# The tree property for non-weakly compact $\kappa$

In my previous question, Weakly-compact cardinals, I was asking about weakly-compact cardinals and equivalent definitions to the basic one, which is $\kappa \to (\kappa)^2_2$.

One of which was that $\kappa$ is inaccessible and has the tree property (that is if any tree of cardinality $\kappa$ for which every level is of cardinality $<\kappa$ then it has a branch (i.e. a maximal chain) of cardinality $\kappa$).

I can understand the property itself and what it means. However, since $\aleph_1$ or $\aleph_\omega$ are clearly not weakly-compact cardinals, there should be a tree which contradicts this property.

How do you build this sort of tree?

• – Andrés E. Caicedo Aug 11 '13 at 2:50
• Joel and Andres seem to have covered the important points, but since you mentioned $\aleph_\omega$ and (unless I overlooked it) they didn't, let me add a triviality. You can get a counterexample to the tree property for $\aleph_\omega$ by building the following tree. Start with a root, and countably many immediate successors of it, say $a_n$. Then attach to $a_n$ a chain of length $\omega_n$. The same idea gives you counterexamples to the tree property for all singular cardinals. – Andreas Blass Aug 11 '13 at 5:15
• @Andreas: Thank you for adding that! – Asaf Karagila Aug 11 '13 at 10:25
• @AndreasBlass Thanks. The two results I didn't mention where that the tree property implies regularity, and Specker's theorem that the tree property has some consequences in cardinal arithmetic. For completeness, they are mentioned in the link above. – Andrés E. Caicedo Aug 11 '13 at 14:42

What you are looking for is the concept of Aronszajn tree. You can read about constructions of Aronszajn trees in any graduate level set theory text, and meanwhile, the Wikipedia page lists a summary of the basic facts:

• König's lemma states that $\aleph_0$-Aronszajn trees do not exist.

• The existence of Aronszajn trees ($=\aleph_1$-Aronszajn trees) was proven by Nachman Aronszajn, and implies that the analogue of König's lemma does not hold for uncountable trees.

• The existence of $\aleph_2$-Aronszajn trees is undecidable (assuming a certain large cardinal axiom): more precisely, the continuum hypothesis implies the existence of an $\aleph_2$-Aronszajn tree, and Mitchell and Silver showed that it is consistent (relative to the existence of a weakly compact cardinal) that no $\aleph_2$-Aronszajn trees exist.

• Jensen proved that $V=L$ implies that there is a κ-Aronszajn tree (in fact a κ-Suslin tree) for every infinite successor cardinal κ.

• Cummings & Foreman (1998) showed (using a large cardinal axiom) that it is consistent that no $\aleph_n$-Aronszajn trees exist for any finite n other than 1.

• If κ is weakly compact then no κ-Aronszajn trees exist. Conversely if κ is inaccessible and no κ-Aronszajn trees exist then κ is weakly compact.

Finally, let me point out a small inaccuracy in your question. The equivalence is that $\kappa$ is weakly compact iff it is inaccessible and has the tree property. It is not correct to drop the inaccessibility part, as you did, since it is consistent that $\aleph_2$ has the tree property.

• Shelah proved (around 1995) that if $\lambda$ has cofinality $\omega$ and is the supremum of strongly compact cardinals, then $\lambda^+$ has the tree property. See

Menachem Magidor, and Saharon Shelah. The tree property at successors of singular cardinals, Archive for Math Logic, 35 (5-6), (1996), 385-404. MR1420265 (97j:03093).

• Neeman proved that, assuming the existence of $\omega$ supercompact cardinals, we can force a model where the tree property holds at all the $\aleph_n$ ($2\le n<\omega$) and at $\aleph_{\omega+1}$. See

Itay Neeman. The tree property up to $\aleph_{\omega+1}$, preprint.

Neeman's result improves previous results, both in terms of the cardinals with the tree property, and in consistency strength: Magidor and Shelah had obtained the tree property at $\aleph_{\omega+1}$ from a huge cardinal with $\omega$ supercompact cardinals above. As mentioned in Joel's answer, Cummings and Foreman had obtained the tree property for the $\aleph_n$ ($2\le n<\omega$), also from $\omega$ supercompact cardinals. At the moment, Neeman's is the best current result in terms of intervals of regular cardinals with the tree property. At least in $\mathsf{ZFC}$.

• Arthur Apter proved (around 2009) that the following is consistent, relative to a proper class of supercompact cardinals: $\mathsf{ZF} + \mathsf{DC} +$ Every successor cardinal is regular and has the tree property, while every limit cardinal is singular. See these slides, and

Arthur W. Apter. A remark on the tree property in a choiceless context, Arch. Math. Logic, 50 (5-6), (2011), 585–590. MR2805298 (2012d:03115).

The conclusion of Apter's result implies determinacy in $L(\mathbb R)$, and more.

• The upper bound in consistency strength for successive cardinals with the tree property is a supercompact cardinal with a weakly compact cardinal above it. Around 1983, Abraham forced, from these assumptions, that $2^{\aleph_0}=\aleph_2$, and both $\aleph_2$ and $\aleph_3$ have the tree property. All results on successive cardinals with the tree property build on Abraham's argument. See

Uri Abraham. Aronszajn trees on $\aleph_2$ and $\aleph_3$, Ann. Pure Appl. Logic, 24 (3), (1983), 213–230. MR0717829 (85d:03100).

• The best known lower bound is due to Foreman, Magidor, and Schindler. They show that if all $\aleph_n$ ($2\le n<\omega$) have the tree property, and $\aleph_\omega$ is strong limit, then $\mathsf{PD}$ holds. See

Matthew Foreman, Menachem Magidor, and Ralf Schindler. The consistency strength of successive cardinals with the tree property, J. Symbolic Logic, 66 (4),(2001), 1837–1847. MR1877026 (2003m:03083).

This result is frustrating in the sense that we expect two successive cardinals with the tree property should give us much more in consistency strength than this, beyond $\mathsf{AD}^{L(\mathbb R)}$, and likely beyond the current reach of descriptive inner model theory. Still, this would be frustratingly short of the best current upper bounds, which experts expect are much closer to the truth.

• It seems that the tree property is a big topic of the Vienna gathering. – Asaf Karagila Sep 18 '13 at 22:28
• (Oh, that was nice. Thanks!) – Andrés E. Caicedo Sep 20 '13 at 5:49
• Well, I was talking to James Cummings a few days ago, and this topic is a main focal point here. I told him that I asked this question a long time ago, and it seemed that you added this answer "when I was ready to understand it" or sort of something like that. But thanks to this I am able to at least understand what's going on around here... I figured that the answer is worth well more than one vote, so I voted it six times! :-) – Asaf Karagila Sep 20 '13 at 7:45