580 -Partition calculus (3)

 

1. Infinitary Jónsson algebras

 

Once again, assume choice throughout. Last lecture, we showed that {\kappa\not\rightarrow(\kappa)^{\aleph_0}} for any {\kappa.} The results below strengthen this fact in several ways.

Definition 1 Let {x} be a set. A function {f:[x]^{\aleph_0}\rightarrow x} is {\omega}-Jónsson for {x} iff for all {y\subseteq x,} if {|y|=|x|,} then {f''[y]^{\aleph_0}=x.}

 

Actually, for {x=\lambda} a cardinal, the examples to follow usually satisfy the stronger requirement that {f''[y]^\omega=\lambda.} In the notation from Definition 16 from last lecture, {\lambda\not\rightarrow[\lambda]^\omega_\lambda.}

The following result was originally proved in 1966 with a significantly more elaborate argument. The proof below, from 1976, is due to Galvin and Prikry.

Theorem 2 (Erdös-Hajnal) For any infinite {x,} there is an {\omega}-Jónsson function for {x.}

 

Proof: It is enough to show that {\lambda\not\rightarrow[\lambda]^\omega_\lambda} for all infinite cardinals {\lambda.}

Define a version of the equivalence relation {E_0} on {[\lambda]^\omega} by setting {x\sim y} iff there is some {\alpha<\sup x} such that {x\setminus\alpha=y\setminus\alpha.} Choose a representative {\pi_\tau} from each {\sim}-equivalence class {\tau.} For {x\in[\lambda]^\omega} let {\tau=[x]_\sim} and let {\alpha_x\in \pi_\tau} be least such that {\pi_\tau\setminus(\alpha_x+1)=x\setminus(\alpha_x+1).}

(Note that {\alpha_x} needs not be in {x,} and that it needs not be the least {\alpha} such that {\pi_\tau\setminus(\alpha+1)=x\setminus(\alpha+1).})

Consider the function {f(x)=\alpha_x.} Although {f} may fail to be {\omega}-Jónsson, we claim that it has the following property: There is an {A\in[\lambda]^\lambda} such that for any {B\in[A]^\lambda,} {f''[B]^\omega\supseteq A.} It is obvious one can now use {f} and a bijection between {A} and {\lambda} to define an {\omega}-Jónsson function for {\lambda.}

To see the claim, assume otherwise and use this to define recursively a sequence {(A_n,\alpha_n:n<\omega)} as follows:

  • {A_0=\lambda.} Pick {A_1\in[\lambda]^\lambda} such that there is some {\alpha_0\in \lambda\setminus f''[A_1]^\omega.}
  • Given {A_{n+1}} (of size {\lambda}) and {\alpha_n,} let {A_{n+2}\in[A_{n+1}\setminus(\alpha_n+1)]^\lambda} be such that there is some {\alpha_{n+1}\in (A_{n+1}\setminus(\alpha_n+1))\setminus f''[A_{n+2}]^\omega.}

It follows that {(\alpha_n:n<\omega)} is strictly increasing, {(A_n:n<\omega)} is strictly decreasing, and for all {k<\omega,} {\{\alpha_n:n\ge k\}\in[A_k]^\omega.}

Now let {y=\{\alpha_n\colon n\in\omega\}} and {\tau=[y]_\sim.} Let {\alpha\in\lambda} be least such that

\displaystyle  \pi_\tau\setminus\alpha=y\setminus\alpha=\{\alpha_n\colon n\ge m\}

for some {m\in\omega,} and fix such {m.} Note that {\{\alpha_n\colon n>m\}\sim \pi_\tau} and {\alpha_m\in \pi_\tau,} so {f(\{\alpha_n:n>m\})=\alpha_m.} It follows that {\alpha_m\in f''[A_{m+1}]^\omega,} contradiction. \Box

In a few particular cases, direct proofs are also possible, and I discuss a few below, before exploring generalizations.

I begin with a result of Solovay. We need a preliminary lemma, generalizing Ulam’s Theorem 13 from lecture II.5 that stationary subsets of {\kappa^+} can be split into {\kappa^+} many stationary subsets:

Theorem 3 (Solovay) Any stationary subset of a regular cardinal {\kappa} can be split into {\kappa} many disjoint stationary sets.

 

Proof: Let { \kappa} be regular and { S\subseteq\kappa} be stationary, and set

\displaystyle  T=\{\alpha\in S:{\rm cf}(\alpha)=\omega or \displaystyle ({\rm cf}(\alpha)>\omega\mbox{ and }S\cap\alpha \mbox{ is not stationary in }\alpha)\}.

We claim that { T} is stationary. To see this, let { C} be an arbitrary club subset of { \kappa.} Then the set { C'} of limit points of { C} is also club (and a subset of { C}), so it meets { S,} since { S} is stationary. Let { \alpha=\min(C'\cap S).} Then either { \alpha} has cofinality { \omega,} so it is in { T,} or else it has uncountable cofinality. In that case, notice that since { \alpha\in C',} it is a limit of points in { C,} so { C\cap\alpha} is club in { \alpha,} and therefore { (C\cap\alpha)'=C'\cap\alpha} is also club in { \alpha.} Were { S\cap\alpha} stationary in { \alpha,} it would meet { C'\cap\alpha,} and this would contradict the minimality of { \alpha.} It follows that { \alpha\in T\cap C,} and therefore { T} is stationary, as wanted.

Let now { \alpha} be an arbitrary point of { T.} If { \alpha} has cofinality { \omega,} it is the limit of an { \omega}-sequence of successor ordinals. Let { f_\alpha} be the increasing enumeration of this sequence, and notice that { {\rm ran}(f_\alpha)\cap T=\emptyset,} since all ordinals in { T} are limit ordinals. Suppose now that { \alpha} has uncountable cofinality, so { S\cap\alpha} is not stationary in { \alpha.} Since { T\subseteq S,} it follows that { T\cap\alpha} is not stationary either, so there is a club subset of { \alpha} disjoint from { T,} and let { f_{\alpha}} be the increasing enumeration of this club set so, again, {{\rm ran}(f_\alpha)\cap T=\emptyset.}

With the sequences { f_\alpha} defined as above for all { \alpha\in T,} we now claim that there is some { \xi<\kappa} such that for all { \eta<\kappa,} the set

\displaystyle  T^\xi_\eta=\{\alpha\in T:\xi\in{\rm dom}(f_\alpha)\mbox{ and } f_\alpha(\xi)\ge\eta\}

is stationary. The proof is by contradiction, assuming that no { \xi<\kappa} is as wanted.

It follows then that for all { \xi<\kappa} there is some {\eta(\xi)<\kappa} such that the set {T_{\eta(\xi)}^\xi} as above is non-stationary. Fix a club { C_\xi} disjoint from it, and let { C} be the club { C=\bigtriangleup_\xi C_\xi.} Let { D=C\cap E,} where { E=\{\alpha<\kappa:\eta''\alpha\subseteq\alpha\}.} Notice that { E} is club, and so is { D.} We claim that { |D\cap T|\le 1.} This contradicts that { T} is stationary, and therefore there must be a { \xi<\kappa} as claimed.

Suppose then that { \gamma<\alpha} are points in { D\cap T.} Since { \alpha\in D,} then { \alpha\in C,} so { \alpha\in C_\xi} (hence, { \alpha\notin T^\xi_{\eta(\xi)}}) for all { \xi<\alpha.} We claim that { \gamma\in{\rm dom}(f_\alpha).} To see this, let { \xi\in\gamma\cap{\rm dom}(f_\alpha).} Then (by definition of { T^\xi_{\eta(\xi)}}), { f_\alpha(\xi)<\eta(\xi).} Since { \gamma\in D,} then { \gamma\in E} and, since { \xi<\gamma,} then { \eta(\xi)<\gamma.} It follows that { \sup{\rm ran}(f_\alpha\upharpoonright\gamma)\le\gamma<\alpha.} Since { f_\alpha} is cofinal in { \alpha,} we must necessarily have { \gamma\in{\rm dom}(f_\alpha).}

Since { f_\alpha} is continuous and { \gamma} is a limit ordinal (since it is in {T}), it follows that { f_\alpha(\gamma)=\sup_{\xi<\gamma}f_\alpha(\xi)\le\gamma.} But, since { f_\alpha} is increasing, then also { f_\alpha(\gamma)\ge\gamma.} Hence, { f_\alpha(\gamma)=\gamma.} We have finally reached a contradiction, because { \gamma\in T,} but the sequence { f_\alpha} was chosen so its range is disjoint from { T.} This proves that { |D\cap T|\le 1,} which of course is a contradiction since { T} is stationary. It follows that indeed there is some { \xi<\kappa} such that all the sets { T_\eta=T_\eta^\xi} are stationary for all {\eta<\kappa.}

Now let { f:T\rightarrow\kappa} be the map

\displaystyle  f(\alpha)=\left\{\begin{array}{cl}f_\alpha(\xi)&\mbox{ if }\xi\in{\rm dom}(f_\alpha),\\ 0&\mbox{ otherwise.}\end{array}\right.

Clearly, { f} is regressive. Also, from the definition of { f,} it follows that

\displaystyle  \{\alpha\in T:f(\alpha)\ge\eta\}=T_\eta

for all { \eta<\kappa,} so { f} is unbounded in { \kappa,} since each { T_\eta} is in fact stationary, as we showed above. Given any { \eta<\kappa,} since { f\upharpoonright T_\eta} is regressive, there is some { \gamma} (necessarily, { \gamma\ge\eta}) such that { \{\alpha\in T_\eta:f(\alpha)=\gamma\}} is stationary, by Fodor’s lemma. A simple induction allows us to define a strictly increasing sequence { (\gamma_\eta:\eta<\kappa)} such that { \{\alpha\in T_{\gamma_\eta+1}:f(\alpha)=\gamma_{\eta+1}\}} is stationary for all { \eta.} Notice that these { \kappa} many subsets of { T} (hence, of { S}) so defined are all disjoint. By adding to one of them whatever (if anything) remains of { S} after removing all these sets, we obtain a partition of { S} into { \kappa} many disjoint stationary subsets, as wanted. \Box

Theorem 4 (Solovay) Let {\mu} be a regular uncountable cardinal. Then {\mu\not\rightarrow[\mu]^\omega_\mu.} In fact, for any infinite regular {\kappa<\mu,} {\mu\not\rightarrow[\mu]^\kappa_\mu,} even if we consider {\kappa} as an order type.

 

Proof: Let {(S_\alpha\colon\alpha<\mu)} be a partition of {S^\mu_\kappa} into stationary sets, and let {f:[\mu]^\kappa\rightarrow\mu} be given by {f(s)=} the unique {\alpha} such that {\sup(s)\in S_\alpha.} Then {f} is as required, i.e., {f''[Y]^\kappa=\mu} for any {Y\in[\mu]^\mu.}

To see this, let {Y\in[\mu]^\mu} and set {Y^*=\{\sup(s)\colon s\in[Y]^\kappa\}.} Then {Y^*} is a {\kappa}-club subset of {\mu} so, for all {\alpha<\mu,} {Y^*\cap S_\alpha\ne0.} Hence {f''[Y]^\kappa=\mu.} \Box

For {\mu=\omega} we can in fact show something stronger:

Theorem 5 ({\mbox{Erd\H os}}-Hajnal) For any infinite cardinal {\kappa,} {\kappa\not\rightarrow[\kappa]^\kappa_{2^\kappa},} i.e., there is a map {f:[\kappa]^\kappa\rightarrow2^\kappa} such that for any {Y\in[\kappa]^\kappa,} {f''[Y]^\kappa=2^\kappa.}

 

Proof: Let {\vec X=(X_\xi\colon\xi<2^\kappa)} enumerate {[\kappa]^\kappa} in such a way that each set is listed {2^\kappa} many times. Inductively define {(Y_\xi\colon\xi<2^\kappa)} so that {Y_\xi\in[X_\xi]^\kappa} and {Y_\xi\ne Y_\eta} for all {\eta<\xi<2^\kappa.}

Let {f(Y_\xi)={\rm ot}(\{\alpha<\xi\colon X_\alpha=X_\xi\}),} and set {f(Y)=0} if {Y\ne Y_\xi} for all {\xi.}

We claim that {f} is as wanted. Because given any {Y\in[\kappa]^\kappa} and {\eta<2^\kappa,} if we let {X_\gamma} be the {(\eta+1)}-st occurrence of {Y} in the sequence {\vec X,} then {f(Y_\gamma)=\eta} and {Y_\gamma\in[Y]^\kappa.} \Box

This can be generalized thanks to the following strengthening of its core idea.

Lemma 6 (Galvin-Prikry) For every infinite cardinal {\kappa} and every {S,} there is an injective function {\varphi:[S]^\kappa\times 2^\kappa\rightarrow[S]^\kappa} such that {\varphi(X,\xi)\subseteq X} for all {X\in[S]^\kappa} and {\xi\in2^\kappa.} (Here, {\kappa} is treated as a cardinal rather than an order type.)

 

Proof: If {|S|\le2^\kappa,} as in Theorem 5, a straightforward transfinite induction suffices to find {\varphi.}

For general {S,} let {(S_\alpha:\alpha<\gamma)} enumerate {[S]^\kappa.} For {\alpha<\gamma} let {\varphi_\alpha:[S_\alpha]^\kappa\times2^\kappa\rightarrow[S_\alpha]^\kappa} be injective and such that {\varphi_\alpha(X,\xi)\subseteq X} for all {X\in[S_\alpha]^\kappa} and {\xi\in2^\kappa.}

Now, given {\xi\in2^\kappa} and {X\in[S]^\kappa,} let {\alpha(X)} be the least {\alpha} such that {|X\cap S_\alpha|=\kappa,} and define

\displaystyle  \varphi(X,\xi)=(X\setminus S_{\alpha(X)})\cup\varphi_{\alpha(X)}(X\cap S_{\alpha(X)},\xi).

Then {\varphi(X,\xi)\in[X]^\kappa} and {\alpha(\varphi(X,\xi))=\alpha(X).} It follows that if {\varphi(X,\xi)=\varphi(Y,\eta)=Z,} say, then there is an ordinal {\alpha} such that {\alpha=\alpha(Z)=\alpha(X)=\alpha(Y).} Therefore {Z\setminus S_\alpha=X\setminus S_\alpha=Y\setminus S_\alpha} and {Z\cap S_\alpha=\varphi_\alpha(X\cap S_\alpha,\xi)=\varphi_\alpha(Y\cap S_\alpha,\eta).} Since {\varphi_\alpha} is 1-1, we conclude that {X\cap S_\alpha=Y\cap S_\alpha} and {\xi=\eta.} But then {X=Y} as well, and {\varphi} is 1-1. \Box

Corollary 7 (Galvin-Prikry) If {\tau} and {\kappa} are cardinals and {\kappa} is infinite, then {\tau\not\rightarrow[\kappa]^\kappa_{2^\kappa}.}

 

Proof: Let {\varphi:[\tau]^\kappa\times 2^\kappa\rightarrow[\tau]^\kappa} be as in Lemma 6 with {S=\tau.} Since {\varphi} is 1-1, we can define {f:[\tau]^\kappa\rightarrow2^\kappa} so that {f(\varphi(X,\xi))=\xi} for all {X\in[\tau]^\kappa} and {\xi\in2^\kappa.} But then {f''[X]^\kappa=2^\kappa.} \Box

The same idea gives us {\omega}-Jónsson functions for some singular cardinals, as shown by Kunen in this argument from 1971.

Theorem 8 Let {{\rm cf}(\lambda)=\omega} for {\lambda} singular strong limit. Then {\lambda\not\rightarrow[\lambda]^\omega_\lambda.}

 

Proof: Let {\{(X_\alpha,\gamma_\alpha)\colon\alpha<2^\lambda\}} enumerate {[\lambda]^\lambda\times\lambda.} By transfinite recursion define {(s_\alpha\colon\alpha<2^\lambda)} such that {s_\alpha\in[X_\alpha]^\omega\setminus\{s_\beta\colon\beta<\alpha\}} for all {\alpha.} This is possible, because there are {\lambda^{\aleph_0}} elements of {[X_\alpha]^\omega} and {\alpha<2^\lambda=\lambda^{\aleph_0}.} Now let {f:[\lambda]^\omega\rightarrow\lambda} be such that {f(s_\alpha)=\gamma_\alpha} for {\alpha<2^\lambda} and {f(s)=0} if {s} is not an {s_\alpha.}

To see that {f} is {\omega}-Jónsson, let {X\in[\lambda]^\lambda.} Let {I=\{\alpha<2^\lambda:X_\alpha=X\},} so

\displaystyle  \lambda=\{\gamma_\alpha: \alpha\in I\}=\{f(s_\alpha):\alpha\in I\}\subseteq f''[X]^\omega.

The result follows. \Box

This gives us another proof of Kunen’s inconsistency Theorem 13 from lectures II.10 and II.12.

Corollary 9 (Kunen) If {j:V\rightarrow M} is elementary, then {V\ne M.}

 

Proof: Argue by contradiction. As usual, let {\lambda} be the first fixed point of {j} past its critical point {\kappa.} If {V=M,} then elementarity implies that {j^n(\kappa)} is a strong limit cardinal for all {n<\omega,} so {\lambda} is a singular strong limit cardinal of cofinality {\omega} and, by Theorem 8, there is an {\omega}-Jónsson function {f:[\lambda]^\omega\rightarrow\lambda.} If {A=j''\lambda\in M,} then by elementarity there is some {s\in[A]^\omega\cap M} such that {j(f)(s)=\kappa.} But then {s=j''t=j(t)} for some {t\in[\lambda]^\omega,} and {\kappa=j(f(t))} is in the range of {j.} Of course, {\kappa} cannot then be the critical point of {j,} and we are done. \Box

For {\kappa=\aleph_0,} one can further strengthen Corollary 7 as follows.

Theorem 10 (Galvin-Prikry) For any set {A} there is a function {f:[A]^{\aleph_0}\rightarrow2^\omega} such that whenever {X,Y\in[A]^{\aleph_0}} and {|X\bigtriangleup Y|<\aleph_0,} then {f(X)=f(Y);} and for any {\omega}-sequence {(P_0,P_1,\dots)} of pairwise disjoint sets in {[A]^2,} and any {t\in 2^\omega,} there is a sequence {(x_0,x_1,\dots)} with {x_n\in P_n} for all {n,} and {f(\{x_n:n\in\omega\})=t.}

 

Proof: In fact, we build {f:[A]^{\aleph_0}\rightarrow2^\omega} so that

  • {f(X)} only depends on {X/{\mathcal P}_\omega(A)} for any {X\in[A]^{\aleph_0},} and
  • Whenever {(P_0,Q_0,P_1,Q_1,\dots)} is a sequence of pairwise disjoint finite subsets of {A} such that {\bigcup_n(P_n\cup Q_n)} is infinite, and {t\in2^\omega,} then there is a sequence {(R_0,R_1,\dots)} with {R_n\in\{P_n,Q_n\}} for all {n,} {\bigcup_n R_n} is infinite, and {f(\bigcup_n R_n)=t.}

To do this, use Zorn’s lemma to find a maximal sequence {(A_\beta:\beta<\alpha)} of pairwise almost disjoint elements of {[A]^{\aleph_0}.} For each {A_\beta,} being countable, it is easy to find a function {f_\beta:[A_\beta]^{\aleph_0}\rightarrow2^\omega} satisfying the two conditions above with {A_\beta} in place of {A.}

Now, given {X\in[A]^{\aleph_0},} set {f(X)=f_\beta(X\cap A_\beta),} where {\beta} is least such that {X\cap A_\beta} is infinite. It is now easy to check that {f} is as wanted. \Box

A variant of Lemma 6 gives us the following:

Lemma 11 (Galvin-Prikry) Given an infinite cardinal {\kappa} and a set {S,} there is an injective function {\varphi:[S]^\kappa\times\kappa\rightarrow[S]^\kappa} such that {\varphi(X,\xi)\subseteq X} and {X\setminus\varphi(X,\xi)} is a singleton for all {X\in[S]^\kappa} and {\xi\in\kappa.}

 

Proof: It suffices to argue when {|S|=\kappa,} and the general case is then handled as in the proof of Lemma 6.

Let {(C_i:i<\tau)} be an injective enumeration of {[S]^\kappa/{\mathcal P}_\omega(S).} Note that {|C_i|=\kappa} for all {i<\tau.} One can easily find {\varphi_i:C_i\times\kappa\rightarrow C_i} such that {\varphi_i} is injective, {\varphi_i(X,\xi)\subseteq X,} and {X\setminus\varphi_i(X,\xi)} is a singleton for all {X\in C_i.}

We can then take {\varphi=\bigcup_i\varphi_i.} \Box

Corollary 12 For all infinite cardinals {\kappa} and all sets {S,} there is a function {f:[S]^\kappa\rightarrow\kappa} such that {\{f(X\setminus\{x\}):x\in X\}=\kappa} for al {X\in[S]^\kappa.}

 

Proof: With {\varphi} as above, let {f:[S]^\kappa\rightarrow\kappa} satisfy {f(\varphi(X,\xi))=\xi} for all {X\in[S]^\kappa} and {\xi\in\kappa.} \Box

We now proceed to show that one can `lift’ exponents and relax the requirements a bit, while still obtaining infinitary Jónsson algebras. We will repeatedly use the following lemma.

Lemma 13 Let {\kappa} be an infinite cardinal and let {S} and {W} be sets, with {W\ne\emptyset.} Given {F:[S]^\kappa\rightarrow[W]^{\le2^\kappa},} there is a function {f:[S]^\kappa\rightarrow W} such that {F(X)\subseteq f''[X]^\kappa} for all {X\in[S]^\kappa.}

 

Proof: Use Lemma 6 to find an injective {\varphi:[S]^\kappa\times 2^\kappa\rightarrow[S]^\kappa} such that {\varphi(X,\xi)\subseteq X} for all {X\in[S]^\kappa} and {\xi\in 2^\kappa.}

Choose injections {h_X:F(X)\rightarrow2^\kappa} for all {X\in[S]^\kappa.}

Now define {f:[S]^\kappa\rightarrow W} so that {f(\varphi(X,h_X(w)))=w} for all {X\in[S]^\kappa} and {w\in F(X).} \Box

Corollary 14 (Hajnal) If {\kappa_0\le\kappa\le\lambda} and {\kappa} is infinite, then {\tau\not\rightarrow[\lambda]^{\kappa_0}_\delta} implies {\tau\not\rightarrow[\lambda]^\kappa_\delta.}

 

Proof: Let {g:[\tau]^{\kappa_0}\rightarrow\delta} witness {\tau\not\rightarrow[\lambda]^{\kappa_0}_\delta.}

Define {F:[\tau]^\kappa\rightarrow[\delta]^{\le2^\kappa}} as follows: Given {X\in[\tau]^\kappa,} let {F(X)=g''[X]^{\kappa_0}.} By Lemma 13, we can then find {f:[\tau]^\kappa\rightarrow\delta} such that {g''[X]^{\kappa_0}\subseteq f''[X]^\kappa} for all {X\in[\tau]^\kappa.} But {\kappa\ge\kappa_0,} so in fact {g''[A]^{\kappa_0}\subseteq f''[A]^\kappa} for all {A\subseteq\tau} with {|A|\ge\kappa.}

Now consider {A\in[\tau]^\lambda.} It follows that {\delta=f''[A]^\kappa,} so {f} witnesses {\tau\not\rightarrow[\lambda]^\kappa_\delta.} \Box

Recall that {\tau\rightarrow[\lambda]^\kappa_{\delta,<\rho}} means that whenever {f:[\tau]^\kappa\rightarrow\delta,} there is some {A\in[\tau]^\lambda} and {|f''[A]^\kappa|<\rho.}

Theorem 15 (Galvin-Prikry) If {\kappa} is infinite, then {\tau\rightarrow[\lambda]^\kappa_\delta} iff {\tau\rightarrow[\lambda]^\kappa_{\delta,<\delta}.}

 

Proof: Clearly, {\tau\rightarrow[\lambda]^\kappa_{\delta,<\delta}} implies {\tau\rightarrow[\lambda]^\kappa_{\delta}.}

Suppose now that {\tau\not\rightarrow[\lambda]^\kappa_{\delta,<\delta},} and let {g:[\tau]^\kappa\rightarrow\delta} be a witness. We need to show that {\tau\not\rightarrow[\lambda]^\kappa_{\delta}.} We may assume that {\lambda\ge\kappa} and that {\delta>2^\kappa,} by Corollary 7.

By Theorem 2 and Corollary 14, {\delta\not\rightarrow[\delta]^\kappa_\delta.} Let {h:[\delta]^\kappa\rightarrow\delta} be a witness, and define {F:[\tau]^\kappa\rightarrow[\delta]^{\le2^\kappa}} by {F(X)=h''[g''[X]^\kappa]^\kappa} for all {X\in[\tau]^\kappa.}

Use Lemma 13 to find a function {f:[\tau]^\kappa\rightarrow\delta} such that {h''[g''[X]^\kappa]^\kappa\subseteq f''[X]^\kappa} for all {X\in[\tau]^\kappa.} Hence the same holds for all {X\in[\tau]^\lambda,} and since {|g''[X]^\kappa|=\delta} for all such {X,} then {h''[g''[X]^\kappa]^\kappa=\delta.} It follows that {f} witnesses {\tau\not\rightarrow[\lambda]^\kappa_\delta.} \Box

Corollary 16 If {\lambda\ge\kappa\ge\aleph_0,} then {\lambda^\kappa\not\rightarrow[\lambda]^\kappa_{\lambda^\kappa}.}

 

Proof: For any {\tau,} any injective {f:[\tau]^\kappa\rightarrow\tau^\kappa} witnesses {\tau\not\rightarrow[\lambda]^\kappa_{\tau^\kappa,<\lambda^\kappa} .} Take {\tau=\lambda^\kappa,} and apply Theorem 15. \Box

Theorem 17 (Galvin-Prikry) If {\kappa} is infinite, {\tau\not\rightarrow[\lambda]^\kappa_{\delta,<\epsilon},} and {\delta\not\rightarrow[\epsilon]^\kappa_{\pi,<\sigma},} then also {\tau\not\rightarrow[\lambda]^\kappa_{\pi,<\sigma}.}

 

Proof: Let {g:[\tau]^\kappa\rightarrow\delta} witness {\tau\not\rightarrow[\lambda]^\kappa_{\delta,<\epsilon},} and {h:[\delta]^\kappa\rightarrow\pi} witness {\delta\not\rightarrow[\epsilon]^\kappa_{\pi,<\sigma}.}

Let {F:[\tau]^\kappa\rightarrow[\pi]^{\le2^\kappa}} be given by {F(X)=h''[g''[X]^\kappa]^\kappa} for all {X\in[\tau]^\kappa,} and find {f:[\tau]^\kappa\rightarrow\pi} with {F(X)\subseteq f''[X]^\kappa} for all such {X.} As before, we also have {h''[g''[X]^\kappa]^\kappa\subseteq f''[X]^\kappa} for all {X\in[\tau]^\lambda.}

But, for any such {X,} {|g''[X]^\kappa|\ge\epsilon} and therefore {|h''[g''[X]^\kappa]^\kappa|\ge\sigma,} and it follows that {f} witnesses {\tau\not\rightarrow[\lambda]^\kappa_{\pi,<\sigma}.} \Box

Galvin and Prikry point out that one could define a relation {\ge_\kappa} by {(\tau,\lambda)\ge_\kappa(\delta,\epsilon)} iff {\tau\not\rightarrow[\lambda]^\kappa_{\delta,<\epsilon},} and then Theorem 17 could be read as saying that {\ge_\kappa} is transitive.

Corollary 18 If {\lambda\ge\kappa\ge\aleph_0} and {\tau^\kappa\not\rightarrow[\lambda^\kappa]^\kappa_{\delta,<\epsilon}} then, in fact, {\tau^\kappa\not\rightarrow[\lambda]^\kappa_{\delta,<\epsilon}.}

 

Proof: As mentioned in the proof of Corollary 14, one always has {\rho\not\rightarrow[\lambda]^\kappa_{\rho^\kappa,<\lambda^\kappa} .} In particular, {\tau^\kappa\not\rightarrow[\lambda]^\kappa_{\tau^\kappa,<\lambda^\kappa}.} By transitivity, {\tau^\kappa\not\rightarrow[\lambda]^\kappa_{\delta,<\epsilon}.} \Box

Corollary 19 If {\kappa\ge\aleph_0} and {\tau\not\rightarrow[\lambda]^\kappa_{\delta,<\epsilon},} then {\tau\not\rightarrow[\lambda]^\kappa_{\delta^\kappa,<\epsilon^\kappa}.}

 

Proof: We may of course assume {\lambda\ge\kappa} and {\delta>1.} By Corollary 7, we may also assume that {\epsilon>2^\kappa.} Clearly, {\delta\not\rightarrow[\epsilon]^\kappa_{\delta^\kappa,<\epsilon^\kappa}.} By transitivity, {\tau\not\rightarrow[\lambda]^\kappa_{\delta^\kappa,<\epsilon^\kappa}.} \Box

Corollary 20 For any infinite cardinal {\kappa,} {\tau\rightarrow[\lambda]^\kappa_\delta} iff {\tau\rightarrow[\lambda]^\kappa_{\delta^\kappa}.}

 

Proof: By Theorem 15 and Corollary 19, {\tau\not\rightarrow[\lambda]^\kappa_\delta} iff {\tau\not\rightarrow[\lambda]^\kappa_{\delta,<\delta},} which implies {\tau\not\rightarrow[\lambda]^\kappa_{\delta^\kappa,<\delta^\kappa}} or, equivalently, {\tau\not\rightarrow[\lambda]^\kappa_{\delta^\kappa}.}

The reverse implication is clear. \Box

Definition 21 (Kunen) For {\lambda} an infinite cardinal, let {P(\lambda)} be the statement that for all functions {f:[\lambda]^{\aleph_0}\rightarrow\lambda} there is an infinite set {A\subseteq\lambda} such that {|A|\not\subseteq f''[A]^{\aleph_0}.}

 

Kunen showed that {P(\lambda)} implies that {\lambda\ge{\mathfrak c}^{++}} and that the assertion that {P(\lambda)} holds for some {\lambda} is of large cardinal character, in the sense that it implies that for all {\alpha<\omega_1} there is an inner model with at least (order type) {\alpha} many measurable cardinals.

Corollary 22 If {\lambda\ge\kappa\ge\aleph_0} and {\tau\rightarrow[\lambda]^\kappa_{\lambda^\kappa},} then {P(\tau)} holds.

 

Proof: By Corollaries 14 and 20, {\tau\rightarrow[\lambda]^\kappa_{\lambda^\kappa}} is equivalent to {\tau\rightarrow[\lambda]^\kappa_\lambda,} which implies {\tau\rightarrow[\lambda]^{\aleph_0}_\lambda,} which implies {P(\tau)}. \Box

It is a question of {\mbox{Erd\H os}} and Hajnal whether {\tau\rightarrow[\lambda]^\kappa_{\lambda^\kappa}} can ever hold. As far as I know, this is still open (I would be glad to hear of any updates). On the other hand, the existence of some {\lambda} such that {P(\lambda)} is consistent.

Theorem 23 (Kunen) If {\kappa} is a {\kappa^+}-supercompact cardinal, then {P(\kappa^+).}

 

Proof: Fix an elementary {j:V\rightarrow M} with critical point {\kappa} such that {{}^{\kappa^+}M\subseteq M,} and let {A=j''\kappa^+.}

Let {F:[\kappa^+]^{\aleph_0}\rightarrow\kappa^+.} As in the proof of Corollary 9, {\kappa\notin (j(F))''[A]^{\aleph_0}.} Also, {A\subset j(\kappa^+).} By elementarity, there is some {X\subset \kappa^+} such that {|X|\not\subseteq F''[X]^{\aleph_0}.} \Box

Theorem 24 (Galvin-Prikry) Let {\kappa} and {\rho} be infinite cardinals, with {\rho} singular. Suppose that {\tau\not\rightarrow[\lambda]^\kappa_\delta} for all {\delta<\rho.} Then also {\tau\not\rightarrow[\lambda]^\kappa_\rho.}

 

Proof: Let {\sigma={\rm cf}(\rho),} and fix a strictly increasing sequence {(\delta_i:i<\sigma)} of cardinals cofinal in {\rho.} Let {g:[\tau]^\kappa\rightarrow\sigma} witness {\tau\not\rightarrow[\lambda]^\kappa_\sigma} and for all {i<\sigma} let {h_i:[\tau]^\kappa\rightarrow\delta_i} witness {\tau\not\rightarrow[\lambda]^\kappa_{\delta_i}.}

Let {F:[\tau]^\kappa\rightarrow[\rho]^{\le2^\kappa}} be given by

\displaystyle  F(X)=\{h_{g(Y)}(Z):Y,Z\in[X]^\kappa\}

for all {X\in[\tau]^\kappa.} Let {f:[\tau]^\kappa\rightarrow\rho} be such that {F(X)\subseteq f''[X]^\kappa} for all such {X.} We claim that {f} witnesses {\tau\not\rightarrow[\lambda]^\kappa_\rho.}

Let {A\in[\tau]^\lambda} and {\xi\in\rho.} Then there is some {i\in\sigma} with {\xi\in \delta_i.} There must then be some {Y\in[A]^\kappa} such that {g(Y)=i,} and some {Z\in[A]^\kappa} such that {h_i(X)=\xi,} and it follows that {\xi=h_{g(Y)}(Z)\in F(X)\subseteq f''[X]^\kappa\subseteq f''[A]^\kappa.} This shows that {f''[A]^\kappa=\rho,} and we are done. \Box

 

Bibliography

 

Here are some references consulted while preparing this note:

 

  • Fred Galvin, Karel Prikry, Borel sets and Ramsey’s theorem, The Journal of Symbolic Logic, 38 (2) (Jun., 1973), 193–198.
  • Fred Galvin, Karel Prikry, Infinitary Jonsson algebras and partition relations, Algebra Universalis, 6 (1976), 367–376.
  • Akihiro Kanamori, The higher infinite, Springer (1994).
  • Kenneth Kunen, Elementary embeddings and infinitary combinatorics, The Journal of Symbolic Logic, 36 (3), (Sep., 1971), 407–413.

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