Kurepa Hypothesis says there is a Kurepa tree, which is a $\omega_1$-tree has at least $\omega_2$ many branches. It is known that beginning from a model with an inaccessible cardinal $\kappa$, after collapes $\kappa$ to $\omega_2$ using the Levy collape, then in the generic extension, Kurepa Hypothesis fails. In above generic extension, $\omega_2$ is equal to $\kappa$ and by a counting argument for nice names, $2^{\omega_1}=\omega_2$. My question is that "is it consistent that Kurepa Hypothesis fails and $2^{\omega_1}>\omega_2$?" (The reason I think this question: The biggest possible value of the number of branches is $2^{\omega_1}$, so in the environment of $2^{\omega_1}=\omega_2$, it is most difficult for the living of a Kurepa tree. So I want to know whether this requiement is necessary.)

  • $\begingroup$ The result is first proved by Keith Devlin in "Kurepa's hypothesis and the continuum. Fund. Math. 89 (1975), no. 1, 23–31" $\endgroup$ Nov 26, 2013 at 3:54

3 Answers 3


The answer is yes, and merely forcing over your model to add additional subsets of $\omega_1$ will pump up the value of $2^{\omega_1}$, while not creating Kurepa trees.

Specifically, let us start in $V$, where $\kappa$ is an inaccessible cardinal, and suppose also that the GCH holds. You mentioned the result of Silver, which shows that in the forcing extension obtained via the Levy collapse making $\kappa=\omega_2$, there are no Kurepa trees. I propose simply to add more subsets to $\omega_1$ over this model, and claim that still no Kurepa trees will be created.

To see this, consider the product forcing $P\times Q$, where $P=\text{Coll}(\omega_1,\lt\kappa)$ is the relevant Levy collapse, and $Q=\text{Add}(\omega_1,\theta)$ is the forcing to add $\theta$ many Cohen subsets of $\omega_1$. Suppose that $V[G][H]$ is $V$-generic for this forcing. Note that because the Levy collapse is countably closed, the forcing $Q$ is $\text{Add}(\omega_1,\theta)$ in both $V$ and in $V[G]$. In particular, the $Q$ forcing is countably closed in $V[G]$ and $\kappa$-c.c. there, so cardinals $\kappa$ and above are all preserved.

By the result of Silver, we know that there are no Kurepa trees in $V[G]$, since this is the Levy collapse you mentioned.

Next, I claim that no $\omega_1$-tree $T$ in $V[G]$ becomes Kurepa in $V[G][H]$. This is because countably closed forcing cannot add any new branches to an $\omega_1$ tree in the ground model. If we had a name for a new branch $\dot b$, then we could decide this name in various incompatible ways, and get a level of $T$ that must have continuum many elements on it, contrary to $T$ being an $\omega_1$ tree. (And this observation was critical to Silver's argument.)

Suppose that there is a Kurepa tree $T$ in $V[G][H]$, created by the $H$ forcing over $V[G]$. Since $T$ has size $\omega_1$ and the $Q$ forcing is $\omega_2$-c.c., it follows that $T$ exists in $V[G][H|A]$ for some subset $A\subset \theta$ of size $\omega_1$. We may rearrange the forcing and assume without loss that $T\in V[G][H|\omega_1]$ is added by the first $\omega_1$ many stages of the forcing. Indeed, since adding $\omega_1$ many subsets to $\omega_1$ is the same as adding just one, we may assume that $T\in V[G][H_0]$, where $H_0$ is the first subset to $\omega_1$ added by $H$. What is more, the rest of the $H$ forcing remains countably closed over this model, and therefore adds no new branches to $T$ that are not already in $V[G][H_0]$. Thus, all the branches of $T$ of $V[G][H]$ are in $V[G][H_0]$. But now, the point is that doing the Levy collapse and then adding one subset of $\omega_1$ is isomorphic to just doing the Levy collapse. The later forcing is absorbed by the Levy collapse. (I can explain this if you like.) Thus, $V[G][H_0]=V[G^\ast]$ for some $V$-generic filter $G^\ast\subset P$. In particular, $T$ is not Kurepa there. So $T$ does not have $\omega_2$ many branches there, and hence does not have $\omega_2$ many branches in $V[G][H]$. So the model $V[G][H]$ has no Kurepa trees, yet $2^{\omega_1}\geq\theta$ there, as desired.

  • $\begingroup$ I suppose another way to go about the argument is simply to perform the Levy collapse itself an enormous number of times. (This is actually forcing equivalent to the forcing I describe.) The point now is that any tree that is added appears in a small number of factors, which is equivalent to just one factor, and the remaining forcing adds no additional branches, so we're done again. $\endgroup$ Jun 22, 2011 at 20:54
  • $\begingroup$ Thank you for your answer, Joel. Another thing is unknown for me: In Silver's model, no new real has been added, and inaccessible $\kappa$ is collaped to $\omega_2$, so Continumm hypothesis holds in this model. But if adding some Cohen reals to this model, maybe some new Kurepa tree are created. This should be my second question: is it consistent that "Kurepa hypothesis fails and $2^{\omega_1}>\omega_2$ and $2^{\omega}>\omega_1$"? $\endgroup$ Jun 23, 2011 at 7:32
  • 1
    $\begingroup$ That is an interesting question; I encourage you to ask it as a separate question. My inclination is simply to add many Cohen reals over a model of KH. Since absolutely c.c. forcing does not add any new branches to a ground model $omega_1$-tree, this reduces the problem to showing that $\text{Add}(\omega,\omega_1)$ does not create a Kurepa tree, which seems likely to be true. But I'll think about it. (And incidentally, if this method works, then it provides another way to answer the current question.) $\endgroup$ Jun 23, 2011 at 12:49
  • $\begingroup$ I find a paper: Random trees under CH James Hirschorn Israel Journal of Mathematics, 2007, Volume 157, Number 1, Pages 123-153 I just read its abstract, he says there is a model of $\neg{KH}$, and we can add many random reals and preserve $\neg{KH}$. But I think $Add(\omega,\omega_1)$ does not add Kurepa tree over Silver model is also a question. $\endgroup$ Jun 24, 2011 at 14:12

This paper contains several results of the kind: Keith Devlin: $\aleph _{1}$-trees, Ann. Math. Logic 13(1978), 267–330.


Another solution can be obtained by adding many reals and using a lemma of Spencer Unger that generalizes the lemma used by Silver.

Lemma (Unger) Suppose $\mu \leq \kappa$ are regular, there is $\tau \leq \mu$ such that $2^\tau \geq \kappa$, and $\mathbb{P}$ is $\mu$-c.c. and $\mathbb{Q}$ is $\mu$-closed. Then $\Vdash_\mathbb{P} \mathbb{Q}$ does not add branches to $\kappa$-trees.

MR2945572 Unger, Spencer. Fragility and indestructibility of the tree property. Arch. Math. Logic 51 (2012), no. 5-6, 635–645.

Start with Silver's model of $\neg KH$, in which an inaccessible $\theta$ is Levy collapsed to $\omega_2$ by $\mathbb{Q} = Col(\omega_1,<\kappa)$. Then add at least $\omega_3$ many Cohen or Random reals, call that $\mathbb{P}$. Any $\omega_1$-tree $T$ in $V^{\mathbb{Q} \times \mathbb{P}}$ is captured in some intermediate extension by a subforcing $\mathbb{Q}_0 \times \mathbb{P}_0$ of size $<\kappa$, where $\mathbb{Q}_0$ is rank initial segment of $\mathbb{Q}$. By the well known fact about Knaster posets $\mathbb{P} / \mathbb{P}_0$ adds no branches to $T$, and by Unger's lemma, $\mathbb{Q} / \mathbb{Q}_0$ does not add branches over $V^{\mathbb{P} \times \mathbb{Q}_0}$. Finally, since $\kappa$ is inaccessible in $V$, the number of branches of $T$ in $V^{\mathbb{Q}_0 \times \mathbb{P}_0}$ is $<\kappa$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.