Set theory - Chapter 8: Stationary Sets

8.1

Let ff be a normal function and C={α<κ:f(α)=α}C=\{\alpha<\kappa:f(\alpha)=\alpha\}. If CC is not unbounded then there is γ<κ\gamma<\kappa such that for all αγ\alpha\geq\gamma, f(α)αf(\alpha)\neq\alpha and because ff is increasing f(α)>αf(\alpha)>\alpha. Define the increasing sequence u0=γu_{0}=\gamma and n<ω,un+1=f(un)\forall n<\omega,u_{{n+1}}=f(u_{n}) . Let beta=limnωunγbeta=\lim_{{n\rightarrow\rightarrow\omega}}u_{n}\geq\gamma. κ\kappa is a regular uncountable cardinal so β<κ\beta<\kappa. Because ff is continuous, β<f(β)=limnωf(un)=limnωf(un)=limnωun+1=limnωun=β\beta<f(\beta)=\lim_{{n\rightarrow\omega}}f(u_{n})=\lim_{{n\rightarrow\omega}}% f(u_{n})=\lim_{{n\rightarrow\omega}}u_{{n+1}}=\lim_{{n\rightarrow\omega}}u_{{n% }}=\beta. A contradiction. Hence CC is unbounded.

Now if (αξ)ξ<γ{(\alpha_{\xi})}_{{\xi<\gamma}} is an increasing sequence of elements of CC (i.e. f(αξ)=αξf(\alpha_{\xi})=\alpha_{\xi}) of length γ<κ\gamma<\kappa, then because ff is continuous, we have f(limξγαξ)=limξγf(αξ)=limξγαξf(\lim_{{\xi\rightarrow\gamma}}\alpha_{\xi})=\lim_{{\xi\rightarrow\gamma}}f(% \alpha_{\xi})=\lim_{{\xi\rightarrow\gamma}}\alpha_{\xi} and so limξγαξC\lim_{{\xi\rightarrow\gamma}}\alpha_{\xi}\in C. So CC is closed.

8.2

Let f:κκf:\kappa\rightarrow\kappa and C={α<κ:ξ<α,f(ξ)<α}C=\{\alpha<\kappa:\forall\xi<\alpha,f(\xi)<\alpha\}. Suppose CC is bounded and let γ\gamma such that for all αγ\alpha\geq\gamma αC\alpha\notin C. We define the closed unbounded (thus stationary) set S={α,αγ}S=\{\alpha,\alpha\geq\gamma\}. For all αS\alpha\in S we let g(α)κg(\alpha)\in\kappa be the least ξ<α\xi<\alpha such that f(ξ)αf(\xi)\geq\alpha. Then gg is regressive on SS and by Fodor’s theorem there is a stationary set TST\subseteq S and an ordinal βκ\beta\in\kappa such that αT,f(β)α\forall\alpha\in T,f(\beta)\geq\alpha. In particular TT does not intersect the closed unbounded set {α<κ:α>f(β)}\{\alpha<\kappa:\alpha>f(\beta)\}. A contradiction. So CC is unbounded.

Now let (αξ)ξ<γ{(\alpha_{\xi})}_{{\xi<\gamma}} be an increasing sequence of elements of CC of length γ<κ\gamma<\kappa and α=limξγαξ\alpha=\lim_{{\xi\rightarrow\gamma}}\alpha_{\xi}. For all β<α\beta<\alpha, there exists some ξ<γ\xi<\gamma such that β<αξC\beta<\alpha_{\xi}\in C. Then f(β)<αξ<αf(\beta)<\alpha_{\xi}<\alpha and so αC\alpha\in C. Hence CC is closed.

8.3

Let CC be a closed unbounded set and let SS be stationary. For any set CC^{{\prime}} closed unbounded CCC\cap C^{{\prime}} is closed unbounded and so (SC)C=S(CC)(S\cap C)\cap C^{{\prime}}=S\cap(C\cap C^{{\prime}})\neq\emptyset. Hence SCS\cap C is stationary.

8.4

Let XκX\subseteq\kappa be a nonstationary set. Let CC be a closed unbounded set such that XC0X\cap C\neq 0. Define αX,f(α)=sup(Cα)α\forall\alpha\in X,f(\alpha)=\sup{(C\cap\alpha)}\leq\alpha. If α\alpha is a successor ordinal then clearly f(α)<αf(\alpha)<\alpha. Otherwise if α\alpha is limit then because CC is closed, f(α)Cf(\alpha)\in C and so we can not have f(α)=αinXf(\alpha)=\alpha inX. Hence ff is regressive on XX.

Let γ<κ\gamma<\kappa and define Yγ={αX,f(α)γ}Y_{\gamma}=\{\alpha\in X,f(\alpha)\leq\gamma\}. Let βC\beta\in C such that β>γ\beta>\gamma. If αX\alpha\in X and α>β\alpha>\beta then βCα\beta\in C\cap\alpha and so f(α)=sup(Cα)β>γf(\alpha)=\sup{(C\cap\alpha)}\geq\beta>\gamma i.e. αYγ\alpha\notin Y_{\gamma}. Thus αYγαβ\alpha\in Y_{\gamma}\implies\alpha\leq\beta and finally YγY_{\gamma} is bounded.

8.5

8.6

Suppose κ\kappa is least inaccessible cardinal such that κ\kappa is the κ\kappa-th inaccessible cardinal. Let S={κα:α<κ}S=\{\kappa_{\alpha}:\alpha<\kappa\} an enumeration of the stationary set of inaccessible cardinal cardinal below κ\kappa. Define for all α<κ,f(κα)=α\alpha<\kappa,f(\kappa_{\alpha})=\alpha. κα\kappa_{\alpha} is the the α\alpha-th innaccessible cardinal and, by minimality of κ\kappa, α<κα\alpha<\kappa_{\alpha}. So ff is regressive on SS and by Fodor’s theorem there is a stationary set TST\subseteq S such that ff is constant on TT, with value γ\gamma. But then we would have T={κγ}T=\{\kappa_{\gamma}\} which does not intersect the closed unbounded set C={α<κ:α>κγ}C=\{\alpha<\kappa:\alpha>\kappa_{\gamma}\}.

8.7

Let κ\kappa be an uncountable limit cardinal. (If κ=0\kappa=\aleph_{0} then κ\kappa is strong limit). Suppose that the set of strong limit cardinal below κ\kappa is unbounded in κ\kappa. Let α<κ\alpha<\kappa be a cardinal and β<κ\beta<\kappa a strong limit cardinal such that α<β\alpha<\beta. Then 2α<β<κ2^{\alpha}<\beta<\kappa and so κ\kappa is strong limit.

If κ\kappa was additionally weakly inaccessible then it is regular and so κ\kappa is actually inaccessible.

If κ\kappa was additionally a weakly Mahlo cardinal then the set of all regular cardinal below κ\kappa is stationary and so κ\kappa is actually Mahlo.

8.8

Let II be a κ\kappa-complete ideal on κ\kappa and F={κX:XI}F=\{\kappa\setminus X:X\in I\} the dual filter. We have XFYF,YXYF,(κX)Y=X\in F\Leftrightarrow\exists Y\in F,Y\subseteq X\Leftrightarrow\exists Y\in F,% (\kappa\setminus X)\cap Y=\emptyset and thus XIκXFYF,XYX\notin I\Leftrightarrow\kappa\setminus X\notin F\Leftrightarrow\forall Y\in F% ,X\cap Y\neq\emptyset.

Let S0IS_{0}\notin I and ff be a regressive function on S0S_{0}. Suppose that for each γ<κ\gamma<\kappa, the set {αS0:f(α)=γ}I\{\alpha\in S_{0}:f(\alpha)=\gamma\}\in I and define Xγ={αS0:f(α)γ}FX_{\gamma}=\{\alpha\in S_{0}:f(\alpha)\neq\gamma\}\in F. Let X=γ<κXγFX=\triangle_{{\gamma<\kappa}}X_{\gamma}\in F. Then S0XS_{0}\cap X\neq\emptyset and we can pick αS0X\alpha\in S_{0}\cap X. On the one hand, f(α)<αf(\alpha)<\alpha and on the other hand αγ<αXγ\alpha\in\bigcap_{{\gamma<\alpha}}X_{\gamma} that is γ<α,f(α)γ\forall\gamma<\alpha,f(\alpha)\neq\gamma. A contradiction. So for some γ<κ\gamma<\kappa, S={αS0:f(α)=γ}IS=\{\alpha\in S_{0}:f(\alpha)=\gamma\}\notin I and we have SS0S\subseteq S_{0} and ff constant on SS.

Conversely, consider (Xα)α<κFκ{(X_{\alpha})}_{{\alpha<\kappa}}\in F^{\kappa} and suppose that XαF\triangle X_{\alpha}\notin F. Let S0=κξ<κXαIS_{0}=\kappa\setminus\triangle_{{\xi<\kappa}}X_{\alpha}\notin I. If αS0\alpha\in S_{0} then αξ<αXξ\alpha\notin\bigcap_{{\xi<\alpha}}X_{\xi} and there is ξ<κ\xi<\kappa such that αXξ\alpha\notin X_{{\xi}}. Define f(α)f(\alpha) to be the least ξ\xi such αXξ\alpha\notin X_{{\xi}}. Then ff takes a constant value γ\gamma on some SS0S\subseteq S_{0}. For all αS\alpha\in S, αXf(α)=Xγ\alpha\notin X_{{f(\alpha)}}=X_{{\gamma}} and so SXγ=S\cap X_{\gamma}=\emptyset. This contradicts SIS\notin I and XγFX_{{\gamma}}\in F. So II is normal.

8.9

Let FF be nonprincipal filter on ω\omega and II its dual. Necessarily, FF is 0\aleph_{0}-complete. Because FF is nonprincipal there is XFX\in F such that {0}X\{0\}\nsubseteq X. Hence {0}F\{0\}\notin F or otherwise =X{0}F\emptyset=X\cap\{0\}\in F. We have S0=ω{0}IS_{0}=\omega\setminus\{0\}\notin I. Define the regressive function ff on S0S_{0} by f(n)=n-1f(n)=n-1.

If FF is normal then by exercise 8.8, there exists SS0S\subseteq S_{0} such that SIS\notin I and ff takes a constant value mm on SS. We have S={m+1}IS=\{m+1\}\notin I. Then for all XFX\in F, SXS\cap X\neq\emptyset that is m+1Xm+1\in X. But because FF is nonprincipal, there is some XFX\in F such that {m+1}X\{m+1\}\nsubseteq X. A contradiction.

8.10

Let κ\kappa be a singular cardinal. Suppose that II is a normal ideal on κ\kappa that contains all bounded sets. Write κ=limαcf(κ)κα\kappa=\lim_{{\alpha\rightarrow\operatorname{cf}(\kappa)}}\kappa_{{\alpha}} where each κα\kappa_{\alpha} is a cardinal less than κ\kappa. Each κα\kappa_{\alpha} is a bounded subset and so καI\kappa_{\alpha}\in I. For all αcf(κ)\alpha\geq\operatorname{cf}(\kappa), set κα=\kappa_{\alpha}=\emptyset. Then for all ξcf(κ)\xi\geq\operatorname{cf}(\kappa), ξκ=α<cf(κ)κα=α<ξκα\xi\in\kappa=\bigcup_{{\alpha<\operatorname{cf}(\kappa)}}\kappa_{\alpha}=% \bigcup_{{\alpha<\xi}}\kappa_{\alpha} that is ξα<ξκαI\xi\in\sum_{{\alpha<\xi}}\kappa_{\alpha}\in I (because II is normal). If ξ<cf(κ)\xi<\operatorname{cf}(\kappa) then ξcf(κ)I\xi\in\operatorname{cf}(\kappa)\in I (because κ\kappa is regular and II contains the bounded subsets). Thus κ=cf(κ)α<ξκαI\kappa=\operatorname{cf}(\kappa)\cup\sum_{{\alpha<\xi}}\kappa_{\alpha}\in I. A contradiction.

8.11

Note: if CC is closed unbounded in β\beta and α<β\alpha<\beta is a limit point of CC then CαC\cap\alpha is closed unbounded in α\alpha.

  1. (1)

    Suppose STS\subseteq T are stationary sets. Let αTr(α)\alpha\in\operatorname{Tr}(\alpha). For all closed unbounded set CC in α\alpha, CTαCSαC\cap T\cap\alpha\supseteq C\cap S\cap\alpha\neq\emptyset. Hence Tr(S)Tr(T)\operatorname{Tr}(S)\subseteq\operatorname{Tr}(T).

  2. (2)

    Let T,ST,S be stationary sets. By the previous point, Tr(TS)Tr(T)Tr(S)\operatorname{Tr}(T\cup S)\supseteq\operatorname{Tr}(T)\cup\operatorname{Tr}(S). Let αTr(TS)\alpha\in\operatorname{Tr}(T\cup S). Suppose for example that αTr(S)\alpha\notin\operatorname{Tr}(S). Let C1C_{1} be a closed unbounded set in α\alpha such that C1Sα=C_{1}\cap S\cap\alpha=\emptyset. For any closed unbounded set C2C_{2} in α\alpha, C2TαC2C1Tα=C2(C1(TS)α)=(C2C1)(TS)αC_{2}\cap T\cap\alpha\supseteq C_{2}\cap C_{1}\cap T\cap\alpha=C_{2}\cap{(C_{1% }\cap{(T\cup S)}\cap\alpha)}={(C_{2}\cap C_{1})}\cap{(T\cup S)}\cap\alpha\neq\emptyset. Hence αTr(T)\alpha\in\operatorname{Tr}(T).

  3. (3)

    Let αTr(Tr(S))\alpha\in\operatorname{Tr}(\operatorname{Tr}(S)) and CC a closed unbounded set in α\alpha and CC^{{\prime}} the closed unbounded set of limit points of CC. We have CTr(S)αC^{{\prime}}\cap\operatorname{Tr}(S)\cap\alpha\neq\emptyset. Let βCTr(S)α\beta\in C^{{\prime}}\cap\operatorname{Tr}(S)\cap\alpha. We have CβC\cap\beta closed unbounded in β\beta. Because βTr(S)\beta\in\operatorname{Tr}(S), (Cβ)Sβ(C\cap\beta)\cap S\cap\beta\neq\emptyset. Let γCSβ\gamma\in C\cap S\cap\beta. We have γ<β<α\gamma<\beta<\alpha. Hence γCSα\gamma\in C\cap S\cap\alpha and so CSαC\cap S\cap\alpha\neq\emptyset. Hence αTr(S)\alpha\in\operatorname{Tr}(S).

  4. (4)

    Suppose STmodINSS\equiv T\mod I_{{\rm{NS}}}. Let C1C_{1} a closed unbounded set such that C1S=C1TC_{1}\cap S=C_{1}\cap T and C1C_{1}^{{\prime}} be the closed unbounded set of limit points of C1C_{1}. If αTr(S)C1\alpha\in\operatorname{Tr}(S)\cap C_{1}^{{\prime}}, then C1αC_{1}^{{\prime}}\cap\alpha is closed unbounded in α\alpha and for all C2C_{2} closed unbounded in α\alpha, we have (C1αC2)(Sα)C2(C1S)α=C2(C1T)αC2Tα\emptyset\neq(C_{1}^{{\prime}}\cap\alpha\cap C_{2})\cap(S\cap\alpha)\subseteq C% _{2}\cap(C_{1}^{{\prime}}\cap S)\cap\alpha=C_{2}\cap(C_{1}^{{\prime}}\cap T)% \cap\alpha\subseteq C_{2}\cap T\cap\alpha and so αTr(T)\alpha\in\operatorname{Tr}(T). By symmetry, Tr(S)C1=Tr(T)C1\operatorname{Tr}(S)\cap C_{1}^{{\prime}}=\operatorname{Tr}(T)\cap C_{1}^{{% \prime}} that is Tr(S)Tr(T)modINS\operatorname{Tr}(S)\equiv\operatorname{Tr}(T)\mod I_{{\rm{NS}}}

8.12

Let λ<κ\lambda<\kappa be an infinite regular cardinal. We shall prove that Tr(Eλκ)={α<κ:cf(α)λ+}\operatorname{Tr}(E_{{\lambda}}^{\kappa})=\{\alpha<\kappa:\operatorname{cf}(% \alpha)\geq\lambda^{+}\}.

Let α<κ\alpha<\kappa such that cf(α)λ+\operatorname{cf}(\alpha)\geq\lambda^{+}. We have cf(α)>ω\operatorname{cf}(\alpha)>\omega. If CC is closed unbounded in α\alpha then because λ<cf(α)α\lambda<\operatorname{cf}(\alpha)\leq\alpha we can construct an increasing sequence (βξ)ξ<λCλ{(\beta_{\xi})}_{{\xi<\lambda}}\in C^{\lambda}. Let β=limξλβξ\beta=\lim_{{\xi\rightarrow\lambda}}\beta_{\xi}. We have βC\beta\in C and cf(β)=cf(λ)=λ\operatorname{cf}(\beta)=\operatorname{cf}(\lambda)=\lambda. So αTr(Eλκ)\alpha\in\operatorname{Tr}(E_{{\lambda}}^{\kappa}).

Conversely, suppose aTr(Eλκ)a\in\operatorname{Tr}(E_{{\lambda}}^{\kappa}). Let f:cf(α)αf:\operatorname{cf}(\alpha)\rightarrow\alpha be a normal function. f-1(αEλκ)f_{{-1}}(\alpha\cap E_{{\lambda}}^{\kappa}) is stationary in cf(α)\operatorname{cf}(\alpha), in particular nonempty. Let βf-1(αEλκ)\beta\in f_{{-1}}(\alpha\cap E_{{\lambda}}^{\kappa}) we have β<cf(α)\beta<\operatorname{cf}(\alpha) so cf(β)<cf(α)\operatorname{cf}(\beta)<\operatorname{cf}(\alpha). If (βξ)ξ<λcf(α)λ{(\beta_{\xi})}_{{\xi<\lambda}}\in{\operatorname{cf}(\alpha)}^{{\lambda}} is an increasing sequence such that β=limξλβξ\beta=\lim_{{\xi\rightarrow\lambda}}\beta_{\xi}, then f(β)=limξλf(βξ)f(\beta)=\lim_{{\xi\rightarrow\lambda}}f(\beta_{\xi}) and so λ=cf(f(β))=cf(β)<cf(α)\lambda=\operatorname{cf}({f(\beta)})=\operatorname{cf}(\beta)<\operatorname{% cf}(\alpha) i.e. cf(α)λ+\operatorname{cf}(\alpha)\geq\lambda^{+}.

8.13

Let μ<λ\mu<\lambda be regular infinite cardinal. By exercise 8.12, Eλκ={α<κ:cf(α)=λ}{α<κ:cf(α)μ+}=Tr(Eμκ)E_{{\lambda}}^{\kappa}=\{\alpha<\kappa:\operatorname{cf}(\alpha)=\lambda\}% \subseteq\{\alpha<\kappa:\operatorname{cf}(\alpha)\geq\mu^{+}\}=\operatorname{% Tr}(E_{{\mu}}^{\kappa}). Hence Eμκ<EλκE_{{\mu}}^{\kappa}<E_{{\lambda}}^{\kappa}.

Let λ0=0,λ1=1,\lambda_{0}=\aleph_{0},\lambda_{1}=\aleph_{1},... and in general λα\lambda_{\alpha} be the α\alpha-th regular infinite cardinal. We show by induction that o(Eλακ)=αo(E_{{\lambda_{\alpha}}}^{\kappa})=\alpha.

Let SS be a stationary set. All the elements of Tr(S)\operatorname{Tr}(S) have uncountable cofinality while all the elements of EωκE_{{\omega}}^{\kappa} have countable cofinality so for any closed unbounded CC, CEωκTr(S)\emptyset\neq C\cap E_{{\omega}}^{\kappa}\nsubseteq\operatorname{Tr}(S). Hence o(Eωκ)=sup=0o(E_{{\omega}}^{\kappa})=\sup\emptyset=0.

Suppose that o(Eλξκ)=ξo(E_{{\lambda_{\xi}}}^{\kappa})=\xi for all ξ<α\xi<\alpha. Then for all ξ<α\xi<\alpha we have Eλξκ<EλακE_{{\lambda_{\xi}}}^{\kappa}<E_{{\lambda_{{\alpha}}}}^{\kappa} and so o(Eλακ)supξ<α{o(Eλξκ)+1}=supξ<α{ξ+1}=αo(E_{{\lambda_{{\alpha}}}}^{\kappa})\geq\sup_{{\xi<\alpha}}\{o(E_{{\lambda_{% \xi}}}^{\kappa})+1\}=\sup_{{\xi<\alpha}}\{\xi+1\}=\alpha.

8.14

8.15

Let aPκ(A)a\in P_{\kappa}(A) and C={xPκ(A):xa}C=\{x\in P_{\kappa}(A):x\supseteq a\}. If xPκ(A)x\in P_{\kappa}(A) and y=xay=x\cup a then yCy\in C (yay\supseteq a and |y||x|+|a|<κ|y|\leq|x|+|a|<\kappa) and yxy\supseteq x. So CC is unbounded. If (xξ)ξ<λ{(x_{\xi})}_{{\xi<\lambda}} is an increasing sequence of elements of CC of length λ<κ\lambda<\kappa then ξ<λxξC{\bigcup_{{\xi<\lambda}}x_{\xi}}\in C (|ξ<λxξ|=λsupξ<λ|xξ|<κ|{\bigcup_{{\xi<\lambda}}x_{\xi}}|=\lambda{\sup_{{\xi<\lambda}}|x_{\xi}|}<\kappa and ξ<λxξx0a{\bigcup_{{\xi<\lambda}}x_{\xi}}\supseteq x_{0}\supseteq a) so CC is closed.

8.16

Let FF be a normal κ\kappa-complete filter on Pκ(A)P_{\kappa}(A). Let XPκ(A)X\subseteq P_{\kappa}(A) be FF-positive. For all xXx\in X, we have g(x)[x]<ωxPκ(A)g(x)\in{[x]}^{{<\omega}}\subseteq x\in P_{\kappa}(A). For all ag(X)a\in g(X), define Xa={xX,g(x)=a}X_{a}=\{x\in X,g(x)=a\} and assume that the XaX_{a} are not FF-positive.

For all ag(X)a\in g(X), define Ya=Pκ(A)XaFY_{a}=P_{\kappa}(A)\setminus X_{a}\in F (by assumption) and for all aAg(X)a\in A\setminus g(X), define Ya=Pκ(A)FY_{a}=P_{\kappa}(A)\in F. We have xaAYaaA,xYaag(X),xYaag(X),(xXg(x)a)xX(ag(X),g(x)a)xXx\in\triangle_{{a\in A}}Y_{a}\Leftrightarrow\forall a\in A,x\in Y_{a}% \Leftrightarrow\forall a\in g(X),x\in Y_{a}\Leftrightarrow\forall a\in g(X),(x% \notin X\vee g(x)\neq a)\Leftrightarrow x\notin X\vee{(\forall a\in g(X),g(x)% \neq a)}\Leftrightarrow x\notin X. Hence Pκ(A)X=aAYaFP_{\kappa}(A)\setminus X=\triangle_{{a\in A}}Y_{a}\in F, contradicting the positivity of XX.

8.17

Let FF be a normal κ\kappa-complete filter on Pκ(A)P_{\kappa}(A). We shall prove that FF contains all closed unbounded sets. By Lemma 8.26, it is enough to prove that CfFC_{f}\in F for all f:[A]<ωPκ(A)f:{[A]}^{{<\omega}}\rightarrow P_{\kappa}(A). Suppose the contratry and let ff such that CfFC_{f}\notin F.

The complement of CfC_{f} is X={xPκ(A):e[x]<ω,f(e)x}X=\{x\in P_{\kappa}(A):\exists e\in{[x]}^{{<\omega}},f(e)\notin x\} and is FF-positive by assumption. Define a function gg on it by picking for each xXx\in X an element g(x)=e[x]<ωg(x)=e\in{[x]}^{{<\omega}} such that f(e)xf(e)\notin x. By exercise 8.16, there exists aAa\in A and a FF-positive set YXY\subseteq X such that xY,f(a)x\forall x\in Y,f(a)\notin x. Let Z={xPκ(A):f(e)x}Z=\{x\in P_{\kappa}(A):f(e)\in x\}. ZFZ\in F by assumption and ZPκ(A)YZ\subseteq P_{\kappa}(A)\setminus Y so Pκ(A)YFP_{\kappa}(A)\setminus Y\in F. This contradicts the positivity of YY.

8.18

Let C={xPκ(A):|x|1}C=\{x\in P_{\kappa}(A):|x|\geq\aleph_{1}\}. Suppose that |A|κ>ω1|A|\leq\kappa>\omega_{1} and let yAy\subseteq A such that |y|=1|y|=\aleph_{1}. If xPκ(A)x\in P_{\kappa}(A) and z=xyz=x\cup y then |z||z|, zxz\supseteq x and 1|y||z||x|+|y|<κ\aleph_{1}\leq|y|\leq|z|\leq|x|+|y|<\kappa so CC is unbounded. If (xξ)ξ<λ{(x_{\xi})}_{{\xi<\lambda}} is an increasing sequence of elements of CC such that λ<κ\lambda<\kappa then 0|x0||ξ<λxξ|=λsupξ<λ|xξ|<κ\aleph_{0}\leq|x_{0}|\leq|{\bigcup_{{\xi<\lambda}}x_{\xi}}|=\lambda{\sup_{{\xi% <\lambda}}|x_{\xi}|}<\kappa so CC is closed.

Let F:[A]<ωAF:{[A]}^{{<\omega}}\rightarrow A and CF={x[A]ω:e[x]<ω,F(e)x}C_{F}=\{x\in{[A]}^{\omega}:\forall e\in{[x]}^{{<\omega}},F(e)\in x\}. If κ>ω1\kappa>\omega_{1} then CF={x[A]ω:e[x]<ω,F(e)x}C_{F}=\{x\in{[A]}^{\omega}:\forall e\in{[x]}^{{<\omega}},F(e)\in x\} is not closed unbounded because it does not intersect the closed unbounded set C={xPκ(A):|x|1}C=\{x\in P_{\kappa}(A):|x|\geq\aleph_{1}\}.

However, if κ=ω1\kappa=\omega_{1} then CFC_{F} is closed unbounded. If xPκ(A)x\in P_{\kappa}(A), then xx is at most countable and we can find y0xy_{0}\supseteq x such that |y0|=0|y_{0}|=\aleph_{0}. Define by induction yn+1=yn{F(e):e[yn]<ω}y_{{n+1}}=y_{n}\cup\{F(e):e\in{[y_{n}]}^{{<\omega}}\}. Let y=n<ωyny=\bigcup_{{n<\omega}}y_{n}. |y0|0|y_{0}|\leq\aleph_{0} and if |yn|0|y_{n}|\leq\aleph_{0} then |yn+1||yn|+|yn|<ω|yn|0|y_{{n+1}}|\leq|y_{n}|+{|y_{n}|}^{{<\omega}}|y_{n}|\leq\aleph_{0} and finally 0=|y0||y|=0supn<ω|yn|0\aleph_{0}=|y_{0}|\leq|y|=\aleph_{0}\sup_{{n<\omega}}|y_{n}|\leq\aleph_{0} that is y[A]ωy\in{[A]}^{{\omega}}. Moreover if e[y]<ωe\in{[y]}^{{<\omega}} then there exists n<ωn<\omega such that eyne\subseteq y_{n} and so F(e)yn+1yF(e)\in y_{{n+1}}\subseteq y. So yCFy\in C_{F} and yy0=xy\supseteq y_{0}=x. Now if (xξ)ξ<λCFλ{(x_{\xi})}_{{\xi<\lambda}}\in C_{F}^{{\lambda}} is an increasing sequence of length λ0\lambda\leq\aleph_{0} then 0=|x0||ξ<λxξ|=λ0=0\aleph_{0}=|x_{0}|\leq|\bigcup_{{\xi<\lambda}}x_{\xi}|=\lambda\aleph_{0}=% \aleph_{0}. Moreover if eξ<λxξe\subseteq\bigcup_{{\xi<\lambda}}x_{\xi} is finite then there exists ξ<λ\xi<\lambda such that exξe\subseteq x_{\xi} and so F(e)xξξ<λxξF(e)\in x_{\xi}\subseteq\bigcup_{{\xi<\lambda}}x_{\xi}. Hence ξ<λxξCF\bigcup_{{\xi<\lambda}}x_{\xi}\in C_{F}.

8.19

Let C={xPκ(λ):xκκ}C=\{x\in P_{\kappa}(\lambda):x\cap\kappa\in\kappa\}. If λ>κ\lambda>\kappa then CC is not unbounded: if α=κ+1={κ}κPκ(λ)\alpha=\kappa+1=\{\kappa\}\cup\kappa\in P_{{\kappa}}(\lambda) and if yαy\supseteq\alpha then κyκα=κ\kappa\cap y\supseteq\kappa\cap\alpha=\kappa so κyC\kappa\cap y\notin C.

Suppose now that λ=κ\lambda=\kappa. Let xPκ(κ)x\in P_{{\kappa}}(\kappa) and define y=supx+1<κy={\sup x}+1<\kappa. If zxz\in x then zsupx<yz\leq{\sup x}<y so xyx\supseteq y. Moreover, yκ=yκy\cap\kappa=y\in\kappa. So yCy\in C. Moreover if (xξ)ξ<λCλ{(x_{\xi})}_{{\xi<\lambda}}\in C^{{\lambda}} is an increasing sequence of length λ<κ\lambda<\kappa then (ξ<λxξ)κ=ξ<λ(xξκ)κ\left(\bigcup_{{\xi<\lambda}}x_{\xi}\right)\cap\kappa=\bigcup_{{\xi<\lambda}}{% (x_{\xi}\cap\kappa)}\in\kappa because each (xξκ)κ(x_{\xi}\cap\kappa)\in\kappa. So ξ<λxξC\bigcup_{{\xi<\lambda}}x_{\xi}\in C. Hence CC is closed unbounded.

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