Set theory - Chapter 9: Combinatorial Set Theory


  1. (1)

    Let SS be an infinite set endowed with a partial ordering. Let F:[S]2{0,1}F:{[S]}^{2}\rightarrow\{0,1\} such that F(x,y)=1F({x,y})=1 if and only if xx and yy are comparable. Then an infinite homogeneous set for FF is either a chain or a set of mutually incomparable elements

  2. (2)

    Let SS be an infinite set endowed with a total ordering. Let {xα:α<|S|}\{x_{\alpha}:\alpha<{|S|}\} an enumeration of SS and define F:[S]2{0,1}F:{[S]}^{2}\rightarrow\{0,1\} as follows: for all α<β\alpha<\beta, F(xα,xβ)=1F({x_{\alpha},x_{\beta}})=1 if and only if xα<xβx_{\alpha}<x_{\beta}. Then an infinite homogeneous set provides an increasing or decreasing sequence xφ(α)x_{{\varphi(\alpha)}}.


Let κ\kappa be an infinite cardinal. Define exp0(κ)=κ\exp_{{0}}(\kappa)=\kappa and for all nn\in\mathbb{N}, expn+1(κ)=2expn(κ)\exp_{{n+1}}(\kappa)=2^{{\exp_{{n}}(\kappa)}}. For all n1n\geq 1, we shall prove the partition property (expn(κ))+(κ+)κn+1{(\exp_{n}(\kappa))}^{+}\rightarrow{(\kappa^{+})}_{\kappa}^{{n+1}} For n=1n=1 we get (2κ)+(κ+)κ2{(2^{\kappa})}^{+}\rightarrow{(\kappa^{+})}_{\kappa}^{2} and in particular (2κ)+(κ+)2{(2^{\kappa})}^{+}\rightarrow{(\kappa^{+})}^{2}. We will denote λn=expn(κ)\lambda_{n}=\exp_{n}(\kappa).

Let n1n\geq 1 and F:[λn+]n+1κF:{[\lambda_{n}^{+}]}^{{n+1}}\rightarrow\kappa. Define for all aλn+a\in\lambda_{n}^{+} the function FaF_{a} on [λn+{a}]n{\left[\lambda_{n}^{+}\setminus\{a\}\right]}^{{n}} by Fa(x)=F(x{a})F_{a}(x)=F(x\cup\{a\}).

First we construct by induction on α<λn-1+\alpha<\lambda_{{n-1}}^{+} a sequence Aαλn+A_{\alpha}\subseteq\lambda_{n}^{+} of sets of size λn\lambda_{{n}}. A0A_{0} is arbitrary and at limit step Aα=β<αAβA_{\alpha}=\bigcup_{{\beta<\alpha}}A_{\beta}. If AαA_{\alpha} is defined then for all C[Aα]<λn-1+C\in{[A_{\alpha}]}^{{<\lambda_{{n-1}}^{+}}} the sets of Fu|C{F_{u}}_{{|C}} such that uλn+Cu\in\lambda_{n}^{+}\setminus C is of size at most κ|C|=κλn-1=κexpn-1(κ)=2(expn-1(κ))=expn(κ)=λn\kappa^{{|C|}}=\kappa^{{\lambda_{{n-1}}}}=\kappa^{{\exp_{{n-1}}(\kappa)}}=2^{{% (\exp_{{n-1}}(\kappa))}}=\exp_{n}(\kappa)=\lambda_{n}. Let VCλn+CV_{C}\subseteq\lambda_{n}^{+}\setminus C of size λn\lambda_{n} such that {Fu|C:uλn+C}={Fv|C:vVC\{{F_{u}}_{{|C}}:u\in\lambda_{n}^{+}\setminus C\}=\{{F_{v}}_{{|C}}:v\in V_{C} and define Aα+1=AαC[Aα]<λn-1+VCA_{{\alpha+1}}=A_{\alpha}\cup\bigcup_{{C\in{[A_{\alpha}]}^{{<\lambda_{{n-1}}^{% +}}}}}V_{C}. We have λn=|Aα||Aα+1|λn+C[Aα]<λn-1+|VC|λn+|[Aα]<λn-1+|λn=λnλn-1λn=2λn-1λn=λn\lambda_{{n}}=|A_{\alpha}|\leq|A_{{\alpha+1}}|\leq\lambda_{{n}}+\sum_{{C\in{[A% _{\alpha}]}^{{<\lambda_{{n-1}}^{+}}}}}{|V_{C}|}\leq\lambda_{{n}}+{\left|{[A_{% \alpha}]}^{{<\lambda_{{n-1}}^{+}}}\right|}\lambda_{n}=\lambda_{n}^{{\lambda_{{% n-1}}}}\lambda_{n}=2^{{\lambda_{{n-1}}}}\lambda_{n}=\lambda_{n}.

Now define A=α<λn-1+AαA=\bigcup_{{\alpha<\lambda_{{n-1}}^{+}}}A_{\alpha}, which is of size at most λn-1+λn=λn<λn+\lambda_{{n-1}}^{+}\lambda_{n}=\lambda_{n}<\lambda_{n}^{+}. In particular, there exists aλn+Aa\in\lambda_{n}^{+}\setminus A. We shall define by induction a set X={xα:α<λn-1+}AX=\{x_{\alpha}:\alpha<\lambda_{{n-1}}^{+}\}\subseteq A. Let x0Ax_{0}\in A be arbitrary. Suppose C={xβ:β<α}C=\{x_{\beta}:\beta<\alpha\} is defined. |C|=|α|<λn-1+|C|=|\alpha|<\lambda_{{n-1}}^{+}. So there is α<λn-1+\alpha<\lambda_{{n-1}}^{+} such that CAαC\subseteq A_{\alpha}. We have aλn-1+Aλn-1+Aαa\in\lambda_{{n-1}}^{+}\setminus A\subseteq\lambda_{{n-1}}^{+}\setminus A_{\alpha} so there is xα+1Aα+1Ax_{{\alpha+1}}\in A_{{\alpha+1}}\subseteq A such that Fxα+1|C=Fa|C{F_{{x_{{\alpha+1}}}}}_{{|C}}={F_{a}}_{{|C}}.

Now for any xα1<xα2<<xαn+1x_{{\alpha_{1}}}<x_{{\alpha_{2}}}<...<x_{{\alpha_{{n+1}}}} elements of XX, we have F({xα1,xα2,,xαn+1})=Fxαn+1({xα1,xα2,,xαn})=Fa({xα1,xα2,,xαn})=G({xα1,xα2,,xαn})F(\{x_{{\alpha_{1}}},x_{{\alpha_{2}}},...,x_{{\alpha_{{n+1}}}}\})=F_{{x_{{% \alpha_{{n+1}}}}}}(\{x_{{\alpha_{1}}},x_{{\alpha_{2}}},...,x_{{\alpha_{{n}}}}% \})=F_{a}(\{x_{{\alpha_{1}}},x_{{\alpha_{2}}},...,x_{{\alpha_{{n}}}}\})=G(\{x_% {{\alpha_{1}}},x_{{\alpha_{2}}},...,x_{{\alpha_{{n}}}}\}) where G:[X]nκG:{[X]}^{n}\rightarrow\kappa is defined by xFa(x)x\mapsto F_{a}(x). By construction |X|=λn-1+|X|=\lambda_{{n-1}}^{+}.

If n=1n=1 then |X|=λn-1+=κ+>κ|X|=\lambda_{{n-1}}^{+}=\kappa^{+}>\kappa and κ+\kappa^{+} regular. So by there is HXH\subseteq X of size κ+\kappa^{+} such that GG is constant on [X]nX{[X]}^{n}\cong X. If n>1n>1 and if the result is true for n-1n-1 then applying this to GG also gives an homogeneous HXH\subseteq X of size κ+\kappa^{+}. Finally, FF is constant on [H]n+1{[H]}^{{n+1}}.


Let A,BA,B be a partition of [ω1]2{[\omega_{1}]}^{2}. For all α<ω1\alpha<\omega_{1} let KαK_{\alpha} be a maximal subset of α\alpha such that [Kα{α}]2B[K_{\alpha}\cup\{\alpha\}]^{2}\subseteq B. Suppose that one of the KαK_{\alpha} is infinite. Then extract an ω\omega-sequence α0<α1<\alpha_{0}<\alpha_{1}<... in KααK_{\alpha}\subseteq\alpha. Then H={αn:n<ω}{α}H=\{\alpha_{n}:n<\omega\}\cup\{\alpha\} is of order type ω+1\omega+1 and [H]2B{[H]}^{2}\subseteq B.

Otherwise, all the KαK_{\alpha} are finite and we can write the element of KαK_{\alpha} βα1<βα2<<βαnα\beta_{{\alpha 1}}<\beta_{{\alpha 2}}<...<\beta_{{\alpha n_{\alpha}}}. Consider the stationary set C=ω1ωC=\omega_{1}\setminus\omega. Then for all αC\alpha\in C we have nα<ωαn_{\alpha}<\omega\leq\alpha. By Fodor’s theorem, there exists a stationary set SCS\subseteq C and nn\in\mathbb{N} such that for all αS\alpha\in S |Kα|=nα=n|K_{\alpha}|=n_{\alpha}=n. Similarly, because βα1<α\beta_{{\alpha 1}}<\alpha we can find a stationary set S1SS_{1}\subseteq S and β1\beta_{1} such that for all αS1\alpha\in S_{1}, we have βα1=β1\beta_{{\alpha 1}}=\beta_{1}. And by induction we get β1,,βn\beta_{1},...,\beta_{n} and SnS_{n} stationary such that for all αSn\alpha\in S_{n} we have Kα={β1,β2,,βn}=KK_{\alpha}=\{\beta_{1},\beta_{2},...,\beta_{n}\}=K.

Now let α,βSn\alpha,\beta\in S_{n}. By maximality of α\alpha we have [K{α}{β}]2B{[K\cup\{\alpha\}\cup\{\beta\}]}^{2}\nsubseteq B. But [K{α}{β}]2=[Kα{α}]2[Kβ{β}]2{{β,α}}{[K\cup\{\alpha\}\cup\{\beta\}]}^{2}={[K_{\alpha}\cup\{\alpha\}]}^{2}\cup{[K_{% \beta}\cup\{\beta\}]}^{2}\cup\{\{\beta,\alpha\}\} and the two first sets are included in BB so {β,α}B\{\beta,\alpha\}\notin B that is {β,α}A\{\beta,\alpha\}\in A. SnS_{n} is unbounded so we can construct by induction for all αω1\alpha\in\omega_{1} γαSn{β:β>supξ<αγξ}\gamma_{\alpha}\in S_{n}\cap\{\beta:\beta>\sup_{{\xi<\alpha}}\gamma_{\xi}\}. Then S={γα:α<ω1}S=\{\gamma_{\alpha}:\alpha<\omega_{1}\} is of order type ω1\omega_{1} and [S]2[Sn]2A{[S]}^{2}\subseteq{[S_{n}]}^{2}\subseteq A.


In this exercise, for any infinite set of ordinal AA and ordinal α\alpha, [A]α{[A]}^{\alpha} denotes the set of all increasing α\alpha-sequence in AA. Let κ\kappa be an infinite cardinal. For all s,t[κ]ωs,t\in{[\kappa]}^{\omega}, let sts\equiv t if and only if {n:s(n)t(n)}\{n:s(n)\neq t(n)\} is finite. Pick a representative for each equivalence class. Define a function FF on [κ]ω{[\kappa]}^{{\omega}} by F(s)=0F(s)=0 if ss differs from the representative in its class at an even number of places and F(s)=1F(s)=1 otherwise. Let HκH\subseteq\kappa of order-type ω\omega. Define by induction on nωn\in\omega an increasing sequence of ordinal by xn=min(H{xi:i<n})x_{n}=\min(H\setminus\{x_{i}:i<n\}). Define an increasing ω\omega-sequence ss of elements of HH by n,s(n)=x2nn\in\mathbb{N},s(n)=x_{{2n}}. Let s¯\bar{s} be the representative of ss. Consider the finite set X={n:s(n)s¯(n)}X=\{n:s(n)\neq\bar{s}(n)\} and pick m>maxXm>\max X. Define t(n)=s(n)=x2nt(n)=s(n)=x_{{2n}} if nmn\neq m and t(m)=x2m+1t(m)=x_{{2m+1}}. Then the set tt is still an increasing ω\omega-sequence of elements of HH. We have {n:t(n)s¯(n)}=X{m}\{n:t(n)\neq\bar{s}(n)\}=X\cup\{m\} finite so t¯=s¯\bar{t}=\bar{s}. Moreover, F(s)|X|mod2F(s)\equiv|X|\mod 2 and F(t)|X{m}||X|+1mod2F(t)\equiv|X\cup\{m\}|\equiv|X|+1\mod 2. So F(s)F(t)F(s)\neq F(t) and FF is not constant on [H]ω{[H]}^{\omega}.


Let TT be a tree of height ω\omega such that all levels are finite. For all xTx\in T, let SxS_{x} be the successors of xx. We construct by induction a sequence x0<x1<x2x_{0}<x_{1}<x_{2}... such that o(xn)=no(x_{n})=n and Tn=SxnT_{n}=S_{{x_{n}}} is infinite.

T={xT:o(x)=0}SxT=\bigcup_{{\{x\in T:o(x)=0\}}}S_{x} is infinite and level 00 is finite so we can find so x0x_{0} at level 00 such that T0=Sx0T_{0}=S_{{x_{0}}} is infinite. If xnx_{n} is constructed, then Tn={xn}{xTn:o(x)=n+1}T_{n}=\{x_{n}\}\cup\bigcup_{{\{x\in T_{n}:o(x)=n+1\}}} is infinite and level nn is finite so we can find xn+1>xnx_{{n+1}}>x_{n} at level n+1n+1 such that Tn+1=Sxn+1T_{{n+1}}=S_{{x_{{n+1}}}} is infinite.

If BB is a maximal chain containing X={xn:n}X=\{x_{n}:n\in\mathbb{N}\} then BB is an infinite branch of TT


Let TT be a normal α\alpha-tree. Given xTx\in T, let’s xξx_{\xi} denote the predecessor of xx at level ξo(x)\xi\leq o(x). Define the application ff on TT by f(x)=(xξ+1)ξ<o(x)f(x)={(x_{{\xi+1}})}_{{\xi<o(x)}} and (¯T)\bar{(}T)=f(T). If x,yTx,y\in T are such that f(x)=f(y)f(x)=f(y) then in particular o(x)=o(y)o(x)=o(y). We show that for all β<α\beta<\alpha, any two x,yTx,y\in T at level β\beta such that f(x)=f(y)f(x)=f(y) satisfy x=yx=y. If β=γ+1\beta=\gamma+1 is a successor ordinal then f(x)=f(y)f(x)=f(y) gives x=xγ+1=yγ+1=yx=x_{{\gamma+1}}=y_{{\gamma+1}}=y. If β\beta is a limit ordinal then f(x)=f(y)f(x)=f(y) only gives xξ=yξx_{{\xi}}=y_{{\xi}} for all successor ordinals ξ<β\xi<\beta. Suppose ξ\xi is the least limit ordinal such that xξyξx_{{\xi}}\neq y_{{\xi}}. Then for all ψ<ξ\psi<\xi we have xψ=yψx_{{\psi}}=y_{{\psi}} and so the predecessors of xξx_{{\xi}} and yξy_{{\xi}} are the same. Because TT is a normal tree and ξ\xi is limit we have xξ=yξx_{{\xi}}=y_{{\xi}}, a contradiction. Hence xξ=yξx_{{\xi}}=y_{{\xi}} for all ξ<β\xi<\beta and again, because TT is normal, we get x=yx=y.

We got a bijection TT¯T\cong\bar{T}. Let’s show that it is actually an order isomorphism (T, <)(T¯, )(T,<)\cong(\bar{T},\subsetneq). If xyx\leq y then o(x)o(y)o(x)\leq o(y) and for all ξo(x)\xi\leq o(x), we have xξ=yξx_{\xi}=y_{\xi}. Hence f(x)=(xξ+1)ξ<o(x)=(yξ+1)ξ<o(x)(yξ+1)ξ<o(y)=f(y)f(x)={(x_{{\xi+1}})}_{{\xi<o(x)}}={(y_{{\xi+1}})}_{{\xi<o(x)}}\subseteq{(y_{{% \xi+1}})}_{{\xi<o(y)}}=f(y). Conversely, if f(x)f(y)f(x)\subseteq f(y) then o(x)o(y)o(x)\leq o(y) and xξ=yξx_{{\xi}}=y_{{\xi}} for all successor ordinal o(x)\leq o(x). As above, we get that this is true for all ordinal o(x)\leq o(x) and in particular x=xo(x)=yo(x)yx=x_{{o(x)}}=y_{{o(x)}}\leq y.

Finally, (T¯, )(T, <)(\bar{T},\subsetneq)\cong(T,<) is a normal α\alpha-tree. If o(f(x))=βo(f(x))=\beta then domf(x)=o(x)=o(f(x))=β\operatorname{dom}f(x)=o(x)=o(f(x))=\beta and so the β\beta-th level is {tT¯:domt=β}\{t\in\bar{T}:\operatorname{dom}t=\beta\}. Suppose tst\subseteq s where sT¯s\in\bar{T} and let xTx\in T such that f(x)=sf(x)=s. We have domtdoms=o(x)\operatorname{dom}t\subseteq\operatorname{dom}s=o(x) Then f(x|t|)=(xξ+1)ξ<|t|=(s(ξ))ξ<|t|=(t(ξ))ξ<|t|=tf(x_{{|t|}})=(x_{{\xi+1}})_{{\xi<|t|}}=(s(\xi))_{{\xi<|t|}}=(t(\xi))_{{\xi<|t|% }}=t so tT¯t\in\bar{T}.


Let TT be a normal ω1\omega_{1} tree that has an uncountable branch BB. Each xBx\in B has infinitely many immediate successor and only one in BB so we can pick zxBz_{x}\notin B immediate successor of xx. Let x<yx<y two elements of BB. Then o(zx)=o(x)+1>o(y)+1=o(zy)o(z_{x})=o(x)+1>o(y)+1=o(z_{y}) so zxzyz_{x}\leq z_{y} is impossible. Suppose zy<zxz_{y}<z_{x}. We also have x<zxx<z_{x}. If zyxz_{y}\leq x then zyBz_{y}\in B and if x<zyx<z_{y} we get o(x)<o(zy)=o(y)+1o(x)o(x)<o(z_{y})=o(y)+1\leq o(x). Both cases are impossible and so x,zyx,z_{y} are incomparable predecessors of zxz_{x}, a contradiction. Finally, zx,zyz_{x},z_{y} are incomparable and so A={zx:xB}A=\{z_{x}:x\in B\} is an uncountable antichain of TT.


We let sup=0\sup\emptyset=0 and we define for all xTx\in T, f(x)=supxf(x)=\sup x If x,yTx,y\in T and x<yx<y then there are α<β\alpha<\beta such that xUαx\in U_{\alpha} and yYβy\in Y_{\beta}. We have xyx\subseteq y and so x=y|αyα+1yx=y_{{|\alpha}}\subsetneq y_{{\alpha+1}}\subseteq y. By construction, supy|α<supy|α+1\sup y_{{|\alpha}}<supy_{{|\alpha+1}} and so f(x)=supx<supy=f(y)f(x)=\sup x<\sup y=f(y).



Let WW be an uncountable set of finite sets. Let {Xα:α<ω1}W\{X_{\alpha}:\alpha<\omega_{1}\}\subseteq W. We have |α<ω1Xα|10=1\left|\bigcup_{{\alpha<\omega_{1}}}X_{\alpha}\right|\leq\aleph_{1}\aleph_{0}=% \aleph_{1}. So we may assume α<ω1Xαω1\bigcup_{{\alpha<\omega_{1}}}X_{\alpha}\subseteq\omega_{1}. For all α<ω1\alpha<\omega_{1}, we have Kα=XαααK_{\alpha}=X_{\alpha}\cup\alpha\subseteq\alpha. As in exercise 9.2, we can find a stationary set SS and a finie set XX such that for all α<ω1\alpha<\omega_{1} we have X=XααXαX=X_{\alpha}\cap\alpha\subseteq X_{\alpha}.

We define by induction on α<ω1\alpha<\omega_{1} a sequence βα\beta_{\alpha}. If βξ\beta_{\xi} is defined for all ξ<α\xi<\alpha then γα=supξ<αXβξ<ω1\gamma_{\alpha}=\sup_{{\xi<\alpha}}X_{{\beta_{\xi}}}<\omega_{1} since α<ω1\alpha<\omega_{1} and |Xβξ|<0|X_{{\beta_{\xi}}}|<\aleph_{0}. Because SS is unbounded, we can find taking βα>γα\beta_{\alpha}>\gamma_{\alpha}.

If α1<α2<ω1\alpha_{1}<\alpha_{2}<\omega_{1} then Xβα1βα2X_{{\beta_{{\alpha_{1}}}}}\subseteq\beta_{{\alpha_{2}}} by construction and so Xβα1Xβα2βα2Xβα2=XXβα1Xβα2X_{{\beta_{{\alpha_{1}}}}}\cap X_{{\beta_{{\alpha_{2}}}}}\subseteq\beta_{{% \alpha_{2}}}\cap X_{{\beta_{{\alpha_{2}}}}}=X\subseteq X_{{\beta_{{\alpha_{1}}% }}}\cap X_{{\beta_{{\alpha_{2}}}}}. Hence Z={Xβα:α<ω1}Z=\{X_{{\beta_{{\alpha}}}}:\alpha<\omega_{1}\} is a Δ\Delta-system.


Let κ\kappa be an infinite cardinal such that 2<κ=κ2^{{<\kappa}}=\kappa. So if S=α<κ{0,1}αS=\bigcup_{{\alpha<\kappa}}{\{0,1\}}^{\alpha} we have |S|=α<κ2|α|=2<κ=κ|S|=\sum_{{\alpha<\kappa}}2^{{|\alpha|}}=2^{{<\kappa}}=\kappa. For all f:κ{0,1}f:\kappa\rightarrow\{0,1\} we define Af={sS:sf}={f|α:α<κ}A_{f}=\{s\in S:s\subseteq f\}=\{f_{{|\alpha}}:\alpha<\kappa\}. If fgf\neq g are functions from κ\kappa to {0,1}\{0,1\} then there is β<κ\beta<\kappa such that f(β)g(β)f(\beta)\neq g(\beta). We have |AfAg||{α<κ:f|α=g|α}|β<κ{|A_{f}\cap A_{g}|}\leq{|\{\alpha<\kappa:f_{{|\alpha}}=g_{{|\alpha}}\}|}\leq% \beta<\kappa. So {Af:f{0,1}κ\{A_{f}:f\in{\{0,1\}}^{\kappa} is a family of 2κ2^{\kappa} almost disjoint subsets of SS.


Let \mathcal{F} be a family of 2\aleph_{2} almost disjoint functions f:ω1ωf:\omega_{1}\rightarrow\omega. Each ff\in\mathcal{F} is regressive on C=ω1ωC=\omega_{1}\setminus\omega so there is a stationary set SfCS_{f}\subseteq C such that ff is constant on SfS_{f} with value nfωn_{f}\in\omega. We get an application ω\mathcal{F}\rightarrow\omega by fnff\mapsto n_{f}. Because ||=2>0=|ω||\mathcal{F}|=\aleph_{2}>\aleph_{0}=|\omega| there is 𝒢\mathcal{G}\subseteq\mathcal{F} of size 2\aleph_{2} and nωn\in\omega such that for all f𝒢f\in\mathcal{G} we have nf=nn_{f}=n. Then for all distinct f,g𝒢f,g\in\mathcal{G}, we have |SfSg||{β<ω1:f(β)=gα(β)}|<ω1|S_{f}\cap S_{g}|\leq|\{\beta<\omega_{1}:f(\beta)=g_{{\alpha}}(\beta)\}|<% \omega_{1}.

So 𝒢\mathcal{G} is an almost disjoint family of 2\aleph_{2} stationary sets. Let {Sfα:α<ω2}\{S_{{f_{\alpha}}}:\alpha<\omega_{2}\} an enumeration of 𝒢\mathcal{G}. We construct by induction on α<ω2\alpha<\omega_{2} a family of pairwise disjoint stationary sets UαSfαU_{\alpha}\subseteq S_{{f_{\alpha}}}.

Let α<ω2\alpha<\omega_{2} and suppose we have constructed the pairwise disjoint stationary sets UξU_{\xi} for all ξ<α\xi<\alpha. Cα={β<ω1:β>α}C_{\alpha}=\{\beta<\omega_{1}:\beta>\alpha\} is closed unbounded so Tα=SfαCαT_{\alpha}=S_{{f_{\alpha}}}\cap C_{\alpha} is stationary. Let xTαx\in T_{\alpha}. By induction hypothesis, there is at most one ξ\xi such that xUξx\in U_{\xi}. Let f(x)=ξf(x)=\xi if xUξx\in U_{\xi} and f(x)=αf(x)=\alpha if xx does not belong to any UξU_{\xi}. Then ff is regressive on TαT_{\alpha} and so is constant with value γ\gamma on some stationary set UαTαU_{\alpha}\subseteq T_{\alpha}. If γ<α\gamma<\alpha then UαUγU_{\alpha}\subseteq U_{\gamma} and |Uα|=|UαUγ||SfαSfγ|<ω1|U_{\alpha}|=|U_{\alpha}\cap U_{\gamma}|\leq|S_{{f_{\alpha}}}\cap S_{{f_{% \gamma}}}|<\omega_{1}, contradicting the fact that UαU_{\alpha} is stationary. So γ=α\gamma=\alpha and UαU_{\alpha} is disjoint from all the UξU_{\xi}, ξ<α\xi<\alpha as wanted.


For all x[ω]<ωx\in{[\omega]}^{{<\omega}} define F(x)=1F(x)=1 if |x|x|x|\in x and F(x)=0F(x)=0 otherwise. Let HωH\subseteq\omega be an infinite set. Let nHn\in H and X,YH{n}X,Y\subseteq H\setminus\{n\} such that |X|=n|X|=n and |Y|=n-1|Y|=n-1. Then F(X)=0F(X)=0 and F(Y{n})=1F(Y\cup\{n\})=1 so FF is not constant on [H]n{[H]}^{n}. Thus ω↛(ω)<ω\omega\not\rightarrow{(\omega)}^{{<\omega}}.

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