Let $\kappa$ and $\lambda$ be cardinals. A thin $(\kappa,\lambda)$-list is a function $L:[\lambda]^{<\kappa}\longrightarrow [\lambda]^{<\kappa}$ such that for all $x\in[\lambda]^{<\kappa}$, $L(x)\subseteq x$ and $\{L(y)\;|\;y\subseteq x\}$ has cardinality $<\kappa$.

Say that $(\kappa,\lambda)$-STP holds iff whenever $L$ is a thin $(\kappa,\lambda)$-list, there is some $b\subseteq\lambda$ such that $\{x\;|\;x\cap b=L(x)\}$ is stationary and that $\kappa$ satisfies the super tree property iff $(\kappa,\lambda)$-STP holds for all $\lambda\geq\kappa$.

Say that $(\kappa,\lambda)$-SSTP holds iff for every sequence $(L_{\alpha})_{\alpha\in\mu}$ (where $\mu<\kappa$), if every $L_{\alpha}$ is a thin $(\kappa,\lambda)$-list, there is a sequence $(b_{\alpha})_{\alpha\in\mu}$ such that $\{x\;|\;\forall\alpha\in\mu(b_{\alpha}\cap x=L_{\alpha}(x))\}$ is stationary. Say that $\kappa$ satisfies the simultaneous super tree property iff $(\kappa,\lambda)$-SSTP holds for all $\lambda\geq\kappa$.

Let $\kappa$ be supercompact, $\lambda\geq\kappa$ and $U$ a normal $\kappa$-complete ultrafilter on $[\lambda]^{<\kappa}$. For a thin $(\kappa,\lambda)$-List $L$ one can show that there is some $b$ such that $\{x\;|\;x\cap b=L(x)\}\in U$, therefore, using $\kappa$-completeness, we have that supercompact cardinals satisfy the simultaneous super tree property, so the simultaneous super tree property is consistent, at least modulo the existence of a supercompact cardinal (and I suspect $\omega_2$ has the simultaneous super tree property if PFA holds). This begs the following questions:

  1. Are the principles $(\kappa,\lambda)$-STP and $(\kappa,\lambda)$-SSTP equivalent for all $\lambda$ (or some fixed $\lambda$ depending on $\kappa$)?
  2. Is the super tree property equivalent to the simultaneous super tree property?

1 Answer 1


The $(\kappa,\lambda)$-STP and $(\kappa,\lambda)$-SSTP are equivalent for any uncountable cardinals $\kappa\leq\lambda$: Let $\mu<\kappa$ and $(L_\gamma)_{\gamma<\mu}$ a sequence of thin $(\kappa,\lambda)$-lists. We can then "amalgamate" these lists into one list $L$ as follows: Let $h:\lambda\times\mu\rightarrow\lambda$ be a bijection. Let $$C=\{x\in[\lambda]^{<\kappa}\mid \mu\subseteq x\wedge\ x \text{ is closed under both } h \text{ and }h^{-1}\}$$ Then $C$ is club in $[\lambda]^{<\kappa}$ so for all our intents and purposes it is enough to define $L$ on $C$. $L$ puts the information of $L_\gamma(x)$ onto the "$\gamma$th slice" of $x$, i.e. for all $\beta, \gamma$: $$h(\beta,\gamma)\in L(x)\Leftrightarrow \beta\in L_\gamma(x)$$ If $x\in C$ then indeed $L(x)\subseteq x$ and every $L_\gamma(x)$, $\gamma<\kappa$ is coded into $L(x)$ in a uniform way. Using $\mu<\kappa$ it should be easy to see that $L$ is a thin $(\kappa,\lambda)$-list (we need here that $\kappa$ is inaccessible but this follows from $(\kappa, \lambda)$-STP, see the remark at the end).

If $(\kappa,\lambda)$-STP holds true, there is $b\subseteq \lambda$ and $S\subseteq C$ stationary with $L(x)=x\cap b$ for all $x\in S$. Now for $\gamma<\mu$ define $b_\gamma$ as the "$\gamma$th slice" of $b$, i.e.: $$b_\gamma=\{\beta<\lambda\mid h(\beta,\gamma)\in b\}$$ We get that for $x\in S$ and $\beta\in x$, $\gamma<\mu$: $$\beta\in x\cap b_\gamma\Leftrightarrow h(\beta,\gamma)\in x\cap b\Leftrightarrow h(\beta, \gamma)\in L(x)\Leftrightarrow \beta\in L_\gamma(x)$$ So $L_\gamma(x)=x\cap b_\gamma$.

It is also worth noting that $(\kappa, \kappa)$-STP (and $(\kappa, \kappa)$-SSTP for that matter) are both equivalent to $\kappa$ being ineffable.


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