MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $\kappa$ is an inaccessible cardinal then the tree property at $\kappa$ is equivalent to weak compactness of $\kappa$, which implies that $\square(\kappa)$ fails---that is, that every coherent sequence of clubs of length $\kappa$ can be threaded.

I am wondering about other implications involving square and the tree property, namely:

  • If $\kappa$ is an inaccessible cardinal and $\square(\kappa)$ fails, must $\kappa$ have the tree property (and therefore be weakly compact?)

  • If $\kappa$ is a regular cardinal, does $\neg \square(\kappa)$ imply that $\kappa$ has the tree property?

  • If $\kappa$ is a regular cardinal with the tree property, must $\square(\kappa)$ fail?

(By the way, I am aware of the relative consistency result that if $\square(\kappa)$ fails for some regular cardinal $\kappa$, then $\kappa$ is weakly compact in $L$ and so in particular it has the tree property in $L$.)

share|cite|improve this question
up vote 7 down vote accepted

The answers to the second and third questions are no and yes, respectively. I don't know the answer to the first question.

For the second question, let $\lambda$ be regular and let $\kappa > \lambda$ be weakly compact. Then forcing with $\mathrm{Coll}(\lambda, <\kappa$) yields a model in which $\kappa = \lambda^+$, $\square(\kappa)$ fails, and, since $\lambda^{<\lambda}=\lambda$, there is a special $\kappa$-Aronszajn tree, so the tree property fails.

For the third question, the usual construction of a special Aronszajn tree from a weak square sequence using minimal walks (see, for example, section 5.1 of Cummings' "Notes on Singular Cardinal Combinatorics") still yields a $\kappa$-Aronszajn tree when applied to a $\square(\kappa)$-sequence when $\kappa$ is regular, so $\square(\kappa)$ implies the failure of the tree property.

share|cite|improve this answer

To complete the accepted answer to my question (which addresses parts 2 and 3) perhaps I should mention here that the answer to part 1 is no: If $\delta$ is a supercompact cardinal and $\kappa$ is the least inaccessible cardinal above $\delta$ then $\square(\kappa)$ fails but $\kappa$ is not weakly compact.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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