A Kurepa tree from diamond-plus

Recall that $T$ is said to be a $\kappa$-Kurepa tree if $T$ is a tree of height $\kappa$, whose levels $T_\alpha$ has size $\le|\alpha|$ for co-boundedly many $\alpha<\kappa$, and such that the set of branches of $T$ has size $>\kappa$.

In this post, we shall remind ourselves of the proof of Jensen’s theorem that $\diamondsuit^+(\kappa)$ entails a $\kappa$-Kurepa tree.

Definition. $\diamondsuit^+(\kappa)$ asserts the existence of a sequence $\langle \mathcal A_\alpha\mid \alpha<\kappa\rangle$ such that $|\mathcal A_\alpha|\le|\alpha|$ for all $\alpha<\kappa$ and such that for every $A\subseteq\kappa$, there exists a club $C\subseteq\kappa$ such that $A\cap\alpha,C\cap\alpha\in\mathcal A_\alpha$ for all $\alpha\in C$.

Proposition (Jensen). $\diamondsuit^+(\kappa)$ entails a family $\mathcal F\subseteq[\kappa]^\kappa$ such that:

  • $\mathcal F$ is dense in $[\kappa]^\kappa$. That is, for every cofinal $A\subseteq\kappa$, there exists $B\in\mathcal F\cap\mathcal P(A)$ such that $B$ is cofinal in $\kappa$;
  • for every stationary $S\subseteq\kappa$, there exists $T\in\mathcal F\cap\mathcal P(S)$ such that $S\setminus T$ is nonstationary;
  • $|\{ X\cap\alpha\mid X\in\mathcal F\}|\le|\alpha|$ for every $\alpha<\kappa$.

Proof. The key notion of this proof is the following:$$A[C]:=\{\delta\in A\mid \delta\in C\text{ or }(\sup(C\cap\delta),\delta)\cap A=\emptyset\}.$$ Let $\langle \mathcal A_\alpha\mid \alpha<\kappa\rangle$ witness $\diamondsuit^+(\kappa)$. We shall let $\mathcal F$ be the collection of all sets of the form $A[C]$ where $A$ is cofinal in $\kappa$, $C$ is a club in $\kappa$, and $A\cap\alpha,C\cap\alpha\in\mathcal A_\alpha$ for all $\alpha\in C$.

Clearly if $S$ is stationary, and $C$ is a club such that $S\cap\alpha,C\cap\alpha\in\mathcal A_\alpha$ for all $\alpha\in C$, then $S[C]$ is a subset of $S$ that lies in $\mathcal F$, and $S\setminus S[C]$ is nonstationary.
Also, if $A$ is a cofinal subset of $\kappa$, and $C$ is a club such that $A\cap\alpha,C\cap\alpha\in\mathcal A_\alpha$ for all $\alpha\in C$, then $A[C]$ is a subset of $A$ that lies in $\mathcal F$. Why is $A[C]$ cofinal in $\kappa$? for given an arbitrary $\beta<\kappa$, we may find $\gamma\in C$ above $\beta$, and then let $\delta:=\min(A\setminus\gamma)$. If $\delta=\gamma$, then $\delta\in A\cap C$, and hence $\delta\in A[C]$. If $\delta>\gamma$, then $(\sup(C\cap\delta),\delta)\subseteq (\gamma,\delta)$ and indeed $(\gamma,\delta)\cap A=\emptyset$, so, $\delta\in A[C]$. In either case, we found $\delta\in A[C]$ above $\beta$.

Fix $\alpha<\kappa$. Denote $\mathcal F_\alpha:=\{ B[D]\cup x\mid \exists\beta\le\alpha(B,D\in\mathcal A_\beta), x\in[\alpha]^{<\omega}\}$. As $|\mathcal F_\alpha|\le|\alpha|$, it suffices to show that $\mathcal F_\alpha\supseteq\{ X\cap\alpha\mid X\in\mathcal F\}$.
Given $X\in\mathcal F$, let $A$ be a cofinal subset of $\kappa$ and $C$ be a club in $\kappa$, such that $A\cap\alpha,C\cap\alpha\in\mathcal A_\alpha$ for all $\alpha\in C$, and $X=A[C]$. Put $\beta:=\sup(C\cap(\alpha+1))$. Denote $B:=A\cap\beta$ and $D:=C\cap\beta$. Clearly, $B,D\in\mathcal A_\beta$, $\beta\le\alpha$, and $X\cap\beta=A[C]\cap\beta=B[D]$. Thus, to prove that $X\cap\alpha$ is in $\mathcal F_\alpha$, it suffices to prove that $x:=(A[C]\cap\alpha)\setminus\beta$ is finite. We consider two cases.

$\blacktriangleright$ if $\beta=\alpha$, then $x=\emptyset$, and we are done.

$\blacktriangleright$ if $\beta<\alpha$, then for every $\delta\in x$, we have $\sup(C\cap\delta)=\beta<\delta$. As $\sup(C\cap\delta,\delta)\cap A=\emptyset$ for all $\delta\in x$, we conclude that $x$ cannot contain more than one element. $\square$

Corollary (Jensen). $\diamondsuit^+(\kappa)$ entails a $\kappa$-Kurepa tree.
Proof. Let $\mathcal F$ be given by the previous proposition. For every $A\subseteq\kappa$, let $\chi_A:\kappa\rightarrow2$ denote the characteristic function of $A$. Put $T:=\{ \chi_A\restriction\alpha\mid \alpha<\kappa, A\in\mathcal F\}$. Then $\mathcal T:=(T,\subseteq)$ is a $\kappa$-tree whose levels $T_\alpha$ has size $\le\alpha$ for all $\alpha<\kappa$.
Why does $\mathcal T$ have more than $\kappa$ many branches? Since $\diamondsuit^+(\kappa)$ holds, we have $\kappa^{<\kappa}=\kappa$, thus let us fix a bijection $\psi:\kappa^{<\kappa}\leftrightarrow\kappa$. Next, given a function $f:\kappa\rightarrow\kappa$, we define $A_f:=\{ \psi(f\restriction\eta)\mid \eta<\kappa\}$. Clearly, $A_f$ has size $\kappa$, and if $f$ and $g$ are distinct elements of ${}^\kappa\kappa$, then $A_f\cap A_g$ has size $<\kappa$. Recalling that $\mathcal F$ is dense in $[\kappa]^\kappa$, we conclude that $|\mathcal F|\ge|\{A_f\mid f\in{}^\kappa\kappa\}|=2^\kappa$. So $\mathcal T$ has $2^\kappa$-many branches. $\square$



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