Kurepa trees and ineffable cardinals

Recall that $T$ is said to be a $κ$ -Kurepa tree if $T$ is a tree of height $κ$ , whose levels $T_{α}$ has size $\leq | α |$ for co-boundedly many $α < κ$ , and such that the set of branches of $T$ has size $> κ$ .

Recall also that an uncountable cardinal $κ$ is said to be ineffable if for every sequence $⟨ A_{α} ∣ α < κ ⟩ \in \prod_{α < κ} P (α)$ , there exists some $A \subseteq κ$ for which the set ${α < κ ∣ A \cap α = A_{α}}$ is stationary. Note that every measurable cardinal is ineffable, and that every ineffable cardinal is weakly compact.

Let us point out that these concepts are mutually contradictory.

Proposition (Jensen-Kunen). If $κ$ is an ineffable cardinal, then there is no $κ$ -Kurepa tree.
Proof. Suppose that $⟨ κ, \leq_{T} ⟩$ is a Kurepa tree, and let $F$ denote the set of its branches. Then $| F | > κ$ , while $C := {α < κ ∣ ⋃_{β < α} T_{β} = α & | {b \cap α ∣ b \in F} | \leq | α |}$ is a club in $κ$ .

For all $α \in C$ , we let ${b_{α}^{i} ∣ i < α}$ be some enumeration of ${b \cap α ∣ b \in F}$ , and then put $X_{α} := ⋃_{i < α} ({i} \times b_{α}^{i})$ . Next, Fix a bijection $π : κ \times κ \leftrightarrow κ$ , and define $A_{α} := {\begin{cases} π [X_{α}] \cap α, & α \in C \\ \emptyset, & otherwise \end{cases} .$

Now, as $κ$ is ineffable, we may find some set $A \subseteq κ$ such that the following set is stationary: $S = {α < κ ∣ A \cap α = A_{α}} .$

For all $i < κ$ , denote $b^{i} := {β < κ ∣ (i, β) \in π^{- 1} [A]} .$ As $| F | > κ$ , let us fix some $b \in F ∖ {b^{i} ∣ i < κ}$ . If follows that we may define a function $f : κ \to κ$ , by letting $f (i) := min {α < κ ∣ b^{i} \cap α \neq b \cap α} .$

Since $S$ is stationary, we may fix an $α \in S \cap C$ such that $π [α \times α] = α$ and $f [α] \subseteq α$ . Let $i < α$ be such that $b_{α}^{i} = b \cap α$ . By the choice of $α$ and since $π$ is a bijection, we get that $π^{- 1} [A] \cap (α \times α) = π^{- 1} [A \cap α] = π^{- 1} [A_{α}] = π^{- 1} [π [X_{α}] \cap π [α \times α]] = X_{α} .$ In particular, $b^{i} \cap α = {β < κ ∣ (i, β) \in X_{α}} = {β < κ ∣ (i, β) \in ⋃_{j < α} ({j} \times b_{α}^{j})} = b_{α}^{i} .$ Altogether, we got that $b^{i} \cap α = b \cap α$ , contradicting the fact that $f (i) < α$ .

9 Responses to Kurepa trees and ineffable cardinals

Joel David Hamkins says:

August 16, 2012 at 7:41 pm

Assaf, what you call a $k a p p a$ -Kurepa tree is what I know as a *slim* $k a p p a$ -Kurepa tree. The forcing to create a slim $k a p p a$ -Kurepa tree is $< k a p p a$ -closed, but as you point out, definitely destroys the measurability of $k a p p a$ . This shows that in Laver's indestructibilty theorem, where a supercompact cardinal $k a p p a$ is made indestructible by all $< k a p p a$ -directed closed forcing, one cannot improve it to indestructibility by all $< k a p p a$ -closed forcing.
- saf says:
  
  August 16, 2012 at 8:05 pm
  
  Thanks for this info, Joel! I wonder what is the more general definition of a $k a p p a$ -Kurepa tree?
  - Joel David Hamkins says:
    
    August 16, 2012 at 8:17 pm
    
    The more general definition that I have seen is just a $k a p p a$ tree (meaning all levels have size less than $k a p p a$ ) with at least $k a p p a^{+}$ many branches. This is trivial when $k a p p a$ is a strong limit, for example, since the full binary tree has the property, and this may be the reason you dont’ consider it. The slimness property I know is essentially the extra property you mention, that the $a l p h a$ -th level should have size $a l p h a$ , when $a l p h a$ is infinite (slight difference from what you say: except for boundedly many $a l p h a$ ).
    - saf says:
      
      August 16, 2012 at 8:21 pm
      
      Indeed, any definition that omits a slimness condition is likely to trivialize the matter.
Joel David Hamkins says:

August 16, 2012 at 7:46 pm

Furthermore, Thomas Johnstone, Vika Gitman and I proved that one may make any strongly unfoldable cardinal indestructible by the forcing to add a slim $k a p p a$ -Kurepa tree. (This was generalized in later work: see http://jdh.hamkins.org/indestructiblestrongunfoldability/.) So one can have a strongly unfoldable cardinal with a slim $k a p p a$ -Kurepa tree, and this provides a bound on improvements to the theorem you mention.
- saf says:
  
  August 16, 2012 at 8:12 pm
  
  That’s very interesting!
  Do you perhaps know of an upper bound for the non-existence of an $a l e p h_{o m e g a + 1}$ -Kurepa tree? Dima Sinapova and myself realized this week that a supercompact cardinal and an almost huge cardinal above it, suffices. (This follows from Foreman’s theorem 4.11 in here.)
Mohammad says:

March 30, 2016 at 1:43 pm

I assume you mean the $a l e p h_{o} m e g a$ and not $a l e p h_{o m e g a + 1}$ .
- saf says:
  
  March 30, 2016 at 1:46 pm
  
  Thanks, Mohammad. I did mean $a l e p h_{o m e g a + 1}$ , but later indeed realized that the classic Silver argument will do the job.
Mohammad says:

March 31, 2016 at 2:32 am

Yes, that was why I asked the questio, as the ordinary Levy collapse works foe successor cardinals.

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Kurepa trees and ineffable cardinals

9 Responses to Kurepa trees and ineffable cardinals

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