Show: If $\kappa$ is regular and $\kappa^{<\kappa}=\kappa$, then there is a $\kappa^+$-tree $T$ with no $\kappa^+$-branch.

Hint: let $T_\alpha$ (the $\alpha$th level of $T$) be one-to-one functions $f:\alpha\rightarrow\kappa$ with ran($f$) nonstationary.

I have seen the construction of the $\omega_1$-tree with no $\omega_1$-branch (Aronszajn tree). The proof I'm trying to imitate is from Ch 12 of Introduction to Set Theory by Hrbacek and Jech:

Construct the levels $T_\alpha, \alpha<\omega_1$ by recursion in such a way that

(i) $T_\alpha\subseteq\omega^\alpha$ and $|T_\alpha|\leq\aleph_0$

(ii) If $f\in T_\alpha$ then $f$ is one-to-one and $\omega-\text{ran}(f)$ is infinite

(iii) If $f\in T_\alpha$ and $\beta<\alpha$ then $f\upharpoonright\beta\in T_\beta$

(iv) For any $\beta<\alpha$, any $g\in T_\beta$, and any finite $X\subseteq \omega-\text{ran}(g)$, there is $f\in T_\alpha$ such that $g\subseteq f$ and ran$(f)\cap X=\varnothing$.

My guess is to use the hint and modify condition (iv) by replacing $X$ finite with $X$ club (and also condition (ii) with $\text{ran}(f)$ nonstationary?), since each nonstationary ran($f$) has that $\text{ran}(f)\cap C=\varnothing$ for some $C$ club set. But I'm not sure how to work out the details.


The proof for $\kappa=\omega_1$ can be generalized to arbitrary $\kappa$ with $\kappa^{<\kappa}=\kappa$ if we

  • Modify (ii) by requiring $\omega- \operatorname{ran} f$ contains a club and
  • Modify (iv) by replacing $X$ finite with $X$ nonstationary.

(Of course, we need more trivial modifications; e.g. replacing $\omega$ with $\kappa$.)

Note that modified (ii) is equivalent to every $f\in T_\alpha$ is one-to-one and has nonstationary image since the complement of a club is nonstationary.

The proof is similar to that of Chapter 12, Theorem 3.5 of Jech & Hrbacek. We must notice that the union of $\kappa$ nonstationary set is also nonstationary.


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