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 ${\mathrm{♢}}^{+}(\kappa )$ entails a $\kappa$-Kurepa tree.

Definition. ${\mathrm{♢}}^{+}(\kappa )$ asserts the existence of a sequence $⟨{\mathcal{A}}_{\alpha }\mid \alpha <\kappa ⟩$ 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). ${\mathrm{♢}}^{+}(\kappa )$ 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:$$A[C]:=\{\delta \in A\mid \delta \in C\text{ or }(sup(C\cap \delta ),\delta )\cap A=\mathrm{\varnothing }\}.$$ Let $⟨{\mathcal{A}}_{\alpha }\mid \alpha <\kappa ⟩$ witness ${\mathrm{♢}}^{+}(\kappa )$. 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 $(\gamma ,\delta )\cap A=\mathrm{\varnothing }$, so, $\delta \in A[C]$. In either case, we found $\delta \in A[C]$ above $\beta$.

Fix $\alpha <\kappa$. Denote ${\mathcal{F}}_{\alpha }:=\{B[D]\cup x\mid \mathrm{\exists }\beta \le \alpha (B,D\in {\mathcal{A}}_{\beta }),x\in [\alpha {]}^{<\omega }\}$. 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.

$▸$ if $\beta =\alpha$, then $x=\mathrm{\varnothing }$, and we are done.

$▸$ if $\beta <\alpha$, then for every $\delta \in x$, we have $sup(C\cap \delta )=\beta <\delta$. As $sup(C\cap \delta ,\delta )\cap A=\mathrm{\varnothing }$ for all $\delta \in x$, we conclude that $x$ cannot contain more than one element. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Corollary (Jensen). ${\mathrm{♢}}^{+}(\kappa )$ entails a $\kappa$-Kurepa tree.
Proof. Let $\mathcal{F}$ be given by the previous proposition. For every $A\subseteq \kappa$, let ${\chi }_A:\kappa \to 2$ denote the characteristic function of $A$. Put $T:=\{{\chi }_A\upharpoonright \alpha \mid \alpha <\kappa ,A\in \mathcal{F}\}$. 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 ${\mathrm{♢}}^{+}(\kappa )$ holds, we have ${\kappa }^{<\kappa }=\kappa$, thus let us fix a bijection $\psi :{\kappa }^{<\kappa }\leftrightarrow \kappa$. Next, given a function $f:\kappa \to \kappa$, we define $A_f:=\{\psi (f\upharpoonright \eta )\mid \eta <\kappa \}$. 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. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$