# Thick Canadian trees

A Canadian tree (also called a weak Kurepa tree) is a tree of height $$\omega_1$$ with levels of cardinality $$\leq \omega_1$$ and at least $$\omega_2$$ many uncountable branches. Let's call a Canadian tree thick if it has $$2^{\omega_1}$$ many uncountable branches. Thick Canadian trees of course exist under CH (just take the complete binary tree of height $$\omega_1$$).

QUESTION: Is the existence of a thick Canadian tree consistent with $$\omega_1 < 2^\omega < 2^{\omega_1}$$?

As noted by Hannes Jakob in the comments, to get even a thick Kurepa tree consistently with $$\omega_1 < 2^\omega=2^{\omega_1}$$ it's enough to add $$\omega_2$$ many Cohen reals to a model of GCH+There is a Kurepa tree.

• Are there any left after the recent fires? Sep 6 at 10:21
• Correct me if i am wrong, but can we not simply start from a model with a Kurepa tree $T$ (e.g. V=L) and force the negation of CH by Adding $\omega_2$ subsets to $\omega$? In the resulting model, $T$ still has height $\omega$ and $\omega_2$ cofinal branches (being a cofinal branch is upwards absolute and we are not collapsing cardinals), $T$ still has countable levels (we are not changing the levels) and $2^{\omega}=\omega_2$, so CH fails. Sep 6 at 12:10
• Also called maples? Sep 6 at 19:30
• @Hannes Jakob You are absolutely right. Such a Kurepa tree would be thick because $2^{\omega_1}=\omega_2$ in the generic extension. However, I forgot to say I also wanted $2^\omega < 2^{\omega_1}$ besides not CH. I've updated the question. Sep 6 at 23:25

## 1 Answer

$$\newcommand{\Add}{\operatorname{Add}}$$Start with a model $$V$$ satisfying $$GCH$$ (or just $$2^{\omega}=\omega_1$$ and $$2^{\omega_1}=\omega_2$$). Force over $$V$$ with the product $$\Add(\omega,\omega_2)\times \Add(\omega_1,\omega_3)$$ to obtain $$V[G]$$. By considering this as an extension first by $$\Add(\omega_1,\omega_3)$$ (which is $$<\omega_1$$-closed and $$\omega_2$$-cc. by $$CH$$) and then by $$\Add(\omega,\omega_2)$$ (which is still ccc.), $$V$$ and $$V[G]$$ have the same cardinals. Now consider the tree $$(2^{<\omega_1})^V$$. This has $$\leq\omega_1$$-sized levels (the $$\alpha$$th level is $$(2^\alpha)^V$$ which has size $$\leq\omega_1$$ because $$CH$$ holds in $$V$$). It also has $$2^{\omega_1}=\omega_3$$ many branches, since every Cohen subset of $$\omega_1$$ introduced by $$\Add(\omega_1,\omega_3)$$ has all of its initial segments in $$V$$, i.e. in $$(2^{<\omega_1})^V$$.