Are very large cardinal axioms like $I_0$, $I_1$, $I_2$ consistent with $CH$ and $GCH$?

4$\begingroup$ I'm pretty sure they're all unaffected by setsized forcing, so certainly consistent with $CH$. $\endgroup$– Noah SchweberJan 22, 2014 at 22:11

1$\begingroup$ @Asaf: quite right, of course e.g. collapsing the cardinal to $\omega$ ruins its large cardinal properties. :P But I think LevySolovay  which says that forcing with a poset of size $<\kappa$ preserves the measurability of $\kappa$  generalizes to these large cardinal axioms (and indeed all known strong large cardinals)  and since $CH$ can be forced with a poset of size $2^{\aleph_0}$, we would need even less. In general, OP, Godel's conjecture is generally regarded as strongly refuted. $\endgroup$– Noah SchweberJan 22, 2014 at 23:03

1$\begingroup$ Jitra, your question is not about Godel's program. It's about three particular large cardinal axioms. If you wanted to ask about Godel's program, you should ask about Godel's program. If Woodin succeeds in his quest for Ultimate L, then this is going to be strongly refuted (as $\sf GCH$ will hold there), but generally all large cardinal axioms are pretty much consistent with $\sf CH$. As @Noah and I remarked (although implicitly) $\sf CH$ is a very local assertion about a very small cardinal compared to these cardinals. $\endgroup$– Asaf Karagila ♦Jan 22, 2014 at 23:10

1$\begingroup$ @NoahS, I think your remarks about inner models are a bit misplaced. The situation is that almost all the large cardinals are known to be consistent with the GCH by the method of forcing, and one doesn't need inner models for this. Almost all of the usual large cardinals are known to be preserved by the canonical forcing of the GCH. In the case of supercompact cardinals and the other cardinals out of reach of the inner model theory, this is the only way that we know relative consistency with GCH. $\endgroup$– Joel David HamkinsJan 23, 2014 at 1:15

2$\begingroup$ @Jitra, nevertheless, Gödel seems just to have been wrong with his idea that large cardinals would settle the CH. It is a firmly established phenomenon, known under the umbrella of the LevySolovay theorem, that large cardinals are invariant by the forcing to force either CH or its negation, and thus none of our large cardinals can settle CH. And similarly, almost all the large cardinals are consistent with both GCH and its negation. $\endgroup$– Joel David HamkinsJan 23, 2014 at 1:25
4 Answers
It is known in the folklore (I could never find a source for the results) that $I_0$, $I_1$, and $I_2$ cardinals are all indestructible by small forcing, meaning that we can force ${\rm CH}$ over a universe with such a large cardinal without destroying it. I have the argument for $I_0$ written up in lecture notes here. It is basically Noah's argument with a few additional details.
Hamkins showed here that ${\rm GCH}$ can be forced over a universe with an $I_1$cardinal without destroying it and very recently Dimonte and Friedman here showed that ${\rm GCH}$ can be forced without destroying an $I_0$, $I_1$, or $I_2$ cardinal. Indeed, they showed much more generally that any weakly increasing class function $F$ on the regular cardinals satisfying $\text{cf}(F(\alpha))>\alpha$ such that $F\upharpoonright\lambda$ is definable over $V_\lambda$ (where $\lambda$ comes from the definition of the $I_0$, $I_1$, $I_2$ cardinal) is consistent with such a large cardinal. Thus, almost any natually defined continuum pattern (on the regular cardinals), such as say $2^\kappa=\kappa^{++}$, is consistent with these large cardinals.

4$\begingroup$ It's nice to see you here, Victoria! $\endgroup$ Jan 23, 2014 at 1:16

$\begingroup$ Very nice answer Victoria. Thank you very much. $\endgroup$– user45878Jan 23, 2014 at 1:22

1$\begingroup$ Wow, I didn't know the GCH question had been answered. Very cool! $\endgroup$ Jan 23, 2014 at 2:17

4$\begingroup$ I think it's worth mentioning that Laver proved the converse direction of LevySolovay for rankintorank cardinals (you mentioned the other direction as folklore) in the same paper that introduces the definability of ground models result you and Johnstone extended recently. $\endgroup$ Jan 23, 2014 at 3:55

$\begingroup$ @EverettPiper: Good point Everett! Thank you very much. $\endgroup$– user45878Jan 23, 2014 at 4:00
Here's a brief sketch of why, assuming $ZFC+I_0$ is consistent, so is $ZFC+CH+I_0$. (This is just LevySolovay.)
Suppose $\lambda$ is $I_0$  that is, there is a nontrivial elementary embedding $j$ of $L(V_{\lambda+1})=M$ into itself with critical point $\kappa<\lambda$. Note that the critical point $\kappa$ of $j$ must be much larger than $2^{\aleph_0}$, and hence that the usual poset $\mathbb{P}$ forcing $CH$ is in $V_\kappa$. This means that $$j(\mathbb{P})=\mathbb{P}\text{ and }j\upharpoonright \mathbb{P}=id_\mathbb{P} .$$ So, taking $G$ a $\mathbb{P}$generic filter, we have $j$"$G=G$ and we can lift the embedding $j:M\rightarrow M $ to an embedding $j^+: M[G]\rightarrow M[G] $ extending $j$. It's now straightforward to check that this $j^+$ witnesses that $\lambda$ is $I_0$ in $V[G]$; in particular, $M[G]=L(V_{\lambda+1})^{V[G]}$, because $G$ is of small rank.
I've written this hastily; please let me know if I screwed something up. EDIT: See Victoria Gitman's comment below.
As to GCH, as I stated in my comment I think this is likely to require inner model theory for the relevant cardinal, which is well above the current limits of inner model theory. But I may be wrong about this.

$\begingroup$ Hm, some weird spacing issues are going on. $\endgroup$ Jan 22, 2014 at 23:36

2$\begingroup$ Noah, your argument is basically correct, but you have to be a little careful because $L(V_{\lambda+1})^{V[G]}$ is not equal to $L(V_{\lambda+1})[G]$ in this case. This does not matter because you can restrict the resulting embedding to $L(V_{\lambda+1})^{V[G]}$. The reason for the inequality is because $V_{\lambda+1}$ cannot be an element of $L(V_{\lambda+1})^{V[G]}$. See here $\endgroup$ Jan 22, 2014 at 23:51

3$\begingroup$ The fact that $V_{\lambda+1}$ of $V$ cannot be an element of $L(V_{\lambda+1})^{V[G]}$ if $\mathbb P$ is any forcing adding a countable sequence to $\lambda$ is a highly nontrivial theorem of Woodin and Cramer. $\endgroup$ Jan 23, 2014 at 1:47

$\begingroup$ @Victoria, very nice point  I hadn't thought of that, and I definitely didn't know that result! $\endgroup$ Jan 23, 2014 at 2:19
There are some candidate axioms that are beginning to surface on the internet that appear to have large cardinal characteristics and could potentially settle questions like the CH. If they are consistent, they live somewhere in the area above $I_1$ but how they relate to $I_0$ is still not known. (At least not to me.)
I say they "appear" to have large cardinal characteristics because what counts as a large cardinal is not really formalized. Maybe it should be for a question about Godel's Program to really get a solid answer. Then again, maybe things are fine with our current understanding of the notion of large cardinal.
I mention them because they are a little uncharacteristic of large cardinals in that they don't seem to line up with the LevySolovay phenomenon in quite the right way. In particular, these axioms are affected (read: "killed") by small forcings like those that can alter the value of CH in the model.
This might not seem that interesting given that Hamkins has shown that certain kinds of large cardinals are already known to be sensitive to (read: "killed") by small forcing, and so they also don't seem to fall into the LevySolovay scheme. These large cardinals, say supercompact, are "prepared" through forcings that are designed to guarantee they remain supercompact in any further large forcing extension (thus they are called "indestructible"). But Hamkins shows that these same large cardinals are also vulnerable to small forcing as a result of the preparation. (There are obviously more details here and my apologies to Prof. Hamkins if I'm misreading some of your results in this area).
However, the large cardinals I allude to seem to be forcingfragile for altogether different reasons. Some of these reasons include the newly discovered fact that ground models can be definable in their extensions, and structural (and maybe also semantic) requirements of the models involved in the relevant embeddings.

$\begingroup$ Very interesting answer Everett. I think @Asaf 's comment about impact of PFA on CH is somehow related. (The equiconsistency of PFA and supercompacts is unknown yet.) $\endgroup$– user45878Jan 23, 2014 at 5:19

2$\begingroup$ Everett, thanks for this answer. I believe that you are referring to the main result of this paper: jdh.hamkins.org/superdestructibility, which is that small forcing kills the indestructibility of any large cardinal. So if we take, say, "indestructibly supercompact" as a large cardinal notion, then this is a violation of the LevySolovay property. The results are extended in jdh.hamkins.org/dual and then later in jdh.hamkins.org/approximationandcoverproperties. $\endgroup$ Jan 23, 2014 at 14:05
I just want to make one simple observation, which I think got lost in the history of the subject. Unfortunately, I do not speak German and so it will be nice if someone who knows German actually reads the sources (and corrects me if I am wrong).
I believe Cantor's original conjecture was that any infinite subset of reals either has the same size as the natural numbers or the same size as the real numbers.
Under AC this becomes the most cited version of CH, namely that $2^\omega=\omega_1$.
However, I believe it is a mistake to interpret Cantor's conjecture under ZFC. To me, one gets the feeling that Cantor's original CH is true by simply playing with actual sets, he himself showed that every closed set satisfies his original CH. And so the proper interpretation of Cantor's original CH under ZFC should be that every definable subset of reals is either countable or has the same size as the reals.
Of course, "definable" is open to interpretation as well, but large cardinals did settle a lot of these. For example, under large cardinals, there is no counterexample to Cantor's original CH among projective sets, sets in $L(\mathbb R)$, etc.
In this sense, Goedel was not wrong at all when he suggested that large cardinals may help settle the continuum hypothesis. In fact, far from it. The large cardinals freeze so much of the definable part of the universe that it is astonishing how right Goedel was. For example, Woodin showed that under large cardinals the theory of Chang model cannot be changed, he also showed that after collapsing the power set of the supercompact, the theory of $L(\Gamma_{uB},\mathbb R)$ cannot be changed, all of these is what Goedel anticipated and it is, at its roots, motivated by his beliefs.

$\begingroup$ mathoverflow.net/questions/135912/… $\endgroup$– Asaf Karagila ♦Apr 7, 2019 at 17:24

$\begingroup$ haha, it is okay, but i really meant Godel, i meant Godel anticipated that LC will settle instances of CH. $\endgroup$– GrigorApr 7, 2019 at 17:25

$\begingroup$ And I've added $\rm\LaTeX$. And a link in the comments to satisfy your historical curiosity. $\endgroup$– Asaf Karagila ♦Apr 7, 2019 at 17:26

$\begingroup$ in general, i do not believe that Cantor has done the awe inspiring part here, it is Godel, Cantor played with sets and stated what he felt the truth should be (i don't mean to strip Cantor of anything here, this is just as great as almost anything that has been done in math), Godel anticipated an entire subject, an entire way of thinking of sets and the laws that govern them, and this is huge. $\endgroup$– GrigorApr 7, 2019 at 17:30

1$\begingroup$ i also disagree with many instances of periods at the end of sentences. $\endgroup$– GrigorApr 7, 2019 at 18:14