# $\kappa^+$ tree with no $\kappa^+$ branch

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.