# What is the shape of large cardinal tree in implication strength order?

There are two natural orders on large cardinal axioms.

(a) Consistency strength order

$\sigma \leq_C \theta \Longleftrightarrow ZFC\vdash Con(ZFC+\theta)\longrightarrow Con(ZFC+\sigma)$

(b) Implication strength order

$\sigma \leq_I \theta \Longleftrightarrow ZFC\vdash \theta\longrightarrow \sigma$

It is well-known that these are not same and many large cardinals (e.g. Woodin cardinals) have different positions in large cardinal tree when we endow it with different orderings.

Question 1. Are there any other important orderings on the tree of large cardinal axioms?

Question 2. What is the shape of large cardinal tree in implication strength order? Is there any diagram somewhere in the texts which summarizes the results on direct implication of large cardinal axioms?

I hope somebody give me an explicit diagram which shows the differences between large cardinal tree in implication and consistency strength orders or at least based on the answers one be able to design a diagram for large cardinal tree in implication order and post it as an answer which summarizes the answers of the others.

• Could you clarify what you are saying about Woodin cardinals? I suspect you refer to the fact that Woodin cardinals need not be measurable (for example.) But the axiom "there is a measurable cardinal" is both consistent relative to, and implied by, the axiom "there is a Woodin cardinal," so this situation does not seem to be an instance of the orderings (a) and (b) of large cardinal axioms being different. – Trevor Wilson Feb 20 '14 at 1:27
• Why do you think it is a tree? Of course, if it is linear, as some set theorists assert, then it will be a tree, but this linearity view seems to be connected with a certain more stringent vision of what counts as a large cardinal axiom. With a more relaxed view, we seem very likely to have a partial order that is not a tree. – Joel David Hamkins Feb 20 '14 at 1:29
• @JoelDavidHamkins: It is a really good point. I don't beleive that large cardinal axioms form a linear order in the sense of $\leq_C$. By the word "tree" I mean a "partially ordered tree" not a "linear cane"! – user45939 Feb 20 '14 at 1:56
• But a tree is a special kind of partial order, one for which the predecessors of every node are linear; so once a node splits, the branches never grow together again. But do you think the large cardinal order is like that? – Joel David Hamkins Feb 20 '14 at 2:08
• @JoelDavidHamkins: Ah! I see what you are saying. It is correct that our knowledge about the "set" of large cardinal axioms doesn't match the mathematical definition of a "tree" yet. Thus using the word "tree" is not appropriate for the set of large cardinal axioms endowed by $\leq_C$. I used the word "tree" just because set theorists usually use it in the literature and I don't mean any further properties of the set of large cardinal axioms by using the word "tree" for it. – user45939 Feb 20 '14 at 2:19

As an answer to the question (1), one can produce a series of natural orderings on large cardinal axioms using the operator $Con(ZFC+\dots)$.

$\sigma \leq_0 \theta \Longleftrightarrow ZFC\vdash \theta \longrightarrow \sigma$

$\sigma \leq_1 \theta \Longleftrightarrow ZFC\vdash Con(ZFC+\theta) \longrightarrow Con(ZFC+\sigma)$

$\sigma \leq_2 \theta \Longleftrightarrow ZFC\vdash Con(ZFC+Con(ZFC+\theta)) \longrightarrow Con(ZFC+Con(ZFC+\sigma))$

$\cdots$

We have $\leq_0 \subseteq \leq_1 \subseteq \leq_2\subseteq \cdots$ and one can think about an animated diagram which shows the change of positions of each large cardinal axiom in the tree when its ordering varies over $\langle \leq_i:i\in \omega\rangle$.

An important property of these orderings on large cardinal axioms is that by the definitions we have $\forall \sigma,\theta \in \text{Large Cardinal Axioms}~~\forall i,j\in \omega~~((i\leq j~~\wedge~~\sigma=_i \theta )\Longrightarrow \sigma =_j \theta)$ and so maybe there is a (possibly ordinal valued) step $k$ such that all main large cardinal axioms are equivalent in the sense of $\leq_k$ ordering. Note that the same phenomenon happens between $WI$ (existence of a weakly inaccessible cardinal) and $SI$ (existence of a strongly inaccessible cardinal) because we have $WI<_0 SI$ but $\forall i\geq 1~~WI=_i SI$.

Also one can ask some natural questions about the interactions of these orderings with each other as follows.

Definition. Let $\sigma,\theta$ be two large cardinal axioms. Define the convergence number of $\sigma, \theta$ ($\alpha_{\sigma, \theta}$) to be $min\{\gamma\in Ord~|~\sigma =_\gamma \theta\}$ if $\{\gamma\in Ord~|~\sigma =_\gamma \theta\}\neq \emptyset$ and $\infty$ if$\{\gamma\in Ord~|~\sigma =_\gamma \theta\}=\emptyset$.

Example. $\alpha_{WI,SI}=1$

Question 1. Is $\alpha_{\sigma,\theta}<\infty$ for each two large cardinal axioms $\sigma, \theta$?

Question 2. Is there an ordinal $\alpha_0$ such that for each two large cardinal axiom $\sigma,\theta$ we have $\alpha_{\sigma,\theta}\leq \alpha_0$?

Question 3. What is the convergence number of "existence of a strongly compact cardinal" and "existence of a superstrong cardinal"?

Question 4: Consider $\mathbb{L}$ be the set of large cardinal axioms. Is there any ordinal $\alpha$ such that the $\langle \mathbb{L}, \leq_{\alpha}\rangle$ be a linear order?

• How do you iterate your Con() operator past $\omega_1^{\mathrm{CK}}$? My understanding is that there's no clearly-defined notion past about that point; mathoverflow.net/questions/153272/… isn't quite the same, but the iterated-consistency operator there should be much stronger than even the one here... – Steven Stadnicki Feb 20 '14 at 2:05
• If $\sigma$ says "there is a strongly inaccessible cardinal" and "$\theta$ says "there is a Mahlo cardinal", then from $\mathsf{ZFC} + \theta$ we can prove $\sigma + \text{Con}(\sigma) + \text{Con}(\sigma + \text{Con}(\sigma)) + \text{Con}(\ldots) + \cdots$, iterated up to any recursive ordinal (and as Steven points out it is not clear what it would mean to continue farther) so the convergence you are talking about does not seem to occur... – Trevor Wilson Feb 20 '14 at 2:53
• Moreover, as far as I know the same thing happens with any pair of large cardinal axioms $\sigma$ and $\theta$ such that $\mathsf{ZFC} + \theta$ proves $\sigma + \text{Con}(\sigma)$. (That is, if such statements have different consistency strength, then the difference in their consistency strength is very large compared to the differences in consistency strength we get by tacking on "Con" statements.) – Trevor Wilson Feb 20 '14 at 2:56
• If $1 \le i < \omega$ then the ordering $\le_i$ on all the large cardinal axioms that I can think of ("natural" large cardinal axioms, not metamathematical statements like consistency statements) seems to be the same as the ordering $\le_1$. Do you have any counterexample to this? – Trevor Wilson Feb 20 '14 at 4:20
• Regarding the possibility of differences between the orderings $\le_i$ for $i \ge 1$: Do you know the proof that $\mathsf{ZFC} + {}$"there is a strongly inaccessible cardinal" implies $\text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC} + \cdots)\cdots)$? Similar things happen higher up in the large cardinal hierarchy, and to me this makes the idea that the orderings are different seem unlikely. In particular, it is not hard to show that "there is a supercompact cardinal" is stronger than any iterated consistency statement obtained from "there is a Woodin cardinal". – Trevor Wilson Feb 20 '14 at 21:58

There is also:

c) interpretability strength order:

$\sigma \leq_F \theta \Longleftrightarrow \exists f\ \forall \psi\ ZFC\vdash Con(ZFC+\theta+f(\psi))\longrightarrow Con(ZFC+\sigma+\psi)$

for some suitable interpretation $f$. This article on Independence and Large Cardinals gives more exposition.

A linear order is also likely here.

If we found $\theta$ and $\phi$ to be incompatible, we would probably say "$\theta$ is a large cardinal axiom, about the size of the set-theoretic universe, and $\phi$ is more about the structure of the universe, so it's not really a large cardinal axiom."