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.)
3 Answers
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 $V[G][HA]$ 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 123153 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. 56, 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 $<\kappa$.