# The Ostaszewski square, and homogeneous Souslin trees

Abstract: Assume GCH and let $\lambda$ denote an uncountable cardinal.
We prove that if $\square_\lambda$ holds, then this may be  witnessed by a coherent sequence $\left\langle C_\alpha \mid \alpha<\lambda^+\right\rangle$ with the following remarkable guessing property:

For every sequence $\langle A_i\mid i<\lambda\rangle$ of unbounded subsets of $\lambda^+$, and every limit $\theta<\lambda$, there exists some $\alpha<\lambda^+$ such that $\text{otp}(C_\alpha)=\theta$, and the $(i+1)_{th}$-element of $C_\alpha$ is a member of $A_i$,  for all $i<\theta$.

As an application, we introduce the first construction of an homogeneous Souslin tree at the successor of a singular cardinal.
In addition, as a by-product, a theorem of Farah and Velickovic (see [FV]), and a theorem of Abraham, Shelah and Solovay (see [AShS:221]) are generalized to cover the case of successors of regulars

Citation information:

A. Rinot, The Ostaszewski square, and homogeneous Souslin trees, Isr. J. Math, 199(2): 975-1012, 2014.

This entry was posted in and tagged , , , , , . Bookmark the permalink.

### 5 Responses to The Ostaszewski square, and homogeneous Souslin trees

1. saf says:

Submitted to Israel Journal of Mathematics, May 2011.
Accepted April 2013.

0 likes

2. saf says:

Correction: In Theorem 1.2, where I wrote “implicit in [17]” – the correct reference is not [17], but this paper.

0 likes

3. saf says:

In a recent paper, Cody and Eskew introduce the following.
Definition. A sequence $\langle a_\alpha \mid \alpha<\kappa\rangle$ is called a fat diamond sequence ($\blacklozenge_\kappa$-sequence) if for every $X\subseteq\kappa$, $\{\alpha<\kappa\mid X\cap\alpha=a_\alpha\}$ is a fat stationary set.

Note that by Theorem D of our paper, if $\lambda$ is singular, $\square_\lambda$ holds and $2^\lambda=\lambda^+$, then there exists a $\blacklozenge_{\lambda^+}$-sequence.
Of course, $2^\lambda=\lambda^+$ is necessary for the existence of a $\blacklozenge_{\lambda^+}$-sequence, but if one just needs a partition of $\lambda^+$ into $\lambda^+$ many pairwise disjoint fat stationary sets, then $\square_\lambda$ suffices (including the case that $\lambda$ is regular), as remarked in here.

0 likes

• Shame for the notation. It should have been preserved for a “black diamond”. It would have gone well with Moti Gitik’s “piste”.

😛

0 likes