# IF $\kappa$ is weakly compact, then $\kappa$ has the tree property.

Let $\kappa$ be weakly compact, and let $(T, <_T)$ be a tree of height $\kappa$ such that each level of $T$ has size $< \kappa$. Assume $T = \kappa$. We extend the partial ordering $<_T$ of $\kappa$ as follows - if $\alpha <_T \beta$ then say $\alpha \prec \beta$. If $\alpha, \beta$ are incomparable then let $\xi$ be the first level where the predecessors of $\alpha, \beta$ differ, and if $\alpha_{\xi} < \beta_{\xi}$ in the usual ordering of $\kappa$, say $\alpha \prec \beta$.

Define $F: [\kappa]^2 \to \{ 0,1\}$ by $F(\{ \alpha, \beta\}) = 1$ if and only if $<$ and $\prec$ agree on $\alpha, \beta$. Assume this is onto and by weak compactness there is a homogeneous set $H \subseteq \kappa$ of size $\kappa$.

Define a set $B$ to be the collection of all $x \in \kappa$ such that $\{ \alpha \in H \mid x <_T \alpha\}$ has size $\kappa$. My question is, why is there an element of $B$ on every level of $T$? It has something to do with the fact that each level of $T$ has size $< \kappa$, but I can't see why. Any help would be appreciated, thank you.

• Now, when you say weakly compact... there are about $\aleph_{\omega_4}$ many useful definitions for weakly compact. Can you specify which one you already know are equivalent (or at least which one you're using...)? – Asaf Karagila Mar 30 '12 at 19:17
• Apologies - I should have said that $\kappa$ is uncountable, and $\kappa \to (\kappa)^2$. – Paul Slevin Mar 30 '12 at 22:26
• I don't think it was needed for my question in the end. It is used to justify my assumption that $T = \kappa$. – Paul Slevin Mar 30 '12 at 22:37

Consider $\operatorname{Lev}_\eta(T)$ for some $\eta<\kappa$. $$\left|\bigcup_{\xi\le\eta}\operatorname{Lev}_\xi(T)\right|<\kappa\;,$$ so $H$ has $\kappa$ members above level $\eta$. For $x\in\operatorname{Lev}_\eta(T)$ let $H(x)=\{y\in H:x<_T y\}$. Then $$\left|\bigcup_{x\in\operatorname{Lev}_\eta(T)}H(x)\right|=\kappa\;,$$ and $|\operatorname{Lev}_\eta(T)|<\kappa$, so there must be some $x\in\operatorname{Lev}_\eta(T)$ with $|H(x)|=\kappa$. But then $x\in B$, as desired.
• Should that read $$\left| \bigcup_{\xi \le \eta} \operatorname{Lev}_{\xi} (T) \right| < \kappa$$ in the first displayed line? – Paul Slevin Mar 30 '12 at 18:27
• @Paul: Yes $-$ fixed. – Brian M. Scott Mar 30 '12 at 18:29