Set theory - Chapter 8: Stationary Sets

Let $f$ be a normal function and $C=\{\alpha<\kappa:f(\alpha)=\alpha\}$. If $C$ is not unbounded then there is $\gamma<\kappa$ such that for all $\alpha\geq\gamma$, $f(\alpha)\neq\alpha$ and because $f$ is increasing $f(\alpha)>\alpha$. Define the increasing sequence $u_{0}=\gamma$ and $\forall n<\omega,u_{{n+1}}=f(u_{n})$ . Let $beta=\lim_{{n\rightarrow\rightarrow\omega}}u_{n}\geq\gamma$. $\kappa$ is a regular uncountable cardinal so $\beta<\kappa$. Because $f$ is continuous, $\beta<f(\beta)=\lim_{{n\rightarrow\omega}}f(u_{n})=\lim_{{n\rightarrow\omega}}f(u_{n})=\lim_{{n\rightarrow\omega}}u_{{n+1}}=\lim_{{n\rightarrow\omega}}u_{{n}}=\beta$. A contradiction. Hence $C$ is unbounded.

Now if ${(\alpha_{\xi})}_{{\xi<\gamma}}$ is an increasing sequence of elements of $C$ (i.e. $f(\alpha_{\xi})=\alpha_{\xi}$) of length $\gamma<\kappa$, then because $f$ is continuous, we have $f(\lim_{{\xi\rightarrow\gamma}}\alpha_{\xi})=\lim_{{\xi\rightarrow\gamma}}f(\alpha_{\xi})=\lim_{{\xi\rightarrow\gamma}}\alpha_{\xi}$ and so $\lim_{{\xi\rightarrow\gamma}}\alpha_{\xi}\in C$. So $C$ is closed.

Let $f:\kappa\rightarrow\kappa$ and $C=\{\alpha<\kappa:\forall\xi<\alpha,f(\xi)<\alpha\}$. Suppose $C$ is bounded and let $\gamma$ such that for all $\alpha\geq\gamma$ $\alpha\notin C$. We define the closed unbounded (thus stationary) set $S=\{\alpha,\alpha\geq\gamma\}$. For all $\alpha\in S$ we let $g(\alpha)\in\kappa$ be the least $\xi<\alpha$ such that $f(\xi)\geq\alpha$. Then $g$ is regressive on $S$ and by Fodor’s theorem there is a stationary set $T\subseteq S$ and an ordinal $\beta\in\kappa$ such that $\forall\alpha\in T,f(\beta)\geq\alpha$. In particular $T$ does not intersect the closed unbounded set $\{\alpha<\kappa:\alpha>f(\beta)\}$. A contradiction. So $C$ is unbounded.

Now let ${(\alpha_{\xi})}_{{\xi<\gamma}}$ be an increasing sequence of elements of $C$ of length $\gamma<\kappa$ and $\alpha=\lim_{{\xi\rightarrow\gamma}}\alpha_{\xi}$. For all $\beta<\alpha$, there exists some $\xi<\gamma$ such that $\beta<\alpha_{\xi}\in C$. Then $f(\beta)<\alpha_{\xi}<\alpha$ and so $\alpha\in C$. Hence $C$ is closed.

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

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

Let $\gamma<\kappa$ and define $Y_{\gamma}=\{\alpha\in X,f(\alpha)\leq\gamma\}$. Let $\beta\in C$ such that $\beta>\gamma$. If $\alpha\in X$ and $\alpha>\beta$ then $\beta\in C\cap\alpha$ and so $f(\alpha)=\sup{(C\cap\alpha)}\geq\beta>\gamma$ i.e. $\alpha\notin Y_{\gamma}$. Thus $\alpha\in Y_{\gamma}\implies\alpha\leq\beta$ and finally $Y_{\gamma}$ is bounded.

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

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

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

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

Let $I$ be a $\kappa$-complete ideal on $\kappa$ and $F=\{\kappa\setminus X:X\in I\}$ the dual filter. We have $X\in F\Leftrightarrow\exists Y\in F,Y\subseteq X\Leftrightarrow\exists Y\in F,(\kappa\setminus X)\cap Y=\emptyset$ and thus $X\notin I\Leftrightarrow\kappa\setminus X\notin F\Leftrightarrow\forall Y\in F,X\cap Y\neq\emptyset$.

Let $S_{0}\notin I$ and $f$ be a regressive function on $S_{0}$. Suppose that for each $\gamma<\kappa$, the set $\{\alpha\in S_{0}:f(\alpha)=\gamma\}\in I$ and define $X_{\gamma}=\{\alpha\in S_{0}:f(\alpha)\neq\gamma\}\in F$. Let $X=\triangle_{{\gamma<\kappa}}X_{\gamma}\in F$. Then $S_{0}\cap X\neq\emptyset$ and we can pick $\alpha\in S_{0}\cap X$. On the one hand, $f(\alpha)<\alpha$ and on the other hand $\alpha\in\bigcap_{{\gamma<\alpha}}X_{\gamma}$ that is $\forall\gamma<\alpha,f(\alpha)\neq\gamma$. A contradiction. So for some $\gamma<\kappa$, $S=\{\alpha\in S_{0}:f(\alpha)=\gamma\}\notin I$ and we have $S\subseteq S_{0}$ and $f$ constant on $S$.

Conversely, consider ${(X_{\alpha})}_{{\alpha<\kappa}}\in F^{\kappa}$ and suppose that $\triangle X_{\alpha}\notin F$. Let $S_{0}=\kappa\setminus\triangle_{{\xi<\kappa}}X_{\alpha}\notin I$. If $\alpha\in S_{0}$ then $\alpha\notin\bigcap_{{\xi<\alpha}}X_{\xi}$ and there is $\xi<\kappa$ such that $\alpha\notin X_{{\xi}}$. Define $f(\alpha)$ to be the least $\xi$ such $\alpha\notin X_{{\xi}}$. Then $f$ takes a constant value $\gamma$ on some $S\subseteq S_{0}$. For all $\alpha\in S$, $\alpha\notin X_{{f(\alpha)}}=X_{{\gamma}}$ and so $S\cap X_{\gamma}=\emptyset$. This contradicts $S\notin I$ and $X_{{\gamma}}\in F$. So $I$ is normal.

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

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

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

Note: if $C$ is closed unbounded in $\beta$ and $\alpha<\beta$ is a limit point of $C$ then $C\cap\alpha$ is closed unbounded in $\alpha$.

1. (1)

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

2. (2)

   Let $T,S$ be stationary sets. By the previous point, $\operatorname{Tr}(T\cup S)\supseteq\operatorname{Tr}(T)\cup\operatorname{Tr}(S)$. Let $\alpha\in\operatorname{Tr}(T\cup S)$. Suppose for example that $\alpha\notin\operatorname{Tr}(S)$. Let $C_{1}$ be a closed unbounded set in $\alpha$ such that $C_{1}\cap S\cap\alpha=\emptyset$. For any closed unbounded set $C_{2}$ in $\alpha$, $C_{2}\cap T\cap\alpha\supseteq C_{2}\cap C_{1}\cap T\cap\alpha=C_{2}\cap{(C_{1}\cap{(T\cup S)}\cap\alpha)}={(C_{2}\cap C_{1})}\cap{(T\cup S)}\cap\alpha\neq\emptyset$. Hence $\alpha\in\operatorname{Tr}(T)$.

3. (3)

   Let $\alpha\in\operatorname{Tr}(\operatorname{Tr}(S))$ and $C$ a closed unbounded set in $\alpha$ and $C^{{\prime}}$ the closed unbounded set of limit points of $C$. We have $C^{{\prime}}\cap\operatorname{Tr}(S)\cap\alpha\neq\emptyset$. Let $\beta\in C^{{\prime}}\cap\operatorname{Tr}(S)\cap\alpha$. We have $C\cap\beta$ closed unbounded in $\beta$. Because $\beta\in\operatorname{Tr}(S)$, $(C\cap\beta)\cap S\cap\beta\neq\emptyset$. Let $\gamma\in C\cap S\cap\beta$. We have $\gamma<\beta<\alpha$. Hence $\gamma\in C\cap S\cap\alpha$ and so $C\cap S\cap\alpha\neq\emptyset$. Hence $\alpha\in\operatorname{Tr}(S)$.

4. (4)

   Suppose $S\equiv T\mod I_{{\rm{NS}}}$. Let $C_{1}$ a closed unbounded set such that $C_{1}\cap S=C_{1}\cap T$ and $C_{1}^{{\prime}}$ be the closed unbounded set of limit points of $C_{1}$. If $\alpha\in\operatorname{Tr}(S)\cap C_{1}^{{\prime}}$, then $C_{1}^{{\prime}}\cap\alpha$ is closed unbounded in $\alpha$ and for all $C_{2}$ closed unbounded in $\alpha$, we have $\emptyset\neq(C_{1}^{{\prime}}\cap\alpha\cap C_{2})\cap(S\cap\alpha)\subseteq C_{2}\cap(C_{1}^{{\prime}}\cap S)\cap\alpha=C_{2}\cap(C_{1}^{{\prime}}\cap T)\cap\alpha\subseteq C_{2}\cap T\cap\alpha$ and so $\alpha\in\operatorname{Tr}(T)$. By symmetry, $\operatorname{Tr}(S)\cap C_{1}^{{\prime}}=\operatorname{Tr}(T)\cap C_{1}^{{\prime}}$ that is $\operatorname{Tr}(S)\equiv\operatorname{Tr}(T)\mod I_{{\rm{NS}}}$

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

Let $\alpha<\kappa$ such that $\operatorname{cf}(\alpha)\geq\lambda^{+}$. We have $\operatorname{cf}(\alpha)>\omega$. If $C$ is closed unbounded in $\alpha$ then because $\lambda<\operatorname{cf}(\alpha)\leq\alpha$ we can construct an increasing sequence ${(\beta_{\xi})}_{{\xi<\lambda}}\in C^{\lambda}$. Let $\beta=\lim_{{\xi\rightarrow\lambda}}\beta_{\xi}$. We have $\beta\in C$ and $\operatorname{cf}(\beta)=\operatorname{cf}(\lambda)=\lambda$. So $\alpha\in\operatorname{Tr}(E_{{\lambda}}^{\kappa})$.

Conversely, suppose $a\in\operatorname{Tr}(E_{{\lambda}}^{\kappa})$. Let $f:\operatorname{cf}(\alpha)\rightarrow\alpha$ be a normal function. $f_{{-1}}(\alpha\cap E_{{\lambda}}^{\kappa})$ is stationary in $\operatorname{cf}(\alpha)$, in particular nonempty. Let $\beta\in f_{{-1}}(\alpha\cap E_{{\lambda}}^{\kappa})$ we have $\beta<\operatorname{cf}(\alpha)$ so $\operatorname{cf}(\beta)<\operatorname{cf}(\alpha)$. If ${(\beta_{\xi})}_{{\xi<\lambda}}\in{\operatorname{cf}(\alpha)}^{{\lambda}}$ is an increasing sequence such that $\beta=\lim_{{\xi\rightarrow\lambda}}\beta_{\xi}$, then $f(\beta)=\lim_{{\xi\rightarrow\lambda}}f(\beta_{\xi})$ and so $\lambda=\operatorname{cf}({f(\beta)})=\operatorname{cf}(\beta)<\operatorname{cf}(\alpha)$ i.e. $\operatorname{cf}(\alpha)\geq\lambda^{+}$.

Let $\mu<\lambda$ be regular infinite cardinal. By exercise 8.12, $E_{{\lambda}}^{\kappa}=\{\alpha<\kappa:\operatorname{cf}(\alpha)=\lambda\}\subseteq\{\alpha<\kappa:\operatorname{cf}(\alpha)\geq\mu^{+}\}=\operatorname{Tr}(E_{{\mu}}^{\kappa})$. Hence $E_{{\mu}}^{\kappa}<E_{{\lambda}}^{\kappa}$.

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

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

Suppose that $o(E_{{\lambda_{\xi}}}^{\kappa})=\xi$ for all $\xi<\alpha$. Then for all $\xi<\alpha$ we have $E_{{\lambda_{\xi}}}^{\kappa}<E_{{\lambda_{{\alpha}}}}^{\kappa}$ and so $o(E_{{\lambda_{{\alpha}}}}^{\kappa})\geq\sup_{{\xi<\alpha}}\{o(E_{{\lambda_{\xi}}}^{\kappa})+1\}=\sup_{{\xi<\alpha}}\{\xi+1\}=\alpha$.

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Let $a\in P_{\kappa}(A)$ and $C=\{x\in P_{\kappa}(A):x\supseteq a\}$. If $x\in P_{\kappa}(A)$ and $y=x\cup a$ then $y\in C$ ($y\supseteq a$ and $|y|\leq|x|+|a|<\kappa$) and $y\supseteq x$. So $C$ is unbounded. If ${(x_{\xi})}_{{\xi<\lambda}}$ is an increasing sequence of elements of $C$ of length $\lambda<\kappa$ then ${\bigcup_{{\xi<\lambda}}x_{\xi}}\in C$ ($|{\bigcup_{{\xi<\lambda}}x_{\xi}}|=\lambda{\sup_{{\xi<\lambda}}|x_{\xi}|}<\kappa$ and ${\bigcup_{{\xi<\lambda}}x_{\xi}}\supseteq x_{0}\supseteq a$) so $C$ is closed.

Let $F$ be a normal $\kappa$-complete filter on $P_{\kappa}(A)$. Let $X\subseteq P_{\kappa}(A)$ be $F$-positive. For all $x\in X$, we have $g(x)\in{[x]}^{{<\omega}}\subseteq x\in P_{\kappa}(A)$. For all $a\in g(X)$, define $X_{a}=\{x\in X,g(x)=a\}$ and assume that the $X_{a}$ are not $F$-positive.

For all $a\in g(X)$, define $Y_{a}=P_{\kappa}(A)\setminus X_{a}\in F$ (by assumption) and for all $a\in A\setminus g(X)$, define $Y_{a}=P_{\kappa}(A)\in F$. We have $x\in\triangle_{{a\in A}}Y_{a}\Leftrightarrow\forall a\in A,x\in Y_{a}\Leftrightarrow\forall a\in g(X),x\in Y_{a}\Leftrightarrow\forall a\in g(X),(x\notin X\vee g(x)\neq a)\Leftrightarrow x\notin X\vee{(\forall a\in g(X),g(x)\neq a)}\Leftrightarrow x\notin X$. Hence $P_{\kappa}(A)\setminus X=\triangle_{{a\in A}}Y_{a}\in F$, contradicting the positivity of $X$.

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

The complement of $C_{f}$ is $X=\{x\in P_{\kappa}(A):\exists e\in{[x]}^{{<\omega}},f(e)\notin x\}$ and is $F$-positive by assumption. Define a function $g$ on it by picking for each $x\in X$ an element $g(x)=e\in{[x]}^{{<\omega}}$ such that $f(e)\notin x$. By exercise 8.16, there exists $a\in A$ and a $F$-positive set $Y\subseteq X$ such that $\forall x\in Y,f(a)\notin x$. Let $Z=\{x\in P_{\kappa}(A):f(e)\in x\}$. $Z\in F$ by assumption and $Z\subseteq P_{\kappa}(A)\setminus Y$ so $P_{\kappa}(A)\setminus Y\in F$. This contradicts the positivity of $Y$.

Let $C=\{x\in P_{\kappa}(A):|x|\geq\aleph_{1}\}$. Suppose that $|A|\leq\kappa>\omega_{1}$ and let $y\subseteq A$ such that $|y|=\aleph_{1}$. If $x\in P_{\kappa}(A)$ and $z=x\cup y$ then $|z|$, $z\supseteq x$ and $\aleph_{1}\leq|y|\leq|z|\leq|x|+|y|<\kappa$ so $C$ is unbounded. If ${(x_{\xi})}_{{\xi<\lambda}}$ is an increasing sequence of elements of $C$ such that $\lambda<\kappa$ then $\aleph_{0}\leq|x_{0}|\leq|{\bigcup_{{\xi<\lambda}}x_{\xi}}|=\lambda{\sup_{{\xi<\lambda}}|x_{\xi}|}<\kappa$ so $C$ is closed.

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

However, if $\kappa=\omega_{1}$ then $C_{F}$ is closed unbounded. If $x\in P_{\kappa}(A)$, then $x$ is at most countable and we can find $y_{0}\supseteq x$ such that $|y_{0}|=\aleph_{0}$. Define by induction $y_{{n+1}}=y_{n}\cup\{F(e):e\in{[y_{n}]}^{{<\omega}}\}$. Let $y=\bigcup_{{n<\omega}}y_{n}$. $|y_{0}|\leq\aleph_{0}$ and if $|y_{n}|\leq\aleph_{0}$ then $|y_{{n+1}}|\leq|y_{n}|+{|y_{n}|}^{{<\omega}}|y_{n}|\leq\aleph_{0}$ and finally $\aleph_{0}=|y_{0}|\leq|y|=\aleph_{0}\sup_{{n<\omega}}|y_{n}|\leq\aleph_{0}$ that is $y\in{[A]}^{{\omega}}$. Moreover if $e\in{[y]}^{{<\omega}}$ then there exists $n<\omega$ such that $e\subseteq y_{n}$ and so $F(e)\in y_{{n+1}}\subseteq y$. So $y\in C_{F}$ and $y\supseteq y_{0}=x$. Now if ${(x_{\xi})}_{{\xi<\lambda}}\in C_{F}^{{\lambda}}$ is an increasing sequence of length $\lambda\leq\aleph_{0}$ then $\aleph_{0}=|x_{0}|\leq|\bigcup_{{\xi<\lambda}}x_{\xi}|=\lambda\aleph_{0}=\aleph_{0}$. Moreover if $e\subseteq\bigcup_{{\xi<\lambda}}x_{\xi}$ is finite then there exists $\xi<\lambda$ such that $e\subseteq x_{\xi}$ and so $F(e)\in x_{\xi}\subseteq\bigcup_{{\xi<\lambda}}x_{\xi}$. Hence $\bigcup_{{\xi<\lambda}}x_{\xi}\in C_{F}$.

Let $C=\{x\in P_{\kappa}(\lambda):x\cap\kappa\in\kappa\}$. If $\lambda>\kappa$ then $C$ is not unbounded: if $\alpha=\kappa+1=\{\kappa\}\cup\kappa\in P_{{\kappa}}(\lambda)$ and if $y\supseteq\alpha$ then $\kappa\cap y\supseteq\kappa\cap\alpha=\kappa$ so $\kappa\cap y\notin C$.

Suppose now that $\lambda=\kappa$. Let $x\in P_{{\kappa}}(\kappa)$ and define $y={\sup x}+1<\kappa$. If $z\in x$ then $z\leq{\sup x}<y$ so $x\supseteq y$. Moreover, $y\cap\kappa=y\in\kappa$. So $y\in C$. Moreover if ${(x_{\xi})}_{{\xi<\lambda}}\in C^{{\lambda}}$ is an increasing sequence of length $\lambda<\kappa$ then $\left(\bigcup_{{\xi<\lambda}}x_{\xi}\right)\cap\kappa=\bigcup_{{\xi<\lambda}}{(x_{\xi}\cap\kappa)}\in\kappa$ because each $(x_{\xi}\cap\kappa)\in\kappa$. So $\bigcup_{{\xi<\lambda}}x_{\xi}\in C$. Hence $C$ is closed unbounded.