Difference between revisions of "Ramsey"

Revision as of 12:04, 26 April 2019

Ramsey cardinals were introduced by Erdős and Hajnal in [1]. A cardinal $\kappa$ is Ramsey if it has the partition property $\kappa\rightarrow (\kappa)^{\lt\omega}_2$.

For infinite cardinals $\kappa$ and $\lambda$, the partition property $\kappa\to(\lambda)^n_\gamma$ asserts that for every function $F:[\kappa]^n\to\gamma$ there is $H\subseteq\kappa$ with $|H|=\lambda$ such that $F\upharpoonright[H]^n$ is constant. Here $[X]^n$ is the set of all $n$-elements subsets of $X$. The more general partition property $\kappa\to(\lambda)^{\lt\omega}_\gamma$ asserts that for every function $F:[\kappa]^{\lt\omega}\to\gamma$ there is $H\subseteq\kappa$ with $|H|=\lambda$ such that $F\upharpoonright[H]^n$ is constant for every $n$, although the value of $F$ on $[H]^n$ may be different for different $n$. Indeed, if $\kappa$ is Ramsey, then $\kappa\rightarrow (\kappa)^{\lt\omega}_\lambda$ for every $\lambda<\kappa$. Ramsey cardinals were named in honor of Frank Ramsey, whose Ramsey theorem for partition properties of $\omega$ motivated the generalizations of these to uncountable cardinals. A Ramsey cardinal $\kappa$ is exactly the $\kappa$-Erdős cardinal.

Ramsey cardinals have a number of other characterizations. They may be characterized model theoretically through the existence of $\kappa$-sized sets of indiscernibles for models meeting the criteria discussed below, as well as through the existence of $\kappa$-sized models of set theory without power set with iterable ultrapowers.

Indiscernibles: Suppose $\mathcal A=(A,\ldots)$ is a model of a language $\mathcal L$ of size less than $\kappa$ whose universe $A$ contains $\kappa$ as a subset.

If a cardinal $\kappa$ is Ramsey, then every such model $\mathcal A$ has a $\kappa$-sized set of indiscernibles $H\subseteq\kappa$, that is, for every formula $\varphi(\overline x)$ of $\mathcal L$ and every pair of tuples $\overline \alpha$ and $\overline \beta$ of elements of $H$, we have $\mathcal A\models\varphi (\overline \alpha)\leftrightarrow \varphi(\overline \beta)$. [2]

Good sets of indiscernibles: Suppose $A\subseteq\kappa$ and $L_\kappa[A]$ denotes the $\kappa^{\text{th}}$-level of the universe constructible using a predicate for $A$. A set $I\subseteq\kappa$ is a good set of indiscernibles for the model $\langle L_\kappa[A],A\rangle$ if for all $\gamma\in I$,

• $\langle L_\gamma[A\cap \gamma],A\cap \gamma\rangle\prec \langle L_\kappa[A], A\rangle$,
• $I\setminus\gamma$ is a set of indiscernibles for the model $\langle L_\kappa[A], A,\xi\rangle_{\xi\in\gamma}$.

A cardinal $\kappa$ is Ramsey if and only if for every $A\subseteq\kappa$, there is a $\kappa$-sized good set of indiscernibles for the model $\langle L_\kappa[A], A\rangle$. [3]

$M$-ultrafilters: Suppose a transitive $M\models {\rm ZFC}^-$, the theory ${\rm ZFC}$ without the power set axiom (and using collection and separation rather than merely replacement) and $\kappa$ is a cardinal in $M$. We call $U\subseteq P(\kappa)^M$ an $M$-ultrafilter if the model $\langle M,U\rangle\models$“$U$ is a normal ultrafilter on $\kappa$”. In the case when the $M$-ultrafilter is not an element of $M$, the model $\langle M,U\rangle$ of $M$ together with a predicate for $U$ often fails to satisfy much of ${\rm ZFC}$. An $M$-ultrafilter $U$ is said to be weakly amenable (to $M$) if for every $A\in M$ of size $\kappa$ in $M$, the intersection $A\cap U$ is an element of $M$. An $M$-ultrafilter $U$ is countably complete if every countable sequence (possibly external to $M$) of elements of $U$ has a non-empty intersection (even if the intersection is not itself an element of $M$). A weak $\kappa$-model is a transitive set $M\models {\rm ZFC}^- $ of size $\kappa$ and containing $\kappa$ as an element. A modified ultrapower construction using only functions on $\kappa$ that are elements of $M$ can be carried out with an $M$-ultrafilter. If the $M$-ultrafilter happens to be countably complete, then the standard argument shows that the ultrapower is well-founded. If the $M$-ultrafilter is moreover weakly amenable, then a weakly amenable ultrafilter on the image of $\kappa$ in the well-founded ultrapower can be constructed from images of the pieces of $U$ that are in $M$. The ultrapower construction may be iterated in this manner, taking direct limits at limit stages, and in this case the countable completeness of the $M$-ultrafilter ensures that every stage of the iteration produces a well-founded model. [4] (Ch. 19)

A cardinal $\kappa$ is Ramsey if and only if every $A\subseteq\kappa$ is contained in a weak $\kappa$-model $M$ for which there exists a weakly amenable countably complete $M$-ultrafilter on $\kappa$. [3]

Ramsey cardinals and the constructible universe

Ramsey cardinals imply that $0^\sharp$ exists and hence there cannot be Ramsey cardinals in $L$. [4]

Relations with other large cardinals

• Measurable cardinals are Ramsey and stationary limits of Ramsey cardinals. [1]
• Ramsey cardinals are unfoldable (using the $M$-ultrafilters characterization) and stationary limits of unfoldable cardinals (as they are stationary limits of $\omega_1$-iterable cardinals).
• Ramsey cardinals are stationary limits of completely ineffable cardinals, they are weakly ineffable but but the least Ramsey cardinal is not ineffable. [5]

Ramsey cardinals and forcing

• Ramsey cardinals are preserved by small forcing. [4]
• Ramsey cardinals $\kappa$ are preserved by the canonical forcing of the ${\rm GCH}$, by fast function forcing, and by the forcing to add a slim $\kappa$-Kurepa tree. [6]
• If $\kappa$ is Ramsey, there is a forcing extension in which $\kappa$ remains Ramsey and $2^\kappa\gt\kappa$. Indeed, if the ${\rm GCH}$ holds and $F$ is a class function on the regular cardinals having a closure point at $\kappa$ and satisfying $F(\alpha)\leq F(\beta)$ for $\alpha<\beta$ and $\text{cf}(F(\alpha))>\alpha$, then there is a cofinality preserving forcing extension in which $\kappa$ remains Ramsey and $2^\delta=F(\delta)$ for every regular cardinal $\delta$. [7]
• If the existence of Ramsey cardinals is consistent with ZFC, then there is a model of ZFC in which $\kappa$ is not Ramsey, but becomes Ramsey in a forcing extension. [6]

Ramsey-like cardinals

There are many Ramsey-like cardinals, most of which can be found in [5].

Completely Romsey cardinals etc.

(All information in this subsection are from [8])

Basic definitions

• $\mathcal{P}(x)$ is the powerset (set of all subsets) of $x$. $\mathcal{P}_k(x)$ is the set of all subsets of $x$ with exactly $k$ elements.
• $f:\mathcal{P}_k(λ) \to λ$ is regressive iff for all $A \in \mathcal{P}_k(λ)$, we have $f(A) < \min(A)$.
• $E$ is $f$-homogenous iff $E \subseteq λ$ and for all $B,C \in \mathcal{P}_k(E)$, we have $f(B) = f(C)$.

$Π_α$-Ramsey and completely Ramsey

Suppose that $κ$ is a regular uncountable cardinal and $I \supseteq \mathcal{P}_{<κ}(κ)$ is an ideal on $κ$. For every $X \subseteq $κ, $X \in \mathcal{R}^+(I)$ iff for every regressive function $f:\mathcal{P}_{<ω}(κ) \to κ$, for every club $C \subseteq κ$, there is a $Y \in I^+f$ such that $Y \subseteq X \cap C$ and $Y$ is homogeneous for $f$.

$\mathcal{R}(I) = \mathcal{P}(κ) - \mathcal{R}^+(I)$

$\mathcal{R}^*(I) = \{ X \subseteq κ : κ - X \in \mathcal{R}(I) \}$

A regular uncountable cardinal $κ$ is Ramsey iff $κ \not\in \mathcal{R}(\mathcal{P}_{<κ}(κ))$. If it is Ramsey, we call $\mathcal{R}(\mathcal{P}_{<κ}(κ))$ the Ramsey ideal on $κ$, its dual $\mathcal{R}^*(\mathcal{P}_{<κ}(κ))$ the Ramsey filter and every element of $\mathcal{R}^+(\mathcal{P}_{<κ}(κ))$ a Ramsey subset of $κ$.

For a regular uncountable cardinal $κ$, we define

• $I_{-2}^κ = \mathcal{P}_{<κ}(κ)$
• $I_{-1}^κ = NS_κ$ (the set of non-stationary subsets of $κ$)
• for $n < ω$, $I_n^κ = \mathcal{R}(I_{n-2}^κ)$
• for $α \geq ω$, $I_{α+1}^κ = \mathcal{R}(I_α^κ)$
• for limit ordinal $γ$, $I_γ^κ = \bigcup_{β<γ} \mathcal{R}(I_β^κ)$

Regular uncountable cardinal $κ$ is $Π_α$-Ramsey iff $κ \not\in I_α^κ$ and completely Ramsey iff for all $α$, $κ \not\in I_α^κ$.

If $κ$ is $Π_α$-Ramsey, we call $I_α^κ$ the $Π_α$-Ramsey ideal on $κ$, its dual the $Π_α$-Ramsey filter and every subset of $κ$ not in $I_α^κ$ a $Π_α$-Ramsey subset. If $κ$ is completely Ramsey, we call $I_{θ_κ}^κ$ the completely Ramsey ideal on $κ$, its dual the completely Ramsey filter and every subset of $κ$ not in $I_{θ_κ}^κ$ a completely Ramsey subset. ($θ_κ$ is the least $α$ such that $I_α^κ = I_{α+1}^κ$ — it must exist before $(2^κ)^+$ for every regular uncountable $κ$ — even if the ideals are trivial)

$α$-hyper completely Ramsey and super completely Ramsey

A sequence $⟨f_α:α<κ^+⟩$ of elements of $^κκ$ is a canonical sequence on $κ$ if both

• for all $α, β\in κ$, $α < β$ implies $f_α < f_β$.
• and for any other sequence $⟨g_α:α<κ^+⟩$ of elements of $κ^κ$ such that $\forall_{α < β < κ^κ} g_α < g_β$, we have $\forall_{α < κ^+} f_α < g_α$.

Note four facts:

• If $⟨f_α:α<κ^+⟩$ and $⟨g_α:α<κ^+⟩$ both are canonical sequences on $κ$, then for all $α < κ^+$ there is a club $C_α \subseteq κ$ such that $\forall_{γ \in C_α} f_α(γ) = g_α(γ)$. (All pairs of corresponding elements of two sequences of functions are equal on a club.)
• There are canonical sequences on each regular uncountable cardinal.
• If $⟨h_α:α<κ^+⟩$ is a canonical sequence on $κ$, then for all $α < κ^+$ there is a club $C_α \subseteq κ$ such that $\forall_{η \in C_α} h_α(η) < |η|^+$. (Each function in a sequence takes on a club values with cardinality not greater then argument's.)
• For all $β < κ^+$ there is a club $C_β \subseteq κ$ such that for all uncountable regular $λ \in C_β$, the set $\{ γ < λ : f^λ_{f^κ_β(λ)}(γ) = f^κ_β(γ) \}$ contains a club in $λ$, where $\vec {f^λ}$ and $\vec {f^κ}$ are canonical sequences on $λ$ and $κ$ respectively.

For a regular uncountable cardinal $κ$, let $\vec f = ⟨f_α:α<κ^+⟩$ be the canonical sequence on $κ$.

• $κ$ is 0-hyper completely Ramsey iff $κ$ is completely Ramsey.
• For $α<κ^+$, $κ$ is $α+1$-hyper completely Ramsey iff $κ$ is $α$-hyper completely Ramsey and there is a completely Ramsey subset $X$ such that for all $λ \in X$, $λ$ is $f_α(λ)$-hyper completely Ramsey.
• For $γ \leq κ^+$, $κ$ is $γ$-hyper completely Ramsey iff $κ$ is $β$-hyper completely Ramsey for all $β<γ$.
• $κ$ is super completely Ramsey iff $κ$ is $κ^+$-hyper completely Ramsey.

Results

All this properties (being Ramsey itself, $Π_α$-Ramsey, completely Ramsey, $α$-hyper completely Ramsey and super completely Ramsey) are downwards absolute to the Dodd-Jensen core model.

Any measurable cardinal is super completely Ramsey and a stationary limit of super completely Ramsey cardinals.

If $κ$ is $Π_n$-Ramsey, then $κ$ is $Π_{n+1}^1$-indescribable. If $X \subseteq κ$ is a $Π_n$-Ramsey subset, then $X$ is $Π_{n+1}^1$-indescribable.

There are stationary many $Π_n$-Ramsey cardinals below each $Π_{n+1}$-Ramsey cardinal.

If $κ$ is $Π_{α+1}$-Ramsey and $α < κ^+$, then the set of $Π_α$-Ramsey cardinals less then $κ$ is in the $Π_{α+1}$-Ramsey filter on $κ$.

Super Ramsey cardinal

Super Ramsey cardinals were introduced by Gitman in [5]. They strengthen one definition of strong Ramseyness.

A weak $\kappa$-model $M$ is a $\kappa$-model if additionally $M^{\lt\kappa}\subseteq M$.

A cardinal $\kappa$ is super Ramsey if and only if for every $A\subseteq\kappa$, there is some $\kappa$-model $M$ with $A\subseteq M\prec H_{\kappa^+}$ such that there is some $N$ and some $\kappa$-powerset preserving nontrivial elementary embedding $j:M\prec N$.

The following are some facts about super Ramsey cardinals:

• Measurable cardinals are super Ramsey limits of super Ramsey cardinals. [5]
• Super Ramsey cardinals are strongly Ramsey limits of strongly Ramsey cardinals. [5]
• Super Ramseyness is downward absolute to $K$. [9]
• The required $M$ for a super Ramsey embedding is stationarily correct. [5]
• The consistency strength of a super Ramsey cardinal is stronger than that of a strongly Ramsey cardinal and weaker than that of a measurable cardinal. [5]

Strongly Ramsey cardinal

Strongly Ramsey cardinals were introduced by Gitman in [5]. They strengthen the $M$-ultrafilters characterization of Ramsey cardinals from weak $\kappa$-models to $\kappa$-models.

A cardinal $\kappa$ is strongly Ramsey if every $A\subseteq\kappa$ is contained in a $\kappa$-model $M$ for which there exists a weakly amenable $M$-ultrafilter on $\kappa$. An $M$-ultrafilter for a $\kappa$-model $M$ is automatically countably complete since $\langle M,U\rangle$ satisfies that it is $\kappa$-complete and it must be correct about this since $M$ is closed under sequences of length less than $\kappa$.

• Measurable cardinals are strongly Ramsey limits of strongly Ramsey cardinals. [5]
• Strongly Ramsey cardinals are Ramsey and stationary limits of Ramsey cardinals. [5]
• The least strongly Ramsey cardinal is not ineffable. [5]
• Forcing related properties of strongly Ramsey cardinals are the same as those of Ramsey cardinals described above. [6]
• The consistency strength of strongly Ramsey cardinals is stronger than that of Ramsey cardinals. [5]
• Strong Ramseyness is downward absolute to $K$. [9]

Virtually Ramsey cardinal

Virtually Ramsey cardinals were introduced by Sharpe and Welch in [10]. They weaken the good indiscernibles characterization of Ramsey cardinals and were motivated by finding an upper bound on the consistency strength of a variant of Chang's Conjecture studied in [10]. For $A\subseteq\kappa$, define that $\mathscr I=\{\alpha<\kappa\mid$ there is an unbounded good set of indiscernibles $I_\alpha\subseteq\alpha$ for $\langle L_\kappa[A],A\rangle\}$. A cardinal $\kappa$ is virtually Ramsey if for every $A\subseteq\kappa$, the set $\mathscr I$ contains a club of $\kappa$.

Virtually Ramsey cardinals are Mahlo and a virtually Ramsey cardinal that is weakly compact is already Ramsey. It is consistent from a strongly Ramsey cardinal that there is a virtually Ramsey cardinal that is not Ramsey. It is open whether such separation is possible from just a Ramsey cardinal and whether virtually Ramsey cardinals are weaker than Ramsey cardinals.[9]

If κ is virtually Ramsey then κ is greatly Erdős.[10]

Almost Ramsey cardinal

An uncountable cardinal $\kappa$ is almost Ramsey if and only if $\kappa\rightarrow(\alpha)^{<\omega}$ for every $\alpha<\kappa$. Equivalently:

• $\kappa\rightarrow(\alpha)^{<\omega}_\lambda$ for every $\alpha,\lambda<\kappa$
• For every structure $\mathcal{M}$ with language of size $<\kappa$, there is are sets of indiscernibles $I\subseteq\kappa$ for $\mathcal{M}$ of any size $<\kappa$.
• For every $\alpha<\kappa$, $\eta_\alpha$ exists and $\eta_\alpha<\kappa$.
• $\kappa=\text{sup}\{\eta_\alpha:\alpha<\kappa\}$

Where $\eta_\alpha$ is the $\alpha$-Erdős cardinal.

Every almost Ramsey cardinal is a $\beth$-fixed point, but the least almost Ramsey cardinal, if it exists, has cofinality $\omega$. In fact, the least almost Ramsey cardinal is not weakly inaccessible, worldly, or correct. However, if the least almost Ramsey cardinal exists, it is larger than the least $\omega_1$-Erdős cardinal. Any regular almost Ramsey cardinal is worldly, and any worldly almost Ramsey cardinal $\kappa$ has $\kappa$ almost Ramsey cardinals below it.

The existence of a worldly almost Ramsey cardinal is stronger than the existence of a proper class of almost Ramsey cardinals. Therefore, the existence of a Ramsey cardinal is stronger than the existence of a proper class of almost Ramsey cardinals. The existence of a proper class of almost Ramsey cardinals is equivalent to the existence of $\eta_\alpha$ for every $\alpha$. The existence of an almost Ramsey cardinal is stronger than the existence of an $\omega_1$-Erdős cardinal.

The existence of an almost Ramsey cardinal is equivalent to the existence of $\eta^n(\omega)$ for every $n<\omega$. On one hand, if a almost Ramsey cardinal $\kappa$ exists, then $\omega<\kappa$. Then, $\eta_\omega$ is less than $\kappa$. Then, $\eta_{\eta_\omega}$ exists and is less than $\kappa$, and so on. On the other hand, if $\eta^n(\omega)$ exists for every $n<\omega$, then $\text{sup}\{\eta^n(\omega):n<\omega\}$ is almost Ramsey, and in fact the least almost Ramsey cardinal. Note that such a set exists by replacement and has a supremum by union.

The Ramsey cardinals are precisely the Erdős almost Ramsey cardinals and also precisely the weakly compact almost Ramsey cardinals.

If $κ$ is a $2$-weakly Erdős cardinal then κ is almost Ramsey.[10]

$\alpha$-iterable cardinal

The $\alpha$-iterable cardinals for $1\leq\alpha\leq\omega_1$ were introduced by Gitman in [9]. They form a hierarchy of large cardinal notions strengthening weakly compact cardinals, while weakening the $M$-ultrafilters characterization of Ramsey cardinals. Recall that if $\kappa$ is Ramsey, then every $A\subseteq\kappa$ is contained in a weak $\kappa$-model $M$ for which there exists an $M$-ultrafilter, the ultrapower construction with which may be iterated through all the ordinals. Suppose $M$ is a weak $\kappa$-model and $U$ is an $M$-ultrafilter on $\kappa$. Define that:

• $U$ is $0$-good if the ultrapower is well-founded,
• $U$ is 1-good if it is 0-good and weakly amenable,
• for an ordinal $\alpha>1$, $U$ is $\alpha$-good, if it produces at least $\alpha$-many well-founded iterated ultrapowers.

Using a theorem of Gaifman [11], if an $M$-ultrafilter is $\omega_1$-good, then it is already $\alpha$-good for every ordinal $\alpha$.

For $1\leq\alpha\leq\omega_1$, a cardinal $\kappa$ is $\alpha$-iterable if every $A\subseteq\kappa$ is contained in a weak $\kappa$-model $M$ for which there exists an $\alpha$-good $M$-ultrafilter on $\kappa$. The $\alpha$-iterable cardinals form a hierarchy of strength above weakly compact cardinals and below Ramsey cardinals.

The $1$-iterable cardinals are sometimes called the weakly Ramsey cardinals.

• $1$-iterable cardinals are weakly ineffable and stationary limits of completely ineffable cardinals. The least $1$-iterable cardinal is not ineffable. [5]
• An $\alpha$-iterable cardinal is $\beta$-iterable and a stationary limit of $\beta$-iterable cardinals for every $\beta<\alpha$. [9]
• A Ramsey cardinal is $\omega_1$-iterable and a stationary limit of $\omega_1$-iterable cardinals. This is already true of an $\omega_1$- Erdős cardinal. [10]
• It is consistent from an $\omega$-Erdős cardinal that for every $n\in\omega$, there is a proper class of $n$-iterable cardinals.
• A $2$-iterable cardinal is a limit of remarkable cardinals. [9]
• A remarkable cardinal implies the consistency of a $1$-iterable cardinal. [9]
• $\omega_1$-iterable cardinals imply that $0^\sharp$ exists and hence there cannot be $\omega_1$-iterable cardinals in $L$. For $L$-countable $\alpha$, the $\alpha$-iterable cardinals are downward absolute to $L$. In fact, if $0^\sharp$ exists, then every Silver indiscernible is $\alpha$-iterable in $L$ for every $L$-countable $\alpha$. [9]
• $\alpha$-iterable cardinals $\kappa$ are preserved by small forcing, by the canonical forcing of the ${\rm GCH}$, by fast function forcing, and by the forcing to add a slim $\kappa$-Kurepa tree. If $\kappa$ is $\alpha$-iterable, there is a forcing extension in which $\kappa$ remains $\alpha$-iterable and $2^\kappa\gt\kappa$. [12]
• For an additively indecomposable ordinal $λ ≤ ω_1$, $η_λ$ (the least $λ$-Erdős cardinal) is a limit of $λ$-iterable cardinals and if there is a $λ + 1$-iterable cardinal, then there is a $λ$-Erdős cardinal below it.[13]
• If $κ$ is 2-iterable, then $V_κ$ is a model of proper class many virtually $C^{(n)}$-extendible cardinals for every $n < ω$ and of proper class many virtually Shelah for supercompactness cardinals.[13]
• Virtually extendible cardinals are 1-iterable limits of 1-iterable cardinals.[13]
• A virtually $n$-huge* cardinal is an $n+1$-iterable limit of $n+1$-iterable cardinals. If $κ$ is $n+2$-iterable, then $V_κ$ is a model of proper class many virtually $n$-huge* cardinals.[13]
• Every element of a club $C$ witnessing that $κ$ is a Silver cardinal is virtually rank-into-rank.[13]
• If $C ∈ V[H]$, a forcing extension by $\mathrm{Coll}(ω, V_κ)$, is a club in $κ$ of generating indiscernibles for $V_κ$ of order-type $κ$ (like in the definition of Silver cardinals), then each $ξ ∈ C$ is $< ω_1$-iterable.[13]

M-rank

M-rank for Ramsey and Ramsey-like cardinals is analogous to Mitchell rank. A difference is that M-rank for Ramsey-like cardinals can be at most $\kappa^+$ and Mitchell rank for measurable cardinals can be at most $(2^\kappa)^+$. Any strongly Ramsey cardinal $κ$ has Ramsey M-rank $κ^+$, any super Ramsey cardinal $κ$ has strongly Ramsey M-rank $κ^+$ and any measurable cardinal $κ$ has super Ramsey M-rank $κ^+$.[14]