weakly compact cardinals and the tree property

A cardinal is weakly compact if and only if it is inaccessible and has the tree property.

Weak compactness implies tree property

Let $\kappa$ be a weakly compact cardinal and let be a $\kappa$ tree with all levels smaller than $\kappa$. We define a theory in $L_{\kappa ,\kappa }$ with for each $x\in T$, a constant $c_x$, and a single unary relation $B$. Then our theory $\mathrm{\Delta }$ consists of the sentences:

• $\mathrm{\neg }\left[B(c_x)\wedge B(c_y)\right]$ for every incompatible $x,y\in T$

• ${\bigvee }_{x\in T(\alpha )}B(c_x)$ for each

It should be clear that $B$ represents membership in a cofinal branch, since the first class of sentences asserts that no incompatible elements are both in $B$ while the second class states that the branch intersects every level.

Clearly $|\mathrm{\Delta }|=\kappa$, since there are $\kappa$ elements in $T$, and hence fewer than $\kappa \cdot \kappa =\kappa$ sentences in the first group, and of course there are $\kappa$ levels and therefore $\kappa$ sentences in the second group.

Now consider any $\mathrm{\Sigma }\subseteq \mathrm{\Delta }$ with . Fewer than $\kappa$ sentences of the second group are included, so the set of $x$ for which the corresponding $c_x$ must all appear in $T(\alpha )$ for some . But since $T$ has branches of arbitrary height, $T(\alpha )\vDash \mathrm{\Sigma }$.

Since $\kappa$ is weakly compact, it follows that $\mathrm{\Delta }$ also has a model, and that model obviously has a set of $c_x$ such that $B(c_x)$ whose corresponding elements of $T$ intersect every level and are compatible, therefore forming a cofinal branch of $T$, proving that $T$ is not Aronszajn.