$.

By gap forcing [GapForcing],
$i\upharpoonright\mathrm{V}:\mathrm{V}\rightarrow\mathrm{N}$ is a class of
$\mathrm{V}$, is closed under $\lambda$-sequences, and is still
$\lambda+2$-strong. It follows that $G\subseteq\lambda^{+}$ is in
$\mathrm{H}_{\lambda^{++}}^{\mathrm{V}[A*G]}=\mathrm{H}_{\lambda^{++}}^{\mathrm{N}[A*i(G)]}$
and hence in $\mathrm{N}[A*i(G)]$, but wasn’t added by
$i(\mathrm{Add}(\lambda^{+},1))$ by $\leq i(\lambda^{+})$-distributivity in
$\mathrm{N}[A]$. Thus $G=\dot{G}_{A}\in\mathrm{N}[A]$ for some
$\mathrm{Add}(\kappa^{+},1)$-name $\dot{G}$ which can be thought of as a
function from $(\lambda^{+})^{\mathrm{V}}=(\lambda^{+})^{\mathrm{N}}$ to
antichains of $\mathrm{Add}(\kappa^{+},1)$, and is therefore in the
hereditarily $<(\lambda^{++})^{\mathrm{N}}=(\lambda^{++})^{\mathrm{V}}$-sized
sets
$\mathrm{H}_{\lambda^{++}}^{\mathrm{N}}=\mathrm{H}_{\lambda^{++}}^{\mathrm{V}}$.
The ability to consider all of this in $\mathrm{V}$ means $G\in\mathrm{V}[A]$,
a contradiction.∎

Without loss of generality, $\mathrm{M}[j(A)]$ has enough agreement with
$\mathrm{V}[A]$ to witness this fact as well. Since
$\kappa<\lambda<j(\kappa)$, this is reflected down into $\mathrm{V}[A]$: there
are arbitrarily large $\delta<\kappa$ such that $\delta$’s $\delta+2$-strength
is destroyed by $\mathrm{Add}(\delta^{+},1)$ in $\mathrm{V}[A]$. It follows
that $\mathrm{Add}(\kappa^{+},1)*\mathrm{Add}(\delta^{+},1)$ is
$<\delta$-directed closed and $\leq\delta$-distributive in $\mathrm{V}$ but
destroys $\delta$’s $\delta+2$-strength. Since $\mathrm{Add}(\kappa^{+},1)$ is
sufficiently distributive, it can’t add to nor destroy $\delta$’s
$\delta+2$-strength. So it must be that $\mathrm{Add}(\delta^{+},1)$ destroyed
$\delta$’s $\delta+2$-strong∎

And this result generalizes to larger degrees of strength with the same proof.

######  Corollary 1.B • 2.

Assume GCH. Let $\mu\in\mathrm{Ord}$. Suppose $\kappa$ is strong, and this is
not destroyed by $\mathrm{Add}(\kappa^{+\mu},1)$. Suppose there is a
$\lambda>\kappa^{+\mu+1}$ that is $\lambda+\mu+1$-strong. Therefore

  1. 1.

There are unboundedly many $\delta<\kappa$ such that $\delta$ is
$\delta+\rho$-strong for some $\rho$, but this is destructible by
$\mathrm{Add}(\delta^{+\rho},1)$.

  2. 2.

If $\mu<\kappa$, then there are arbitrarily large $\delta<\kappa$ such that
$\delta$’s $\delta+\mu+1$-strength is destructible by
$\mathrm{Add}(\delta^{+\mu},1)$.

In particular, one cannot have universal weak indestructibility for _all_
degrees of strength with two strong cardinals. This also easily generalizes to
supercompactness in place of strength.

The contrapositive to hamkinsoopsy details one aspect of what goes wrong if
one naïvely tries to continue the preparation from either [apterhamkins] or
[aptersargsyan].

######  Corollary 1.B • 3.

Assume GCH. Suppose $\kappa$ is strong and $\kappa^{+\sharp\sharp}$ exists.
Suppose every $\delta<\kappa$ that is $\delta+2$-strong has this
indestructible by $\mathrm{Add}(\delta^{+},1)$. Therefore $\kappa$’s strength
is destroyed by $\mathrm{Add}(\kappa^{+},1)$.

All of this is just to justify that one simultaneously cannot have UWISS with
any reasonable amount of indestructibility for any large degrees of strength.
Moreover, hamkinsoopsy and hamkinsoopsyv2 aren’t unique to weak
indestructibility as noted in the original Theorems 10, 11, and 12 of
[apterhamkins], which hold for full indestructibility. Really what
hamkinsoopsyv2 and hamkinsoopsy tell us is that the only way to ensure
universal indestructibility for _all_ degrees of strength with a strong
cardinal $\kappa$ is to ensure $\kappa^{+\sharp\sharp}$ doesn’t exist. And
this is exactly what [apterhamkins] and [aptersargsyan] do.

Let us now show why—assuming the equiconsistency of maingoal—hyperstrongs are
insufficient to establish UWISS with a proper class of (weakly destructible)
strongs. To do this, we need to unfortunately look at the fine details of
calculating hyperstrongs and strongs reflecting strongs inside the cumulative
hierarchy. A straightforward, somewhat tedious induction gives the following,
recalling hyperstrongdef for the definition of a hyperstrong cardinal.

######  Corollary 1.B • 4.

Fix $\kappa,\alpha\in\mathrm{Ord}$ with $\kappa$ infinite. Therefore the
following are equivalent.

  1. 1.

$\kappa$ is $<\alpha$-hyperstrong.

  2. 2.

for all limit $\delta\geq|\kappa+\alpha|^{+}$,
$\mathrm{V}\mid\delta\vDash\text{``}\kappa\text{ is
}<\alpha\text{-hyperstrong}\text{''}$.

######  Lemma 1.B • 5.

Let $\kappa$ be strong such that $\kappa^{+\P}$ exists. Therefore,

  1. 1.

If $\kappa$ is $<\kappa^{+\P}$-hyperstrong, then $\kappa$ is hyperstrong.

  2. 2.

If $\mathrm{V}\mid{\kappa^{+\P}}\vDash\text{``}\kappa\text{ is
hyperstrong}\text{''}$, then $\kappa$ is $<\kappa^{+\P}$-hyperstrong and hence
hyperstrong by (1).

###### Proof.

  1. 1.

Suppose $\kappa$ is not hyperstrong. By reflection,

$\mathrm{V}\mid\delta\vDash\text{``}\exists\gamma\ (\kappa\text{ is not
}\gamma\text{-hyperstrong})\wedge\text{there is no largest cardinal}\text{''}$

for some $\delta>\alpha$. By Σ2ref, we get arbitrarily large
$\delta<{\kappa^{+\P}}$ such that ($*$) holds. Since there is no largest
cardinal in $\mathrm{V}\mid\delta$, such $\delta$ satisfy
$\delta>|\kappa+\gamma|^{+}$ for any $\gamma$ as in ($*$). strhypstr therefore
implies $\kappa$ is not $\gamma$-hyperstrong for some $\gamma<{\kappa^{+\P}}$,
a contradiction.

  2. 2.

Suppose $\kappa$ is not $\alpha$-hyperstrong for some $\alpha<{\kappa^{+\P}}$.
This fact is reflected to some level $\mathrm{V}\mid\delta$, and by Σ2ref, to
arbitrarily large initial segments $\mathrm{V}\mid\delta$ for
$\delta<{\kappa^{+\P}}$. In particular, we get some $\delta_{0}$ with
$|\kappa+\alpha|^{+}<\delta_{0}<{\kappa^{+\P}}$ such that
$\mathrm{V}\mid{\delta_{0}}\vDash$ “$\kappa$ is not $\alpha$-hyperstrong”. But
applying strhypstr in $\mathrm{V}\mid{{\kappa^{+\P}}}\vDash\textup{ZFC}$, by
the hypothesis,
$(\mathrm{V}\mid{{\kappa^{+\P}}})\mid\delta=\mathrm{V}\mid\delta\vDash\text{``}\kappa\text{
is }\alpha\text{-hyperstrong}\text{''}$ for _all_
$\delta>|\kappa+\alpha|^{+}$, a contradiction with the hypothesis on
$\delta_{0}$. Hence $\kappa$ must be $\alpha$-hyperstrong for each
$\alpha<{\kappa^{+\P}}$, and so hyperstrong by (1).∎

This lemma allows us to show that the existence of an srs (with a strong above
it) is strictly stronger than a proper class of hyperstrongs.

######  Result 1.B • 6.

Let $\kappa$ be srs such that $\kappa^{+\P}$ exists. Therefore $\kappa$ is
hyperstrong. In fact, any single ${\kappa^{+\P}}$-srs embedding witnesses all
degrees of hyperstrength of $\kappa$. Moreover, in $\mathrm{V}\mid\kappa$,
there is a proper class of hyperstrongs.

###### Proof.

Let $\kappa$ be srs. $\kappa$ is already $0$-hyperstrong. There is then a
${\kappa^{+\P}}+1$-srs embedding $j:\mathrm{V}\rightarrow\mathrm{M}$ with
$\operatorname{cp}(j)=\kappa$. We proceed by induction on $\alpha$ to show
$\kappa$ is $<\alpha$-hyperstrong in $\mathrm{V}$ and $\mathrm{M}$, with the
base case of $\alpha=1$ already true. The limit case is trivial, so we
consider only the successor case: showing $\alpha$-hyperstrength in
$\mathrm{V}$ and $\mathrm{M}$. By srsishypercl (1), we may assume
$\alpha<{\kappa^{+\P}}$.

Suppose $\kappa$ is $<\alpha$-hyperstrong in $\mathrm{V}$ and $\mathrm{M}$ so
that $\kappa$ is $\alpha$-hyperstrong in $\mathrm{V}$. If there is some
$\lambda$ with no $\lambda$-strong extender in $\mathrm{M}$ witnessing
$\alpha$-hyperstrength of $\kappa$, then by reflection, some
$\mathrm{M}\mid\delta$ (correctly) satisfies there’s no such extender. By
Σ2ref, we can assume $\delta,\lambda<{\kappa^{+\P}}$ so that
$\mathrm{M}\mid\delta=\mathrm{V}\mid\delta$ also has no such extender.
strhypstr tells us $\mathrm{V}$ must satisfy $\kappa$ isn’t
$\alpha$-hyperstrong, a contradiction.

We also can show that $\kappa$ is the limit of hyperstrongs, giving
$\mathrm{V}\mid\kappa$ as a model of a proper class of hyperstrongs. Suppose
not: $\kappa$ is the least hyperstrong above some $\lambda$. By elementarity,
$j(\kappa)$ is still the least hyperstrong above $j(\lambda)=\lambda$ in
$\mathrm{M}$ but
$\mathrm{V}\mid{{\kappa^{+\P}}}={\mathrm{M}}\mid{\kappa^{+\P}}\vDash\text{``}\kappa\text{
is hyperstrong}\text{''}$. So by srsishypercl (2), $\kappa$ is hyperstrong in
$\mathrm{M}$. But $\lambda<\kappa<j(\kappa)$ contradicts that $j(\kappa)$ is
the least hyperstrong above $\lambda$ in $\mathrm{M}$.∎

##  2 Strongs Reflecting Strongs in the Core Model

We begin with proving the “easy” direction of maingoal through core model
techniques. We show (2) implies (1), meaning that UWISS with a proper class of
strongs results in a proper class of srs cardinals. Firstly, note that a
Woodin cardinal implies the existence of a proper class of srs cardinals by an
easy proof.

######  Lemma 2 • 1.

The existence of a Woodin cardinal implies
$\operatorname{Con}(\textup{ZFC}\mathrel{+}$ “There is a proper class of srs
cardinals”$)$.

###### Proof.

Let $\delta$ be Woodin and $A=\\{\kappa<\delta:\kappa\text{ is strong}\\}$. By
Woodin-ness and [Kanamori], there is an unbounded (in fact, stationary) set

$\\{\kappa<\delta:\kappa\text{ is }\lambda\text{-strong reflecting }A\text{
for all }\lambda<\delta\\}\subseteq\mathrm{V}\mid\delta\text{.}$

But this just means we get $\lambda$-strong extenders
$E_{\lambda}\in\mathrm{V}\mid\delta$ for $\lambda<\delta$ on each $\kappa$ in
this set such that the resulting embedding
$j_{\lambda}:\mathrm{V}\rightarrow\mathrm{M}_{\lambda}$ has
$j_{\lambda}(A)\cap\mathrm{V}\mid\lambda=A\cap\mathrm{V}\mid\lambda$, i.e.
$\\{\alpha<\lambda:\mathrm{M}\vDash\text{``}\alpha\text{ is
strong}\text{''}\\}=\\{\alpha<\lambda:\alpha\text{ is strong}\\}$. Hence
$\mathrm{V}\mid\delta$ witnesses the consistency statement.∎

Hence we may assume without loss of generality that there is no inner model
with a Woodin and thus may work with the core model $\mathrm{K}$ below a
Woodin as presented in [Steelfine] and [Steelcore]. There are a few standard
facts about $\mathrm{K}$ that we will use.

######  Lemma 2 • 2.

Suppose there is no inner model with a Woodin cardinal. Therefore the core
model $\mathrm{K}$ is such that

  1. 1.

(Local definability) For every regular $\kappa>\aleph_{1}$,
$\mathrm{K}^{\mathrm{H}_{\kappa}}=\mathrm{K}\cap\mathrm{H}_{\kappa}$.

  2. 2.

(Generic absoluteness) For every poset $\mathbb{P}\in\mathrm{V}$, and every
$\mathbb{P}$-generic $G$ over $\mathrm{V}$,
$\mathrm{K}^{\mathrm{V}}=\mathrm{K}^{\mathrm{V}[G]}$.

  3. 3.

(Initial segment condition) If $E$ is an extender on the sequence of
$\mathrm{K}$, then for every $\alpha<\operatorname{lh}(E)$,
$E\upharpoonright\alpha\in\mathrm{K}$ is on the sequence of $\mathrm{K}$.

When working with $\mathrm{K}$, it will be useful to know when an extender is
on the sequence of $\mathrm{K}$, so we make use of the following lemma.

######  Lemma 2 • 3.

Let $E$ be a $(\kappa,\lambda)$-extender with strength $\lambda=\kappa+\delta$
such that $(\kappa^{+\delta})^{\mathrm{V}}$ is regular. Therefore
$F=E\cap\mathrm{K}\in\mathrm{K}$ is on the sequence of $\mathrm{K}$ and in
fact, $\mathrm{K}\vDash$ “$F$ is a $(\kappa,\lambda)$-extender with
$\mathrm{str}(F)=\lambda$”. Moreover, for each regular
$\mu<(\kappa^{+\delta})^{\mathrm{V}}$ in $\mathrm{K}$, there is an extender
$F_{\mu}$ such that $\mathrm{Ult}_{F_{\mu}}(\mathrm{K})$ and $\mathrm{K}$
agree on $\mathrm{H}_{\mu}$.

###### Proof.

Let $\mathrm{M}=\operatorname{Ult}_{E}(\mathrm{V})$ so that
${}^{\omega}M\subseteq M$. Let $j:\mathrm{V}\rightarrow\mathrm{M}$ be the
canonical ultrapower map. The hypotheses of [Schindler] are satisfied and
hence $\mathrm{K}^{\mathrm{M}}$ (i.e. $j(\mathrm{K})$) is an iterate of
$\mathrm{K}$. Moreover $j\upharpoonright\mathrm{K}=\pi^{\mathcal{T}}$ for some
normal iteration tree $\mathcal{T}$ on $\mathrm{K}$ of successor length where
$\pi^{\mathcal{T}}$ is the iteration map for the main branch of $\mathcal{T}$,
and $\mathcal{T}$ has last model $\mathrm{K}^{\mathrm{M}}$. More explicitly,
let $\operatorname{lh}(\mathcal{T})=\gamma+1$. Let $\xi$ be such that the
$\mathcal{T}$-predecessor of $\xi+1$ is $0$ and $\xi+1$ lies on the branch of
$\gamma$. We have models
$\langle\mathcal{M}_{\alpha}^{\mathcal{T}}:\alpha<\operatorname{lh}(\mathcal{T})\rangle$
with extender sequence $\langle
E_{\alpha}:\alpha<\operatorname{lh}(\mathcal{T})-1\rangle$, iteration maps
$i_{\alpha,\beta}:\mathcal{M}_{\alpha}^{\mathcal{T}}\rightarrow\mathcal{M}_{\beta}^{\mathcal{T}}$
for $\alpha\leq_{\mathcal{T}}\beta$, and the following branch

$\mathrm{K}=\mathcal{M}_{0}^{\mathcal{T}}\rightarrow_{E_{\xi}}\mathcal{M}_{\xi+1}^{\mathcal{T}}\rightarrow\cdots\rightarrow\mathcal{M}_{\gamma}^{\mathcal{T}}=\mathrm{K}^{\mathrm{M}}\text{.}$

We now show that $E_{\xi}\upharpoonright\lambda=F$. [Schindler] tells us that
$i_{0,\gamma}=j\upharpoonright\mathrm{K}$. Since
$\mathrm{V}\mid\lambda=\mathrm{M}\mid\lambda$ and $\lambda=\kappa+\delta$,
$\mathrm{H}_{\kappa^{+\delta}}^{\mathrm{V}}=\mathrm{H}_{\kappa^{+\delta}}^{\mathrm{M}}$
so by local definability,
$\mathrm{K}^{\mathrm{M}}\cap\mathrm{H}_{\kappa^{+\delta}}^{\mathrm{M}}=\mathrm{K}\cap\mathrm{H}_{\kappa^{+\delta}}$.
It follows that
$\operatorname{str}(E_{\xi})=\operatorname{lh}(E_{\xi})\geq\lambda$. By
normality of $\mathcal{T}$,
$\operatorname{cp}(i_{\xi+1,\gamma})>\operatorname{lh}(E_{\xi})\geq\lambda$.
So for any given $\langle
a,A\rangle\in[\lambda]^{<\omega}\times[\kappa]^{|a|}$ in $\mathrm{K}$,

$\displaystyle\langle a,A\rangle\in F$ $\displaystyle\quad\text{iff}\quad a\in
j(A)=i_{0,\gamma}(A)$ $\displaystyle\quad\text{iff}\quad
i_{\xi+1,\gamma}(a)\in i_{\xi+1,\gamma}(i_{0,\xi+1}(A))$
$\displaystyle\quad\text{iff}\quad a\in
i_{0,\xi+1}(A)\quad\text{iff}\quad\langle a,A\rangle\in E_{\xi}$

Hence $E_{\xi}$ and $F$ agree up to $\lambda$. By the initial segment
condition, $E_{\xi}\upharpoonright\lambda=F$ ensures $F$ is on the sequence of
$\mathrm{K}$. The strength of $F$ being $\lambda$ in $\mathrm{K}$ follows from
$\mathrm{str}(E_{\xi})=\operatorname{lh}(E_{\xi})\geq\lambda$ from the
normality of $\mathcal{T}$. Moreover, restricting $E_{\xi}$ appropriately
yields $F_{\mu}$ as in the statement, with these in $\mathrm{K}$ again by the
initial segment condition.∎

Simply put, if $\kappa$ is $\kappa+\delta$-strong in $\mathrm{V}$, then
$\kappa$ is $<\kappa^{+\delta}$-strong in $\mathrm{K}$ according to the
$\mathrm{H}$-hierarchy: $\mathrm{K}\vDash$ “there are extenders $F_{\mu}$ such
that $\mathrm{H}_{\mu}\in\operatorname{Ult}_{F_{\mu}}(\mathrm{K})$ for each
regular $\mu<\kappa^{+\delta}$”.

In particular, any $\kappa$ strong in $\mathrm{V}$ is strong in $\mathrm{K}$:
to get $\kappa$ as $\lambda$-strong in $\mathrm{K}$, we just need a
sufficiently large $\mu$ such that
$\mathrm{K}\mid\lambda\subseteq\mathrm{H}_{\mu}^{\mathrm{K}}$. Then $\kappa$
being $\mu^{+}$-strong in $\mathrm{V}$ implies being $\lambda$-strong in
$\mathrm{K}$.

But we can actually say much more than $\mathrm{V}$-strongs being
$\mathrm{K}$-strong assuming UWISS: any cardinal stronger than a measurable in
$\mathrm{V}$ will be strong in $\mathrm{K}$, as we will prove below. As a
result, any $\mathrm{V}$-strong cardinal is a limit of $\mathrm{K}$-strong
cardinals, for example. Indeed, we will show that any $\mathrm{V}$-strong
cardinal is srs in $\mathrm{K}$. These ideas will be central for the (2)
implies (1) direction of maingoal.

######  Theorem 2 • 4.

(2) implies (1) where these stand for

  1. 1.

$\operatorname{Con}(\textup{ZFC}+\text{``}\text{There is a proper class of
strongs reflecting strongs}\text{''})$;

  2. 2.

$\operatorname{Con}(\textup{ZFC}+\text{``}\text{There is a proper class of
strong cardinals}\text{''}+\textup{UWISS})$.

###### Proof.

We’re done if there is an inner model with a Woodin cardinal, so assume
otherwise. Let $\mathrm{V}\vDash\textup{ZFC}+\textup{UWISS}$ have a proper
class of strong cardinals. It will be useful to understand what cardinals are
strong in the core model $\mathrm{K}$.

######  Claim 1.

Suppose $\kappa$ is $\kappa+2$-strong in $\mathrm{V}$. Therefore $\kappa$ is
strong in $\mathrm{K}$.

###### Proof.

Let $\delta>\kappa^{+}$ be arbitrarily large. Write
$\lambda=\beth_{\delta}^{+}\geq\delta$. Consider
$\mathbb{P}=\mathrm{Col}(\kappa^{+},\lambda)$. By indestructibility, $\kappa$
is still $\kappa+2$-strong in $\mathrm{V}^{\mathbb{P}}$ as witnessed by some
$(\kappa,\kappa+2)$-extender $E$ and notice
$\lambda<(\lambda^{+})^{\mathrm{V}}=(\kappa^{++})^{\mathrm{V}^{\mathbb{P}}}$.
By generic absoluteness,
$\mathrm{K}^{\mathrm{V}^{\mathbb{P}}}=\mathrm{K}^{\mathrm{V}}$ and so
$\lambda$ is regular in $\mathrm{K}$, $\mathrm{Ult}_{F}(\mathrm{K})$. By
calcKseq, there is an extender $F=F_{\lambda}\in\mathrm{K}$ such that
$\mathrm{K}$ and $\operatorname{Ult}_{F}(\mathrm{K})$ agree on
$\mathrm{H}_{\lambda}$. As $\beth_{\delta}<\lambda$ is a strong limit in
$\mathrm{V}$, this also holds in $\mathrm{K}$, so that
$\mathrm{K},\mathrm{Ult}_{F}(\mathrm{K})\vDash$
“$\mathrm{H}_{\lambda}\supseteq\mathrm{V}_{\delta}$”. This means
$\mathrm{K}\mid\delta=\mathrm{Ult}_{F}(\mathrm{K})\mid\delta$ so that $\kappa$
is $\delta$-strong in $\mathrm{K}$.∎

We now aim to show any $\kappa$ strong in $\mathrm{V}$ is actually srs in
$\mathrm{K}$. Let $\kappa$ be strong in $\mathrm{V}$ and let the
$\lambda$-strength of $\kappa$, for some strong $\lambda\geq\kappa^{+\P}$, be
witnessed by a $(\kappa,\lambda)$-extender $E\in\mathrm{V}$ and embedding
$j_{E}:\mathrm{V}\rightarrow\operatorname{Ult}_{E}(\mathrm{V})$. Note that
$j_{E}$ restricts down to
$j_{E}\upharpoonright\mathrm{K}:\mathrm{K}\rightarrow\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}$.
Consider $F=E\cap\mathrm{K}$ which is in $\mathrm{K}$ and on the
$\mathrm{K}$-sequence by calcKseq. So consider the resulting ultrapower
$\mathrm{Ult}_{F}(\mathrm{K})$ via
$j_{F}:\mathrm{K}\rightarrow\operatorname{Ult}_{F}(\mathrm{K})$ which then
factors $j_{E}=k_{F}\circ j_{F}$ via
$k_{F}:\operatorname{Ult}_{F}(\mathrm{K})\rightarrow\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}$
as seen below.

${\mathrm{K}}$${\mathrm{K}^{\operatorname{Ult}_{E}(\mathrm{V})}}$${\operatorname{Ult}_{F}(\mathrm{K})}$$\scriptstyle{j_{E}\upharpoonright\mathrm{K}}$$\scriptstyle{j_{F}}$$\scriptstyle{k_{F}}$
Figure 1: Factoring ultrapower embeddings with $\mathrm{K}$

Note that $\operatorname{cp}(k_{F})\geq\lambda$ so that for any cardinal
$\xi<\lambda$,

$\mathrm{Ult}_{F}(\mathrm{K})\vDash\text{``}\xi\text{ is
strong}\text{''}\quad\text{iff}\quad\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}\vDash\text{``}k_{F}(\xi)=\xi\text{
is strong}\text{''}\text{.}$

Thus it suffices to show $\mathrm{K}$ and
$\mathrm{K}^{\operatorname{Ult}_{E}(\mathrm{V})}$ agree on strongs below
$\lambda$, because then all three would agree and so $F$ would witness a
$\lambda$-srs embedding for $\kappa$ in $\mathrm{K}$. By local definability,
$\mathrm{K}\mid\lambda=\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}\mid\lambda$
and hence we get that $\mathrm{K}$ and
$\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}$ agree on $\text{``}\xi\text{ is
}<\lambda\text{-strong}\text{''}$ whenever $\xi<\lambda$. So it suffices to
show

$\mathrm{K},\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}\vDash\text{``}\forall\xi<\lambda\
(\xi\text{ is }<\lambda\text{-strong}\rightarrow\xi\text{ is
strong})\text{''}\text{.}$

Since $\lambda$ is strong (in $\mathrm{V}$), $\lambda$ is a limit of cardinals
$\xi$ that are $\xi+2$-strong. By the strength of $E$, if $\xi$ is
$\xi+2$-strong in $\mathrm{V}$, then $\xi$ is $\xi+2$-strong in
$\mathrm{Ult}_{E}(\mathrm{V})$, and hence $\lambda$ is a limit of cardinals
$\xi$ that are $\xi+2$-strong in $\mathrm{Ult}_{E}(\mathrm{V})$. So by
maincorecl1, $\lambda$ is a limit of $\mathrm{K}$-strongs and
$\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}$-strongs. As a result, if
$\xi<\lambda$ is not $\beta$-strong in either $\mathrm{K}$ or
$\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}$ for some $\beta$, then by Σ2ref,
this is reflected by some strong below $\lambda$ in $\mathrm{K}$ or
$\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}$, showing ($*$) holds. It follows
that $\mathrm{K}$ and $\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}$ agree on
strongs below $\lambda$, and since $\mathrm{K}^{\mathrm{Ult}_{E}(\mathrm{V})}$
and $\mathrm{Ult}_{F}(\mathrm{K})$ agree on strongs below $\lambda$, we get
that $F\in\mathrm{K}$ witnesses that $\kappa$ is $\lambda$-srs. A proper class
of strongs in $\mathrm{V}$ then gives a proper class of srs cardinals in
$\mathrm{K}$.∎

##  3 The Forcing Direction

Now we show the harder direction of maingoal. We show we can force a proper
class of strongs with UWISS from a proper class of srs cardinals. The general
idea behind the poset, as with most indestructibility results, is a trial by
fire to kill all degrees of strength.

The result will be that the srs cardinals remain strong after forcing with the
preparation, and small degrees of strength are, by virtue of surviving the
trial by fire, weakly indestructible. We do not get universal
indestructibility for _all_ degrees of strength both because that’s impossible
by hamkinsoopsy and because it’s possible for the tail poset to resurrect
degrees of strength that were destroyed, something avoided in [aptersargsyan]
and [apterhamkins] by cutting off the universe and declaring success anytime
this might happen.

Our trial proceeds with appropriate posets via a lottery in an Easton
iteration, that is to say “reverse” Easton in the sense of many of Hamkins’
papers and in [aptersargsyan].333I call an iteration’s support _Easton_ iff
direct limits are taken at (weakly) inaccessible stages and inverse limits
elsewhere. What posets are appropriate? Well, the ones that destroy the
strength of a cardinal $\kappa$ and also are simultaneously
$<\kappa$-strategically closed and $\leq\kappa$-distributive, basically
violating UWISS.

######  Definition 3 • 1.

For $\delta$ a cardinal of strength $\geq\rho$, we say a poset $\mathbb{Q}$ is
_$\delta,\rho$ -appropriate_ iff

  1. 1.

$\mathbb{Q}$ is $<\delta$-strategically closed;

  2. 2.

$\mathbb{Q}$ is $\leq\delta$-distributive; and

  3. 3.

$\mathbb{Q}$ destroys the $\rho$-strength of $\delta$.

If just (1) and (2) hold, we say $\mathbb{Q}$ is _$\delta$ -appropriate_.

So the _destructibility_ of a cardinal $\delta$’s degree of strength $\rho$ by
a $<\delta$-strategically closed and $\leq\delta$-distributive poset is
obviously equivalent to the existence of a $\delta,\rho$-appropriate poset.
Hence we can restate UWISS as the lack of any $\delta,\delta+2$-appropriate
posets for any $\delta$. We will show that the existence of appropriate posets
is equivalent to the existence of “small” appropriate posets, basically
meaning that we can bound the rank of $\mathbb{Q}$ and $\rho$ by
$\delta^{+\P}$.

In examining appropriatedef, we have the following easy, useful result.

######  Corollary 3 • 2.

Let $\mathbb{P}\in\mathrm{V}\mid\alpha$ be a poset. Therefore

  * •

$\mathbb{P}$ is $\delta,\rho$-appropriate iff for some (any)
$\lambda\geq|\alpha+\rho|^{+}$, $\mathrm{V}\mid{\lambda}\mathrel{\vDash}$
“$\mathbb{P}$ is $\delta,\rho$-appropriate”.

  * •

For $\dot{\mathbb{Q}}$ a $\mathbb{P}$-name for a poset,
$\mathbb{P}\Vdash\text{``}\mathbb{Q}\text{ is
}\check{\delta},\check{\rho}\text{-appro\\-priate with rank
}\check{\beta}\text{''}$ iff $\mathrm{V}\mid\lambda$ satisfies this for
$\lambda$ greater than $|\max(\alpha,\rho,\beta)|^{+}$.

###### Proof.

It suffices to show the second since the first follows from it in the case
that $\mathbb{P}$ is trivial. Let $\kappa=|\max(\alpha,\rho,\beta)|^{+}$.
$\mathbb{Q}$ is forced to be a subset of $(\mathrm{V}\mid\beta)^{\mathbb{P}}$.
So without loss of generality, $\dot{\mathbb{Q}}$ is a nice $\mathbb{P}$-name
for a subset of $(\mathrm{V}\mid\beta)^{\mathbb{P}}$ which therefore has rank
$<\kappa$. $<\delta$-strategic closure and $\delta$-distributivity only make
claims about $\mathcal{P}(\mathbb{Q})$. Nice names again mean that we only
need access to $\mathbb{P}$-names of rank $<\kappa$, meaning these concepts
are absolute between $\mathrm{V}$ and $\mathrm{V}\mid\lambda$ for
$\lambda\geq\kappa$. So the remaining concepts we need absoluteness for are
related to the (non)-existence of extenders and inaccessibles: we need

$\mathbb{P}\Vdash\text{``}\dot{\mathbb{Q}}\Vdash\text{``}\check{\delta}\text{
is no longer }\check{\rho}\text{-strong}\text{''}\text{''}\text{.}$

Working in $\mathrm{V}^{\mathbb{P}}$, nice $\mathbb{Q}$-names for
$(\delta,\rho^{\prime})$-extenders will have rank
$<\rho^{\prime}+\beta+\omega<\kappa$. So in $\mathrm{V}$, we only need to
consider nice names for subsets of
$(\mathrm{V}\mid{\rho+\beta+\omega})^{\mathbb{P}}$, which all have rank
$<\kappa$.∎

The search for a $\delta,\rho$-appropriate poset—equivalently the weak
destructibility of $\delta$’s strength—can be bounded in the presence of lots
of strong cardinals: if there is some $\delta,\rho$-appropriate poset, then we
can choose $\rho$ and the rank of the poset to be less than $\delta^{+\P}$
just as in [aptersargsyan]. This isn’t too difficult to show, and helps us in
defining our forcing preparation later.

######  Lemma 3 • 3.

Let $\delta\in\mathrm{Ord}$. Let $\mathbb{R}\in\mathrm{V}\mid\alpha$ be a
poset. Suppose there’s a $\delta,\rho^{**}$-appropriate
$\mathbb{Q}^{**}\in\mathrm{V}^{\mathbb{R}}$. Therefore, there’s a
$\delta,\rho$-appropriate
$\mathbb{Q}\in\mathrm{V}^{\mathbb{R}}\mid(\max(\delta,\alpha)^{+\P})^{\mathrm{V}}$
where $\rho<(\max(\delta,\alpha)^{+\P})^{\mathrm{V}}$ and $\rho\leq\rho^{**}$.

###### Proof.

Let $\mathbb{Q}^{**}$ have rank greater than
$\gamma=(\max(\delta,\alpha)^{+\P})^{\mathrm{V}}$ in
$\mathrm{V}^{\mathbb{P}_{\delta}}$. appropriatereflect tells us for
sufficiently large $\lambda$, $\mathbb{Q}^{**}$ is
$\delta,\rho^{**}$-appropriate in $\mathrm{V}^{\mathbb{R}}\mid\lambda$. More
precisely, we get a name $\dot{\mathbb{Q}}^{**}\in\mathrm{V}\mid\lambda$ such
that

$\mathrm{V}\mid\lambda\vDash\text{``}\mathbb{R}\Vdash\text{``}\dot{\mathbb{Q}}^{**}\text{
is
}\check{\delta},\check{\rho}^{**}\text{-appro\\-priate}\text{''}\text{''}\text{.}$

Let $j:\mathrm{V}\rightarrow\mathrm{M}$ witness the $\lambda$-strength of
$\gamma$ in $\mathrm{V}$. It follows that

$\mathrm{V}\mid\lambda=\mathrm{M}\mid\lambda\vDash\text{``}\mathbb{R}\Vdash\text{``}\dot{\mathbb{Q}}^{**}\text{
is
}\check{\delta},\check{\rho}^{**}\text{-appro\\-priate}\text{''}\text{''}\text{.}$

Since $\rho^{**}$ and the rank of $\mathbb{Q}^{**}$ are below $\lambda\leq
j(\gamma)$, $\mathrm{M}$ believes there’s an $\mathbb{R}$-name for a poset
$\dot{\mathbb{Q}}^{*}$ in $\mathrm{M}\mid{j(\gamma)}$ that is
$\delta,\rho^{*}$-appropriate for some $\rho^{*}<j(\gamma)$ and in particular,
$\rho^{*}=\rho^{**}\leq j(\rho^{**})$. Elementarity then gives a name
$\dot{\mathbb{Q}}$ for a $\delta,\rho$-appropriate poset in
$\mathrm{V}\mid\gamma$ with $\rho<\gamma$ and $\rho\leq\rho^{**}$. The rank of
this poset in $\mathrm{V}^{\mathbb{R}}$ is therefore below $\gamma$.∎

In particular, if $\alpha<\delta^{+\P}$ then $\dot{\mathbb{Q}}$ will have rank
$<\delta^{+\P}$.

###  3.A Forcing UWISS

We now attempt to prove the following, one of the directions from maingoal.

######  Theorem 3.A • 1.

(1) implies (2) where these stand for

  1. 1.

$\operatorname{Con}(\textup{ZFC}\mathrel{+}$ “There is a proper class of srs
cardinals”$)$.

  2. 2.

$\operatorname{Con}(\textup{ZFC}\mathrel{+}$ “There is a proper class of
strongs” $\mathrel{+}\textup{UWISS})$.

Suppose $\mathrm{V}\vDash\textup{ZFC}+\text{``}\text{there is a proper class
of srs cardinals}\text{''}$. Without loss of generality, also assume that
$\mathrm{V}\vDash\textup{GCH}$.444This can be done if necessary by forcing
with the Easton support iteration that adds a cohen subset to each successor
cardinal. This will preserve srs cardinals (and many other large cardinal
properties).

The basic idea is a trial by fire where anything that emerges with some degree
of strength should have its degree of strength weakly indestructible. A slight
hiccup with this is that we don’t actually have much control over what the
resulting strength is, so rather than all of its degrees of strength being
ewakly indestructible, just its small ones are. A broad outline of the proof
is as follows. Here
$\mathbb{P}_{\kappa}=\operatorname*{{\smash{\scalebox{2.5}{\raisebox{-3.87495pt}{$*$}}}}}_{\alpha<\kappa}\dot{\mathbb{Q}}_{\alpha}$,
the whole preparation is $\mathbb{P}=\mathbb{P}_{\mathrm{Ord}}$, and the tail
forcings from $\delta$ up to $\gamma$ are
$\dot{\mathbb{R}}_{[\delta,\gamma)}\cong\operatorname*{{\smash{\scalebox{2.5}{\raisebox{-3.87495pt}{$*$}}}}}_{\delta\leq\alpha<\gamma}\dot{\mathbb{Q}}_{\alpha}$.

  1. i.

Show that if $\kappa$ is $\kappa+2$-strong in $\mathrm{V}^{\mathbb{P}}$, then
this is weakly indestructible, showing
$\mathrm{V}^{\mathbb{P}}\vDash\textup{UWISS}$. This is done by showing

     * •

if we are able to destroy $\kappa$’s $\kappa+2$-strength, then
$\dot{\mathbb{Q}}_{\kappa}$ should be non-empty and would have already
destroyed this.

  2. ii.

Show that if $\kappa$ is $\lambda$-srs in $\mathrm{V}$ then $\kappa$ is
$\lambda$-strong in $\mathrm{V}^{\mathbb{P}}$, implying that any
$\mathrm{V}$-srs cardinal is $\mathrm{V}^{\mathbb{P}}$-strong, and there is a
proper class of both. This is done by showing $\kappa$ is $\lambda$-strong in
$\mathrm{V}^{\mathbb{P}_{\lambda}}$ for large $\lambda$ as follows.

     * •

Take a $\lambda$-srs embedding $j:\mathrm{V}\rightarrow\mathrm{M}$ and in
$\mathrm{V}[G]=\mathrm{V}^{\mathbb{P}_{\lambda}}$, find a generic $H$ for the
tail forcing $\dot{\mathbb{R}}_{[\lambda,j(\lambda))}$ over a specific hull
$\mathrm{N}[G]\preccurlyeq\mathrm{M}[G]$.

     * •

Then we show this is generic over $\mathrm{M}[G]$ using the strategic closure
of the tail _within_ $\mathrm{M}[G]$, not $\mathrm{V}[G]$, to find
representations for dense sets of $\mathrm{M}[G]$ inside $\mathrm{N}[G]$.

     * •

Then we lift $j$ to the $\lambda$-strong
$j^{+}:\mathrm{V}[G]\rightarrow\mathrm{M}[G*H]$ within $\mathrm{V}[G]$,
showing $\kappa$ is still $\lambda$ strong in $\mathrm{V}[G]$.

A necessary consequence of hamkinsoopsy is that srs cardinals have their
strengths destroyed by $\mathbb{P}_{\kappa+1}$, but will have their
$\lambda$-strength resurrected by stage $\lambda$ (whenever $\lambda$ is a
non-measurable, inaccessible limit of strongs).

More generally, showing that degrees of strength aren’t resurrected is not as
simple a task as it might seem, and disregarding hamkinsoopsy, this is partly
why we can’t use this preparation to get indestructible strength for large
strengths. Let me describe the situation in a little more detail to show what
goes wrong. The basic idea is that these cardinals are not going through this
“trial by fire” alone, and a later cardinal can act like a medic.

Naïvely, one will proceed destroying as much strength as possible: if $\kappa$
is $\rho$-strong in $\mathrm{V}^{\mathbb{P}_{\kappa}}$, we try to destroy this
with $\kappa$-appropriate posets if we can, and this constitutes the forcing
done at stage $\kappa$. If we destroy $\kappa$’s strength down to be $<\rho$,
but some non-trivial forcing is done at stage $\delta<\rho$, we might
accidentally resurrect $\kappa$’s $\rho$-strength, as below with Figure 2.

strength of
$\kappa$$\mathrm{V}^{\mathbb{P}_{\kappa}}$$\kappa$$\mathrm{V}^{\mathbb{P}_{\kappa+1}}$$\kappa$destroyed$\mathrm{V}^{\mathbb{P}_{\delta+1}}$$\kappa$$\delta$resurrected
Figure 2: Interaction resurrecting strength

Moreover, there is a problem if a cardinal’s strength is merely playing dead.
For suppose $\kappa$ is $\lambda$-strong in $\mathrm{V}$. As we approach
$\kappa$, $\kappa$ might reduce its strength so that $\kappa$ is merely
$\rho$-strong in $\mathrm{V}^{\mathbb{P}_{\kappa}}$ and weakly indestructible
there. In that case, we would do nothing, for we don’t see any strength to
destroy. But then, it may be that $\rho$ is large enough that the next non-
trivial stage of forcing occurs at some $\delta<\rho$, in which case, $\kappa$
might “wake up” and we accidentally resurrect $\kappa$’s $\lambda$-strength.
Because we never think to return to previously dealt with stages, $\kappa$
might remain $\lambda$ strong in $\mathrm{V}^{\mathbb{P}}$ but not be weakly
indestructible.

Any attempted solution to this will ultimately either result in no resulting
large strength or further forcing that might again resurrect strength by
hamkinsoopsy. For example, we might try to collapse destroyed degrees of
strength below the next stage of forcing. This is due to the general fact that
if $\kappa$’s strength is $\rho<\kappa^{+\sharp\sharp}$, then the subsequent
stages won’t resurrect degrees of strength by distributivity of the tail
forcing proven with prepmainforcingtail, expressed with Figure 3 below.

$\mathrm{V}^{\mathbb{P}_{\kappa}}$$\kappa$$\mathrm{V}^{\mathbb{P}_{\kappa+1}}$$\kappa$destroyed$\mathrm{V}^{\mathbb{P}_{\kappa^{+\sharp\sharp}+1}}$$\kappa$$\kappa^{+\sharp\sharp}$$\begin{array}[]{c}\text{cannot
be}\\\ \text{resurrected}\end{array}$ Figure 3: Non-interaction leaving
strength destroyed

The forcing at $\kappa^{+\sharp\sharp}$ shouldn’t affect smaller degrees of
strength by distributivity. But the collapse we’re considering done at stage
$\kappa$ _could_ then resurrect degrees of strength. So this gives the
following result, the proof of which is given later with
cohenresurrectdegreeslater.

######  Result 3.A • 2.

It’s consistent (relative to two srs cardinals and proper class of strong
cardinals) that we can force a $\kappa$ that is not $\rho$-strong to be
$\rho$-strong by a collapse $\mathrm{Col}(\kappa^{+},\rho)$

This problem isn’t an issue when we only care about finding _one_ strong or
supercompact cardinal with indestructibility properties: if we ever have a
measurable between a cardinal and its remaining strength, we could have just
cut off the universe and end up in the desired model; no need to deal with the
resurrection. This is essentially the argument given in [aptersargsyan] and
[apterhamkins]. Unfortunately for us, we can’t just stop our preparation and
declare success, and this is partially why resurrected degrees of strength
remain a fact of life.

Our partial ordering is a modification of the one found in the proof of
[aptersargsyan]. Specifically, we define an iteration (of proper class length)
$\operatorname*{{\smash{\scalebox{2.5}{\raisebox{-3.87495pt}{$*$}}}}}_{\xi\in\mathrm{Ord}}\dot{\mathbb{Q}}_{\xi}$
which begins by adding a Cohen subset of $\omega$ to make use of gap forcing
arguments. In other words, $\mathbb{P}_{\omega+1}\cong\mathrm{Add}(\omega,1)$.
All other nontrivial stages of forcing can only occur at inaccessible $\delta$
that are $\delta+2$-strong in $\mathrm{V}$.

######  Definition 3.A • 3 (The Preparation).

Suppose
$\mathbb{P}_{\delta}=\operatorname*{{\smash{\scalebox{2.5}{\raisebox{-3.87495pt}{$*$}}}}}_{\xi<\delta}\dot{\mathbb{Q}}_{\xi}$
has been defined for $\delta>\omega$. We aim to define
$\dot{\mathbb{Q}}_{\delta}$.

  1. 1.

If no $p\in\mathbb{P}_{\delta}$ forces that $\delta$ is inaccessible, then
$\dot{\mathbb{Q}}_{\delta}=\dot{\textbf{1}}$ is trivial.

  2. 2.

Otherwise, suppose $\delta$ is forced by some $p\in\mathbb{P}_{\delta}$ to be
$<\lambda$-strong for some (maximal) $\lambda$ (allowing $\lambda=0$ if
$\delta$ isn’t measurable, and $\lambda=\mathrm{Ord}$ if strong), and work
below $p$.

    1. (a)

If the $<\lambda$-strength of $\delta$ is weakly indestructible via posets of
rank $<(\delta^{+\P})^{\mathrm{V}}$, then define
$\dot{\mathbb{Q}}_{\delta}=\dot{\textbf{1}}$.

    2. (b)

Otherwise, we let $\rho<\lambda$ be the minimal degree of $\delta$’s strength
that is weakly destructible and let $\dot{\mathbb{Q}}_{\delta}$ be the lottery
sum of all (what are forced to be) posets that take the form
$\dot{\mathbb{B}}*\dot{\mathbb{C}}$ where the following happen.

    3. (c)

$\dot{\mathbb{B}}$ is an $\delta,\rho$-appropriate poset of rank
$<(\delta^{+\P})^{\mathrm{V}}$, and

    4. (d)

In $\mathrm{V}^{\mathbb{P}_{\delta}*\dot{\mathbb{B}}}$, if
$\rho<\delta^{+\sharp\sharp}\leq|\dot{\mathbb{B}}|$, then $\dot{\mathbb{C}}$
is a name for $\mathrm{Col}(\rho^{+\sharp},|\dot{\mathbb{B}}|)$. Otherwise
$\dot{\mathbb{C}}$ is trivial.

Using Easton support for limit stages, we write
$\mathbb{P}=\operatorname*{{\smash{\scalebox{2.5}{\raisebox{-3.87495pt}{$*$}}}}}_{\xi\in\mathrm{Ord}}\dot{\mathbb{Q}}_{\xi}$
for the class iteration and
$\dot{\mathbb{R}}_{[\delta,\lambda)}\cong\operatorname*{{\smash{\scalebox{2.5}{\raisebox{-3.87495pt}{$*$}}}}}_{\delta\leq\xi<\lambda}\dot{\mathbb{Q}}_{\xi}$
for the tail forcing of $\mathbb{P}_{\lambda}$ whenever $\delta<\lambda$.

Two remarks about this preparation: firstly, note that each
$\dot{\mathbb{Q}}_{\delta}$ is $\delta,\rho$-appropriate for some $\rho$
whenever it’s non-trivial. Secondly, it’s not hard to see that we can regard
$\mathbb{P}_{\delta}\subseteq\mathrm{V}\mid\delta^{+\P}$ for any $\delta$. In
fact, we can regard $\mathbb{P}_{\delta}\subseteq\mathrm{V}\mid\delta$
whenever $\dot{\mathbb{Q}}_{\delta}$ is non-trivial, since $\delta$ is
therefore inaccessible and not collapsed by any previous stage, meaning every
previous stage had a smaller cardinality and thus smaller rank.

The collapsing poset used for each poset in the lottery is used to give better
control over the tail forcing, and in particular to show $\mathbb{P}_{\delta}$
is $\delta$-cc whenever $\delta$ is still inaccessible there. Moreover, the
collapse allows us to ensure that once we collapse a _small_ degree of
strength with a small poset, the strength stays collapsed and won’t be
resurrected. Larger degrees, however, we make no promises about.

There are several standard facts of tail forcings we need to use in order to
prove that the tail forcings of the preparation are strategically closed. The
use of this will be in showing that $\dot{\mathbb{R}}_{[\delta,\lambda)}$ is
$\delta$-appropriate and thus can be used in arguments with
$\delta,\rho$-appropriate posets since it is close to being one. One
background result used is the following [cummings].

######  Theorem 3.A • 4.

Let
$\mathbb{P}_{\lambda}^{\prime}=\operatorname*{{\smash{\scalebox{2.5}{\raisebox{-3.87495pt}{$*$}}}}}_{\xi<\lambda}\dot{\mathbb{Q}}_{\xi}^{\prime}$
be a $\lambda$-stage iteration such that for some $\delta<\lambda$ and some
$\mu$,

  1. 1.

inverse limits or direct limits are taken at every limit stage;

  2. 2.

inverse limits are taken at every limit stage $\geq\delta$ of cofinality
$<\mu$;

  3. 3.

$\dot{\mathbb{Q}}_{\xi}^{\prime}$ is (forced to be) $<\mu$-strategically
closed for every $\xi$ with $\delta\leq\xi<\lambda$; and

  4. 4.

$\mathbb{P}_{\delta}^{\prime}$ is $\mu$-cc.

Therefore the tail forcing $\dot{\mathbb{R}}_{[\delta,\lambda)}$ is (forced to
be) $<\mu$-strategically closed.

######  Corollary 3.A • 5.

Let $\delta<\lambda$. Therefore, the tail forcing of our preparation,
$\dot{\mathbb{R}}_{[\delta,\lambda)}$ is (forced to be)
$<\delta$-strategically closed and $\leq\delta$-distributive.

###### Proof.

If $\dot{\mathbb{Q}}_{\delta}$ is non-trivial, then below the appropriate
conditions, $\delta$ is inaccessible in $\mathrm{V}^{\mathbb{P}_{\delta}}$ so
we take a direct limit at stage $\delta$. It follows that $\delta$ was not
collapsed at some previous stage by a $\kappa,\rho$-appropriate poset
$\mathbb{Q}$ where $\kappa<\delta\leq\max(\rho,|\mathbb{Q}|)$. So for each
$\kappa<\delta$, $\dot{\mathbb{Q}}_{\kappa}$ is equivalent to a poset with
rank $<\delta$ and in fact we can regard
$\mathbb{P}_{\delta}\subseteq\mathrm{V}\mid\delta$ and so
$\mathbb{P}_{\delta}$ is $\delta$-cc. Using iterationtailclosure, Easton
support iterations clearly take inverse or direct limits everywhere, and only
direct limits at certain regular stages. So at any
$\xi\geq\delta>\operatorname{cof}(\xi)$, we take an inverse limit. Thus (1)
and (2) hold from iterationtailclosure. By hypothesis, (3) holds since lottery
sums of $<\kappa$-strategically closed posets are $<\kappa$-strategically
closed. (4) holds since $\mathbb{P}_{\delta}$ is $\delta$-cc and thus
$\dot{\mathbb{R}}_{[\delta,\lambda)}$ is $<\delta$-strategically closed.

If $\dot{\mathbb{Q}}_{\delta}$ is trivial, then below the appropriate
conditions, the next non-trivial stage (if there is one)
$\dot{\mathbb{Q}}_{\mu}$ has by the above argument that
$\dot{\mathbb{R}}_{[\mu,\lambda)}\cong\dot{\mathbb{R}}_{[\delta,\lambda)}$ is
$<\mu$-strategically closed and hence $<\delta$-strategically closed. If there
is no non-trivial stage above $\delta$, then the tail forcing is trivial and
hence $<\delta$-strategically closed.

For $\leq\delta$-distributivity, note that
$\dot{\mathbb{R}}_{[\delta,\lambda)}\cong\dot{\mathbb{Q}}_{\delta}*\dot{\mathbb{R}}_{(\delta,\lambda)}$
is the two-step iteration of two $\leq\delta$-distributive posets.∎

Given that measurability is already indestructible, the next non-trivial stage
of forcing after $\delta$ occurs at stage $\delta^{+\sharp\sharp}$ at the
earliest.

######  Lemma 3.A • 6.

The first non-trivial stage of forcing after any $\kappa\in\mathrm{Ord}$ is at
least $(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$

###### Proof.

To have any degree of $\delta$’s strength weakly destructible over
$\mathrm{V}^{\mathbb{P}_{\delta}}$, we require $\delta$ to be at least
$\delta+2$-strong. Hence the least $\delta$ such that
$\mathbb{P}_{\delta}\cong\mathbb{P}_{\kappa+1}$ but
$\dot{\mathbb{Q}}_{\delta}$ is non-trivial must have that $\delta$ is
$\delta+2$-strong in
$\mathrm{V}^{\mathbb{P}_{\delta}}=\mathrm{V}^{\mathbb{P}_{\kappa+1}}$ and
hence
$\delta\geq(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$. ∎

Hence all the degrees of $\kappa$’s strength below
$(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$ remain
indestructible even if larger degrees are accidentally resurrected. The basic
proof of this is that if we _could_ destroy something, we would have destroyed
it already.

######  Result 3.A • 7.

Let $\kappa$ be such that $\mathbb{P}_{\kappa}$ and
$\mathbb{P}_{\kappa+1}\Vdash$ “$\check{\kappa}$ is $\check{\rho}$-strong for
$\check{\rho}<\kappa^{+\sharp\sharp}$”. Therefore $\mathbb{P}\Vdash$
“$\check{\kappa}$’s $\rho$-strength is weakly indestructible”.

###### Proof.

By downward absoluteness, in $\mathrm{V}^{\mathbb{P}_{\kappa}}$, $\kappa$ is
still inaccessible. Hence we are in case (2) of theprepdef. As “small” posets
and by gap forcing [GapForcing], $\delta^{+\sharp}$, $\delta^{+\sharp\sharp}$,
and $\delta^{+\P}$ are all the same in $\mathrm{V}$ and
$\mathrm{V}^{\mathbb{P}_{\kappa}}$. (Such cardinals retain their large
cardinal status by small forcing, and no new such cardinals are added above
$\delta$ by [GapForcing].)

As a result, because $\rho<\kappa^{+\sharp\sharp}$ in
$\mathrm{V}^{\mathbb{P}_{\kappa+1}}$, the tail forcing after $\kappa$ is
sufficiently distributive such that the $\rho$-strong extender on $\kappa$ in
$\mathrm{V}^{\mathbb{P}_{\kappa+1}}$ is still $\rho$-strong in
$\mathrm{V}^{\mathbb{P}}$. Thus it suffices to show weak indestructibility for
this degree of strength. So suppose $\dot{\mathbb{Q}}$ is
$\kappa,\rho$-appropriate in $\mathrm{V}^{\mathbb{P}}$. By distributivity of
the tail forcing, $\dot{\mathbb{Q}}\in\mathrm{V}^{\mathbb{P}_{\lambda}}$ for
some $\lambda>\kappa$. The tail forcing $\dot{\mathbb{R}}_{(\kappa,\lambda)}$
is therefore $\kappa$-appropriate by prepmainforcingtail.

This tells us that $\dot{\mathbb{Q}}_{\kappa}$ must be non-trivial. To see
this, otherwise $\kappa$ must be $\rho$-strong in
$\mathrm{V}^{\mathbb{P}_{\kappa+1}}=\mathrm{V}^{\mathbb{P}_{\kappa}}$. It
follows that $\dot{\mathbb{R}}_{[\kappa,\lambda)}*\dot{\mathbb{Q}}$ is
$\kappa,\rho$-appropriate in $\mathrm{V}^{\mathbb{P}_{\kappa}}$ so that by
appropriatesmall, we can find a $\kappa,\rho$-appropriate poset of small size
and so $\dot{\mathbb{Q}}_{\kappa}$ should be non-trivial, a contradiction.

Since $\dot{\mathbb{Q}}_{\kappa}$ is non-trivial, by theprepdef, we forced
with some poset $\dot{\mathbb{B}}*\dot{\mathbb{C}}$ at stage $\kappa$ where
$\dot{\mathbb{B}}$ is $\kappa,\rho_{\kappa}$-appropriate in
$\mathrm{V}^{\mathbb{P}_{\kappa}}$ with minimal $\rho_{\kappa}$. If
$\rho<\rho_{\kappa}$, the $\kappa,\rho$-appropriate
$\dot{\mathbb{R}}_{[\kappa,\lambda)}*\dot{\mathbb{Q}}$ would violate
minimality of $\rho_{\kappa}$ (via appropriatesmall to ensure we stay below
$(\delta^{+\P})^{\mathrm{V}}$ in $\mathrm{V}^{\mathbb{P}_{\kappa}}$). So
$\rho\geq\rho_{\kappa}$. We now break into cases.

  * •

Suppose $\dot{\mathbb{C}}$ is non-trivial. Thus
$\rho_{\kappa}<\kappa^{+\sharp\sharp}\leq|\dot{\mathbb{B}}|$ and
$\dot{\mathbb{C}}$ collapses $|\dot{\mathbb{B}}|$ to be $\rho^{+\sharp}$
(which is $<\kappa^{+\sharp\sharp}$) in a way that is
$<\rho^{+\sharp}$-distributive. In particular, this preserves the lack of
$\rho_{\kappa}$-strong extenders, and so the lack of $\rho$-strong extenders
in $\mathrm{V}^{\mathbb{P}_{\kappa}*\dot{\mathbb{B}}}$ to
$\mathrm{V}^{\mathbb{P}_{\kappa}*\dot{\mathbb{B}}*\dot{\mathbb{C}}}$,
contradicting that $\kappa$ is $\rho$-strong in
$\mathrm{V}^{\mathbb{P}_{\kappa+1}}$

  * •

Suppose $\dot{\mathbb{C}}$ is trivial so that because $\dot{\mathbb{B}}$ is
$\kappa,\rho_{\kappa}$-appropriate, $\kappa$ is not $\rho_{\kappa}$-strong in
$\mathrm{V}^{\mathbb{P}_{\kappa}*\dot{\mathbb{B}}}=\mathrm{V}^{\mathbb{P}_{\kappa+1}}$
and hence not $\rho$-strong there, a contradiction.∎

Unfortunately, this does not guarantee weak indestructibility for $\kappa$’s
degrees of strength $<\kappa^{\sharp\sharp}$ in $\mathrm{V}^{\mathbb{P}}$, but
merely for strength
$<(\kappa^{\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$ and so in
particular $\kappa+2$-strength, $\kappa^{+\sharp}$-strength, and
$<\lambda$-strength for the first $\lambda$ a measurable limit of measurables
in $\mathrm{V}^{\mathbb{P}}$.

######  Corollary 3.A • 8 (Forcing UWISS).

Suppose $\mathrm{V}\vDash$ “there is a proper class of strongs”. Therefore

  * •

the preparation $\mathbb{P}$ is well defined and
$\mathbb{P}\Vdash\textup{UWISS}$.

  * •

In fact, if $\kappa$ is $\rho$-strong for
$\rho<(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$ in
$\mathrm{V}^{\mathbb{P}}$ then this strength is weakly indestructible in
$\mathrm{V}^{\mathbb{P}}$.

  * •

In particular, any cardinal $\kappa$’s strength that is below the least
measurable limit of measurables above $\kappa$ is weakly indestructible, e.g.
$\kappa+2$-strength, $\kappa^{+\sharp}$-strength,
$(\kappa^{+\sharp})^{+\sharp}$-strength, etc.

###### Proof.

If $\kappa$ is measurable in $\mathrm{V}^{\mathbb{P}}$, then this degree of
strength is weakly indestructible. So suppose $\kappa$ is stronger than a
measurable, but there is some $\dot{\mathbb{Q}}\in\mathrm{V}^{\mathbb{P}}$
that is $\kappa,\rho$-appropriate for some
$\rho<(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$, which
has some rank below an inaccessible
$\gamma>\rho,\operatorname{rank}(\dot{\mathbb{Q}})$. The tail forcing
$\dot{\mathbb{R}}_{[\gamma,\infty)}$ is $\leq\gamma$-distributive by
prepmainforcingtail and hence
$\dot{\mathbb{Q}}\in\mathrm{V}^{\mathbb{P}_{\gamma}}$ and
$\mathrm{V}^{\mathbb{P}_{\gamma}}\mid\gamma=\mathrm{V}^{\mathbb{P}}\mid\gamma$.
So by appropriatereflect, $\dot{\mathbb{Q}}$ is $\kappa,\rho$-appropriate in
$\mathrm{V}^{\mathbb{P}_{\gamma}}$. The tail forcing
$\dot{\mathbb{R}}_{(\kappa,\gamma)}$ is sufficiently distributive by
nextnontriv to show $\kappa$ was $\rho$-strong in
$\mathrm{V}^{\mathbb{P}_{\kappa+1}}$ and so
$\dot{\mathbb{R}}_{(\kappa,\lambda)}*\dot{\mathbb{Q}}$ is
$\kappa,\rho$-appropriate there, contradicting mainforcingwss.

By nextnontriv, measurables between $\kappa$ and
$(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$ are preserved
by the tail forcing by nextnontriv. So it follows that every cardinal below
the least measurable limit of measurables above $\kappa$—being below
$(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$—is
indestructible.∎

We may not get better than this, since we could have the following: in
$\mathrm{V}^{\mathbb{P}_{\kappa+1}}$, $\kappa$ is $\rho$-strong for
$\rho>\kappa^{+\sharp\sharp}$. Then at stage
$\mu=(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}_{\kappa+1}}}$ we do some
non-trivial forcing that destroys $\mu$’s $\mu+2$-strength, and subsequently
resurrects $\kappa$’s $\rho^{\prime}$-strength for $\rho^{\prime}>\mu$ in a
way that the new
$(\kappa^{+\sharp\sharp})^{\mathrm{V}^{\mathbb{P}}}>\rho^{\prime}$ but
$\kappa$’s $\rho^{\prime}$-strength is now destructible. Nevertheless, this
doesn’t affect somewhat small degrees of strength as the above shows.

###  3.B A proper class of strongs

So all that remains is to show that srs cardinals of $\mathrm{V}$ are strong
in $\mathrm{V}^{\mathbb{P}}$. We do this by working with partial degrees of
reflecting strongs. The usefulness of reflecting strongs allows us to properly
calculate the preparation up to limits of strongs.

######  Lemma 3.B • 1.

Let $\lambda$ be a limit of strongs. Let $j:\mathrm{V}\rightarrow\mathrm{M}$
be at least $\lambda$-strong such that $\mathrm{V}$ and $\mathrm{M}$ agree on
strongs $<\lambda$. Therefore
$\mathbb{P}_{\lambda}^{\mathrm{V}}=j(\mathbb{P})_{\lambda}=\mathbb{P}^{\mathrm{M}}_{\lambda}$.

###### Proof.

Let $\kappa=\operatorname{cp}(j)$. $j$ will not screw with the Easton support
below $\lambda$, so it suffices to show
$\dot{\mathbb{Q}}_{\delta}^{\mathrm{V}}=\dot{\mathbb{Q}}_{\delta}^{\mathrm{M}}$
for all cardinals $\delta<\lambda$. Proceed by induction on $\delta$. In case
(1) of theprepdef, $\delta$’s non-inaccessibility is easily absolute between
$\mathrm{V}$ and $\mathrm{M}$ since inductively
$\mathbb{P}_{\delta}=\mathbb{P}_{\delta}^{\mathrm{M}}$.

Note that since $\lambda$ is a limit of strongs, if $\delta<\lambda$, then
$(\delta^{+\P})^{\mathrm{V}}=(\delta^{+\P})^{\mathrm{M}}$ and so we can
unambiguously write $\delta^{+\P}$ in such cases. Note also if $\delta$ is
$<\lambda_{0}$-strong in $\mathrm{V}$, then either
$\lambda_{0}=\delta^{+\P}<\lambda$ implies $\delta$ is strong in both since
the models agree on strongs below $\lambda$, or else
$\lambda_{0}<\delta^{+\P}<\lambda$ and so the lack of $\lambda_{0}$-strong
extenders in $\mathrm{V}\mid\lambda$ matches with $\mathrm{M}\mid\lambda$.
Note also that all (names for) collapses we consider will exist in
$\mathrm{V}\mid\delta^{+\P}=\mathrm{M}\mid\delta^{+\P}$ and thus have the same
interpretation in both models.

In case (2), by appropriatereflect, the existence of
$\rho<\lambda_{0}\leq\delta^{+\P}<\lambda$ and a $<\delta$-strategically
closed, $\leq\delta$-distributive
$\dot{\mathbb{B}}\in\mathrm{V}\mid\delta^{+\P}$ such that $\delta$ isn’t
$\rho$-strong after forcing with $\mathbb{P}_{\delta}*\dot{\mathbb{B}}$ can be
calculated in $\mathrm{V}\mid\lambda=\mathrm{M}\mid\lambda$. Hence the two
share the same such posets, and moreover, the minimal $\rho$ witnessing this
is the same for both. The calculation of $\delta^{+\sharp\sharp}$ in both will
be below $\delta^{+\P}$ and easily the same in both models. The collapse (also
being below $\delta^{+\P}$) will also be the same, meaning
$\dot{\mathbb{Q}}_{\delta}$ is the same in both.∎

In particular, if $j:\mathrm{V}\rightarrow\mathrm{M}$ is a $\lambda$-srs
embedding, then
$\mathbb{P}_{\lambda}^{\mathrm{V}}=\mathbb{P}_{\lambda}^{\mathrm{M}}$ whenever
$\lambda$ is a limit of strongs. This gives us the edge over hyperstrength in
generalizing [aptersargsyan] which generally only gets agreement for
$\mathbb{P}_{\operatorname{cp}(j)}$, but the resulting argument is adapted
from [aptersargsyan].

######  Lemma 3.B • 2.

Let $\kappa$ be strong and $\lambda$-srs where $\lambda$ is a limit of
strongs. Therefore $\mathbb{P}_{\lambda}\Vdash$ “$\kappa$ is
$\lambda$-strong”.

###### Proof.

Let $j:\mathrm{V}\rightarrow\mathrm{M}$ be $\lambda$-srs with
$\operatorname{cp}(j)=\kappa$ so that
$\mathrm{M}=\operatorname{Ult}_{E}(\mathrm{V})$ for some
$(\kappa,\lambda)$-extender $E$. We can factor by mainforcingcl2

$\displaystyle j(\mathbb{P}_{\lambda})=\mathbb{P}_{j(\lambda)}^{\mathrm{M}}$
$\displaystyle\cong(\mathbb{P}_{\kappa}*\dot{\mathbb{Q}}_{\kappa}*\dot{\mathbb{R}}_{(\kappa,\lambda)})^{\mathrm{M}}*\dot{\mathbb{R}}_{[\lambda,j(\kappa))}^{\mathrm{M}}*\dot{\mathbb{R}}_{[j(\kappa),j(\lambda))}^{\mathrm{M}}$
$\displaystyle\cong(\mathbb{P}_{\kappa}*\dot{\mathbb{Q}}_{\kappa}*\dot{\mathbb{R}}_{(\kappa,\lambda)})^{\mathrm{V}}*\dot{\mathbb{R}}_{[\lambda,j(\kappa))}^{\mathrm{M}}*\dot{\mathbb{R}}_{[j(\kappa),j(\lambda))}^{\mathrm{M}}$
$\displaystyle\cong\mathbb{P}_{\lambda}^{\mathrm{V}}*\dot{\mathbb{R}}_{[\lambda,j(\kappa))}^{\mathrm{M}}*\dot{\mathbb{R}}_{[j(\kappa),j(\lambda))}^{\mathrm{M}}\text{.}$

Without loss of generality, $\lambda$ isn’t $\lambda+2$-strong in $\mathrm{M}$
(otherwise just taking another ultrapower by the Mitchell-least measure on
$\lambda$) so that in $\mathrm{M}^{\mathbb{P}_{\lambda}}$, $\lambda$ is at
most measurable with therefore indestructible degrees of strength and its
original strength already small:
$\dot{\mathbb{Q}}_{\lambda}^{\mathrm{M}}=\dot{\textbf{1}}$. Thus
$\dot{\mathbb{R}}_{[\lambda,j(\lambda))}^{\mathrm{M}}$ is actually
$\lambda^{+}$-strategically closed in $\mathrm{M}$ (and indeed much more).

So let $G=G_{0}*G_{1}$ be
$\mathbb{P}_{\kappa}*\dot{\mathbb{R}}_{[\kappa,\lambda)}$-generic over
$\mathrm{V}$ such that $\mathrm{V}[G]\vDash$ “$\kappa$ is not
$\lambda$-strong”. Our goal is to lift $j$ to
$j^{+}:\mathrm{V}[G]\rightarrow\mathrm{M}[G*H]$ in $\mathrm{V}[G]$ for some
$H=H_{0}*H_{1}\in\mathrm{V}[G]$ that is
$(\dot{\mathbb{R}}_{[\lambda,j(\kappa))}^{\mathrm{M}}*\dot{\mathbb{R}}_{[j(\kappa),j(\lambda))}^{\mathrm{M}})_{G}$-generic
over $\mathrm{M}[G]$ such that $j^{+}$ remains $\lambda$-strong. This requires
examining $j"G$, and beyond this there are only a couple steps in this lift-up
argument: first building $H_{0}$ arbitrarily and then generating $H_{1}$ from
$j"G_{1}$.

######  Claim 1.

For $j(p)\in j"G$, $j(p)\upharpoonright[\lambda,j(\kappa))$—and in fact
$j(p)\upharpoonright[\kappa,j(\kappa))$—is trivial. Hence the only non-trivial
information $j"G$ encodes occurs before $\lambda$ and after $j(\kappa)$.

###### Proof.

Let $p\in G$ be arbitrary. Since we take a direct limit at stage $\kappa$,
there is some $\alpha<\kappa$ where $p\upharpoonright[\alpha,\kappa)$ is just
a sequence of $\dot{\textbf{1}}$s: $p\upharpoonright[\alpha,\kappa)$ is
trivial in $\dot{\mathbb{R}}_{[\alpha,\kappa)}$. By elementarity, $j(p)$ is
similarly trivial from $j(\alpha)=\alpha<\kappa$ to $j(\kappa)$. In
particular, $j(p)\upharpoonright[\kappa,j(\kappa))$ is trivial.∎

As a result, _any_ $H_{0}$ that is
$\mathbb{R}_{[\lambda,j(\kappa))}^{\mathrm{M}[G]}=(\dot{\mathbb{R}}_{[\lambda,j(\kappa))}^{\mathrm{M}})_{G}$-generic
over $\mathrm{M}[G]$ has $j"G\upharpoonright j(\kappa)$ contained in
$G*H_{0}$. So we merely need to find such a generic over $\mathrm{M}[G]$ in
$\mathrm{V}[G]$ to get $H_{0}$. Then we find $H_{1}$. scrmainforcingmainlemcl1
is partly why we need to break up the iteration as we do. The general idea is
that $j"G$ might have conditions with potentially unbounded support in
$j(\lambda)$. So we must generate the generic over the end tail.
scrmainforcingmainlemcl1 ensures that the middle is left unaffected since
$j"G$ has no real information about it.

As an ultrapower, we can regard
$\mathrm{M}=\\{j(f)(r):r\in[\lambda]^{<\omega}\wedge
f:[\kappa]^{<\omega}\rightarrow\mathrm{V}\\}$ so that there are elements of
$r_{0},r_{1}\in[\lambda]^{<\omega}$ and functions from $[\kappa]^{<\omega}$ to
$\mathrm{V}$ that represent $\mathbb{P}_{\kappa}^{\mathrm{V}}$,
$\dot{\mathbb{R}}_{(\kappa,j(\lambda))}^{\mathrm{M}}$. So now we consider the
elementary submodel $\mathrm{N}\preccurlyeq\mathrm{M}$ defined by

$\mathrm{N}=\\{j(f)(r_{0}\cup
r_{1},\kappa,\lambda):f\colon[\kappa]^{<\omega}\rightarrow\mathrm{V}\\}\ni\mathbb{P}_{\kappa}^{\mathrm{V}},\mathbb{P}_{\lambda}^{\mathrm{V}},\dot{\mathbb{Q}}_{\kappa},\dot{\mathbb{R}}_{(\lambda,j(\kappa))}^{\mathrm{M}},\kappa,\lambda\text{,}$

It’s also not hard to see that
$\mathrm{V}\vDash\text{``}{}^{\kappa}\mathrm{N}\subseteq\mathrm{N}\text{''}$.
We write $\mathrm{N}[G]$ for $\\{\tau_{G}:\tau\in\mathrm{N}\\}$.

######  Claim 2.

In $\mathrm{V}[G]$, there is an $H_{0}$
$\mathbb{R}_{(\lambda,j(\kappa))}^{\mathrm{M}[G]}$-generic over
$\mathrm{N}[G]$.

###### Proof.

$\mathbb{R}_{(\lambda,j(\lambda))}^{\mathrm{M}[G]}$ is
$\lambda^{+}$-strategically closed in $\mathrm{M}[G]$ and this translates to
being merely $\kappa^{+}$-strategically closed in $\mathrm{V}[G]$ by the
closure conditions of $\mathrm{M}$ and $\mathrm{N}$ in $\mathrm{V}$:
$\mathrm{N}[G]\preccurlyeq\mathrm{M}[G]$ is closed under $\kappa$-sequences.
Hence $\mathbb{R}_{(\lambda,j(\lambda))}^{\mathrm{M}[G]}\cap\mathrm{N}[G]$ is
still $\kappa^{+}$-strategically closed in $\mathrm{V}[G]$. Since dense sets
in $\mathrm{N}[G]$ can be identified with
$f:[\kappa]^{<\omega}\rightarrow\mathrm{V}_{\kappa}$, a simple counting
argument shows that in $\mathrm{V}[G]$, there are at most
$2^{j(\kappa)}=\kappa^{+}$-many antichains of the poset in $\mathrm{N}[G]$.
Thus we can find in $\mathrm{V}[G]$ an $H$ that is
$\mathbb{R}_{(\kappa,j(\kappa))}^{\mathrm{M}[G]}\cap\mathrm{N}[G]$-generic
over $\mathrm{N}[G]$ by standard techniques (extending into dense open sets
one by one and leaving the other player in the strategic closure game to clean
up our mess at limit stages).∎

Now we must show that $H_{0}$ is actually generic over $M[G]$. Let $D$ be a
dense open set of $\mathbb{R}_{(\lambda,j(\kappa))}^{\mathrm{M}[G]}$. We can
represent $D$ by $j(f)(r)$ for some
$f:[\kappa]^{<\omega}\rightarrow\mathbb{P}_{\kappa}$ and
$r\in[\lambda]^{<\omega}$. So in $\mathrm{M}[G]$, consider the set
$\\{j(f)(s)\text{ dense open}:s\in[\lambda]^{<\omega}\\}$. Since
$\mathrm{M}[G]$ thinks that the tail
$\mathbb{R}_{(\lambda,j(\kappa))}^{\mathrm{M}[G]}$ is
$\lambda^{+}$-strategically closed, the intersection of all of these sets
$E=\bigcap_{s\in[\lambda]^{<\omega}}j(f)(s)$ is definable just with $f$ and so
in $\mathrm{N}[G]$, meaning $E\cap H_{0}\neq\emptyset$ and showing that it’s
generic over $\mathrm{M}[G]$.

######  Claim 3.

Let $H_{1}$ be the filter generated by $j"G_{1}$. Therefore $H_{1}$ is
$\mathbb{R}_{[j(\kappa),j(\lambda))}^{\mathrm{M}[G*H_{0}]}$-generic over
$\mathrm{M}[G*H_{0}]$ and as a result of scrmainforcingmainlemcl1,
$j"G\subseteq G*H_{0}*H_{1}$.

###### Proof.

Let $D\in\mathrm{M}[G*H_{0}]$ be open dense with name $\dot{D}$. It follows
that we can represent $D$ as $j(d)(r)_{G*H_{0}}$ for some
$d:[\kappa]^{<\omega}\rightarrow\mathrm{V}$ and $r\in[\lambda]^{<\omega}$
where each $d(s)$ is a name for a dense open set in
$\dot{\mathbb{R}}_{[\kappa,\lambda)}$. In $\mathrm{V}[G_{0}]$, since the tail
forcing is $\leq\kappa$-distributive by prepmainforcingtail, we can intersect
all of these dense open sets $\bigcap_{s\in[\kappa]^{<\omega}}d(s)_{G_{0}}$
and get another dense open set that intersects $G_{1}$: there is a $p\in
G_{1}\cap d(s)_{G_{0}}$ for every $s\in[\kappa]^{<\omega}$. Thus some $q\in
G_{0}$ has in $\mathrm{M}$ that $j(q)\Vdash$ “$j(p)\in j(d)(s)$ for every
$s\in[j(\kappa)]^{<\omega}$” and in particular
$j(q)\Vdash\text{``}j(p)\in\dot{D}\text{''}$. Since $q\in
G_{0}\subseteq\mathbb{P}_{\kappa}$, $j(q)$ is just $q$ with a bunch of
$\dot{\textbf{1}}$s appended, meaning $j(q)\in G*H_{0}$ and so indeed $j(p)\in
j"G_{1}\cap D\subseteq H_{1}\cap D$ in $\mathrm{M}[G*H_{0}]$.∎

Thus $j"G\subseteq G*H_{0}*H_{1}$ and so $j:\mathrm{V}\rightarrow\mathrm{M}$
lifts to $j^{+}:\mathrm{V}[G]\rightarrow\mathrm{M}[G*H]$. It’s not hard to see
that $j^{+}$ is still $\lambda$-strong since
$G\subseteq\mathrm{V}\mid\lambda=\mathrm{M}\mid\lambda$ and $H$ adds no sets
of rank $<\lambda$.∎

More generally what this shows is the following.

######  Corollary 3.B • 3.

If $j:\mathrm{V}\rightarrow\mathrm{M}$ is $\lambda$-srs for $\lambda$ a limit
of strongs such that $\lambda$ isn’t stronger than a measurable in
$\mathrm{M}$, then we can lift $j$ to
$j^{+}:\mathrm{V}^{\mathbb{P}_{\lambda}}\rightarrow\mathrm{M}^{j(\mathbb{P}_{\lambda})}$.

And this gives the desired result.

######  Result 3.B • 4.

Assume there is a proper class of srs cardinals and GCH holds. Therefore
$\mathbb{P}\Vdash\textup{UWISS}+\text{``}\text{there is a proper class of
strongs}\text{''}$

###### Proof.

That $\mathbb{P}\Vdash\textup{UWISS}$ follows from PforcesUWISS. For a proper
class of strongs, let $\kappa$ be srs and let $\lambda>\kappa$ be strong in
$\mathrm{V}$. By mainforcingmainlem, in $\mathrm{V}^{\mathbb{P}_{\lambda}}$,
$\kappa$ is $\lambda$-strong and this degree of strength is weakly
indestructible, meaning $\kappa$ is still $\lambda$-strong in all later stages
(since, again, the tail forcings will be $\leq\kappa$-distributive and
$<\kappa$-strategically closed) and so $\lambda$-strong in
$\mathrm{V}^{\mathbb{P}}$. Since there are a proper class of
$\mathrm{V}$-strongs above $\kappa$, it follows that $\kappa$ is strong in
$\mathrm{V}^{\mathbb{P}}$. Hence any $\kappa$ srs in $\mathrm{V}$ is strong in
$\mathrm{V}^{\mathbb{P}}$, and we have a proper class of both.∎

This completes the proof of mainforcing, repeated below, and so in conjunction
with maincore, maingoal holds.

######  Corollary 3.B • 5.

(1) implies (2) and (3) where these stand for

  1. 1.

$\operatorname{Con}(\textup{ZFC}\mathrel{+}$ “There is a proper class of srs
cardinals”$)$.

  2. 2.

$\operatorname{Con}(\textup{ZFC}\mathrel{+}$ “There is a proper class of
strongs” $\mathrel{+}\textup{UWISS})$.

  3. 3.

$\operatorname{Con}(\textup{ZFC}\mathrel{+}$ “There is a proper class of
strongs” $+$ “Every $\lambda$-strong $\kappa$ has weakly indestructible
$\lambda$-strength whenever $\lambda$ is below the least measurable limit of
measurable cardinals larger than $\kappa$”.

Again, we can go further than even what (3) states; beyond the next measurable
limit of measurable limits of measurables, and so on. But at some point,
saying this just becomes silly. The point is that we get much more than weak
indestructibility for $\kappa+2$-strength, up to the next cardinal $\lambda$
that has at least as many measurables below it as a $\lambda+2$-strong
cardinal should (since it _was_ $\lambda+2$-strong in
$\mathrm{V}^{\mathbb{P}_{\kappa+1}}$).

The above results allow us to also prove cohenresurrectdegrees, restated
below.

######  Result 3.B • 6.

It’s consistent (relative to two srs cardinals and proper class of strong
cardinals) that we can force a $\kappa$ that is not $\rho$-strong to be
$\rho$-strong by a collapse $\mathrm{Col}(\kappa^{+},\rho)$

###### Proof.

Consider the Easton support iteration
$\mathbb{P}_{\kappa}=\operatorname*{{\smash{\scalebox{2.5}{\raisebox{-3.87495pt}{$*$}}}}}_{\alpha<\kappa}\dot{\mathbb{Q}}_{\alpha}$
where $\dot{\mathbb{Q}}_{\kappa}$ is defined by the following.

  1. 1.

At every stage $\kappa$ that is (forced to be) $\kappa+2$-strong, we force
with the lottery of $\kappa,\rho$-appropriate posets for minimal $\rho$ (if
any exist) of rank $<\kappa^{+\P}$.

  2. 2.

Then with $\mathrm{Col}(\kappa^{+},\rho)$.

  3. 3.

If there is no such $\rho$, and $\kappa$ was originally $<\lambda$-strong in
$\mathrm{V}$ where $\kappa^{+\sharp\sharp}\leq\lambda<\kappa^{+\P}$, then
force with $\mathrm{Col}(\kappa^{+},\lambda)$.

  4. 4.

Otherwise force trivially.

If this collapses never resurrect degrees of strength, then the proof below
goes through to show that the resulting srs cardinals are strong. Moreover,
weak indestructibility will hold for _all_ degrees of strength, simply because
the tail will be appropriate (by prepmainforcingtail) and

  * •

Anytime we destroy $\kappa$’s $\rho$-strength, we make sure the situation in
Figure 2 does not happen and Figure 3 does: the next stage of forcing occurs
far beyond $\rho$, so the lack of $\rho$-strong extenders continues through to
the end of the preparation.

  * •

If we _don’t_ destroy $\kappa$’s strength with a lottery at stage $\kappa$,
then $\kappa$’s strength in $\mathrm{V}^{\mathbb{P}_{\kappa}}$ indestructible
at that stage. The collapse, if there is one, in (3) doesn’t add any degrees
of strength by hypothesis. Thus $\kappa$’s strength would be weakly
indestructible in $\mathrm{V}^{\mathbb{P}_{\kappa+1}}$ (if there were some
$\kappa,\rho$-appropriate $\dot{\mathbb{Q}}$ then in
$\mathrm{V}^{\mathbb{P}_{\kappa}}$, we could force with the
$\kappa,\rho$-appropriate $\dot{\mathbb{Q}}_{\kappa}*\dot{\mathbb{Q}}$). The
collapse will again ensure the next stage of forcing occurs far beyond
$\lambda$ as in Figure 3, so the lack of $\lambda$-strong extenders continues
through to the end of the preparation.

The same proof as liftembedding with the previous preparation in theprepdef
tells us that the srs cardinals become strong in $\mathrm{V}^{\mathbb{P}}$.
But two strong cardinals and weak indestructibility for all degrees of
strength contradicts hamkinsoopsy.∎

##  4 Small Side Results

It’s not hard to see that the forcing preparation $\mathbb{P}$ from theprepdef
can be generalized and restricted to the following theorem when combined with
the original proof from [aptersargsyan]. The idea really is that
[aptersargsyan] only gives UWISS and the only reason there is a weakly
indestructible strong cardinal in the end is that all of its degrees of
strength are small: there is no measurable cardinal above the strong.

######  Theorem 4 • 1.

Let $\alpha\in\mathrm{Ord}$. Therefore the following are equiconsistent.

  1. 1.

$\textup{ZFC}+\text{``}\text{there are (at least) }\alpha\text{ srs cardinals
with a hyperstrong above them}\text{''}$;

  2. 2.

$\textup{ZFC}+\textup{UWISS}+\text{``}\text{there are (at least)
}\alpha+1\text{ strong cardinals}\text{''}$. In fact, the last of those
$\alpha+1$-strong cardinals is weakly indestructible for all of its degrees of
strength.

###### Proof.

That (2) is relatively consistent relative to (1) follows from the techniques
of [aptersargsyan]: forcing GCH, and cutting off the universe at
$\kappa^{+\sharp}$ whenever $\kappa$ has weakly indestructible
$\kappa^{+\sharp}$-strength in $\mathrm{V}^{\mathbb{P}_{\kappa}}$. This always
ends at or before the final hyperstrong. Such a $\kappa$ is strong in the
final model. The lower srs cardinals become strong after forcing with
$\mathbb{P}_{\kappa}$ as noted with mainforcingmainlem, and so there are
$\alpha+1$-strong cardinals after forcing with $\mathbb{P}_{\kappa}$ and
clearly $\mathbb{P}_{\kappa}\cong\mathbb{P}_{\mu}$ forces UWISS over
$\mathrm{V}\mid\mu$.

That (1) is relatively consistent to (2) follows again from the techniques of
[aptersargsyan]: if $\mathrm{V}\vDash\textup{ZFC}+\textup{UWISS}$, any
$\mathrm{V}$-strong cardinal $\kappa$ will be hyperstrong in $\mathrm{K}$ and
if there’s a $\mathrm{V}$-strong above $\kappa$, then $\kappa$ will be srs in
$\mathrm{K}$.∎

Reflecting strongs is just one kind of reflection, but in principle, we could
also enforce that we reflect more. This essentially converges onto the idea of
a Woodin cardinal. The following results are proven in my upcoming thesis
[thesis].

######  Theorem 4 • 2.

Let $\mathbb{P}$ be as in theprepdef. Let $\delta$ be a Woodin cardinal.
Therefore $\mathbb{P}_{\delta}\Vdash\text{``}\delta\text{ is
Woodin}\text{''}$. Hence the existence of a Woodin cardinal $\delta$ gives the
consistency of a Woodin cardinal $\delta$ where
$\mathrm{V}\mid\delta\vDash\textup{UWISS}$.

This gives the consistency of UWISS with a proper class of a large variety of
large cardinal notions relative to the existence of a Woodin cardinal.

While on the topic of Woodins, consider the interaction of universal weak
indestructibility for _large_ degrees of strength with Woodin cardinals. For
example, we used a cardinal that is strong and reflects strongs to get some
weak indestructibility results for strength. We cannot have the same
indestructibility for reflection properties.

######  Definition 4 • 3.

An inaccessible cardinal $\delta$ is _Woodin by weak indestructibility_ iff
for every $A\subseteq\mathrm{V}\mid\delta$, there is a $\kappa<\delta$ that is
$<\delta$-strong reflecting $A$ such that this strength and reflection is
weakly indestructible by $<\kappa$-strategically closed posets.

######  Result 4 • 4.

It is not possible to have a cardinal that is Woodin for weak
indestructibility.

###### Proof.

Suppose not: let $\delta$ be Woodin for weak indestructibility. Let $W$ be the
set of all cardinals with weakly indestructible $<\delta$-strength in
$\mathrm{V}$. There is therefore an (arbitrarily large) $\kappa<\delta$ that
is $<\delta$-strong reflecting $W$ such that this is weakly indestructible. In
particular, after adding a Cohen subset
$A\subseteq\mathrm{Add}(\kappa^{+},1)$, $\kappa$ is still $<\delta$-strong
reflecting $W$. Let $j:\mathrm{V}[A]\rightarrow\mathrm{M}[j(A)]$,
$\operatorname{cp}(j)=\kappa$ witness this for some large $\lambda<\delta$. In
$\mathrm{M}[j(A)]$, $j(W)\cap\lambda=W\cap\lambda\ni\kappa$. Consider the
least element $\mu\in W$ above $\kappa$ that is $<\delta$-strong, and assume
without loss of generality that $\lambda$ is large enough that $\mu<\lambda$.
By [hamkinssuper] in $\mathrm{V}[A]$, since $\mathrm{Add}(\kappa^{+},1)$ is
small relative to $\mu$, $\mu$’s strength is weakly destructible by
$\mathrm{Add}(\mu^{+},1)$. Reflected in $\mathrm{V}[A]$, this means
arbitrarily large cardinals $\hat{\mu}\in W$ have their $<\delta$-strength
destroyed by $\mathrm{Add}(\hat{\mu}^{+},1)$. Hence
$\mathrm{Add}(\kappa^{+},1)*\mathrm{Add}(\hat{\mu}^{+},1)$ destroys
$\hat{\mu}$’s $<\delta$-strength in $\mathrm{V}$, contradicting that
$\hat{\mu}\in W$.∎

Despite this result, we can still have a Woodin cardinal such that the
cardinals witnessing this can always be chosen to have weakly indestructible
strength.

######  Definition 4 • 5.

A cardinal $\delta$ is _Woodin witnessed by (weak) indestructibility_ iff
$\delta$ is Woodin, and every $<\delta$-strong cardinal $\kappa<\delta$ has
(weakly) indestructible strength.

Using strongs reflecting strongs again, it’s possible to force any Woodin
cardinal $\delta$ to be witnessed by weak indestructibility. Stated
differently, we get an equiconsistency.

######  Corollary 4 • 6.

The following are equiconsistent with ZFC.

  1. 1.

There is a Woodin cardinal.

  2. 2.

There is a Woodin witnessed by weak indestructibility.

  3. 3.

There is a Woodin cardinal $\delta$ such that
$\mathrm{V}\mid\delta\vDash\textup{UWISS}$.

This suggests there is a delicate balance between (weak) indestructibility and
reflection: one can have weakly indestructible strength with relative ease,
but too much of this precludes too much weakly indestructible reflection
properties by nowoodinforwi. What kinds of non-trivial reflection can be
indestructible is a subject of further research. For example, is it possible
to have a proper class of strongs, and a strong reflecting strongs with this
strength and reflection weakly indestructible? Below a Woodin witnessed by
weak indestructibility, $\delta$, the answer is no, because in
$\mathrm{V}\mid\delta$, the weakly indestructible strong cardinals are just
all of the strong cardinals. But in a more general setting the answer is not
as obvious to me.

Some of the open problems stated above are collected here for convenience.

######  Questions 4 • 7.

All of the questions below can also be rephrased in terms of supercompactness.

  1. 1.

Is it possible to have every $\kappa$’s $<\kappa^{+\sharp\sharp}$-strength
weakly indestructible in the presence of multiple strong cardinals?

  2. 2.

To what extent can the reflection properties in the embeddings of a measurable
cardinal be (weakly) indestructible?

  3. 3.

Is it possible to have a strong reflecting strongs cardinal (with a strong
above it) such that this strength and reflection of (ground model) strongs is
weakly indestructible?

  4. 4.

To what extent can we control the resurrection of degrees after destroying
degrees of strength in a preparation like theprepdef?

  5. 5.

If a poset is $\leq\kappa$-strategically closed, is it
$<\kappa^{+}$-strategically closed?

## References