Set Theory: Tree Property

Question

Why does the tree property hold for regular cardinals but not singular cardinals? (I.e. There exists a tree of height $\kappa$ with countable levels and no cofinal branch for $\kappa$ a singular cardinal)

Answer 1

There are a few issues here.

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First of all, the tree property doesn't hold of regular cardinals in general. For example, it provably fails for $\kappa=\omega_1$ (see https://en.wikipedia.org/wiki/Aronszajn_tree), and getting the tree property at $\omega_2$ has high consistency strength relative to ZFC - it is equiconsistent with the existence of a weakly compact cardinal. In fact, in $ZFC+V=L$, every successor cardinal fails to have the tree property - thus, in $ZFC+V=L+$"there are no inaccessibles," the tree property never holds of any uncountable cardinal at all.

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Second, you've stated the tree property wrong: the point is not that the levels are countable, but rather that they are small (of size $<\kappa$). (Also, we usually demand that the tree itself have cardinality $\kappa$, if I remember correctly, but this actually doesn't matter: if I have a tree of height $\kappa$ with levels of size $<\kappa$ and no $\kappa$-length branches, I can prune it down to such a tree with size $\kappa$ - just fix a set of $\kappa$-many nodes, the supremum of whose heights is $\kappa$, and look at the induced subtree.)

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Finally, here's why the tree property can never hold of $\kappa$ singular. Let $\lambda=cf(\kappa)<\kappa$, let $\{\alpha_\eta: \eta\in\lambda\}$ be a cofinal subset of $\kappa$, and consider the following tree on $\kappa$:

$$T=\{f\in \kappa^{<\kappa}: f(0)<\lambda\mbox{ and }\vert f\vert<\alpha_{f(0)}\mbox{ and } f(\eta)=0 \mbox{ for all }0<\eta<\vert f\vert\}\cup\{\emptyset\}.$$ Basically, $T$ consists of $\lambda$-many branches, with the $\eta$th such branch of height $\alpha_\eta$. Then the height of $T$ is clearly $\sup\{\alpha_\eta: \eta<\lambda\}=\kappa$, but each level is of cardinality at most $\lambda$.