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.

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    $\begingroup$ Are there any left after the recent fires? $\endgroup$ Sep 6 at 10:21
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    $\begingroup$ 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. $\endgroup$ Sep 6 at 12:10
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    $\begingroup$ Also called maples? $\endgroup$ Sep 6 at 19:30
  • $\begingroup$ @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. $\endgroup$ Sep 6 at 23:25

1 Answer 1


$\newcommand{\Add}{\operatorname{Add}}$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 $(2^{<\omega_1})^V$.


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