The following definition is well known ($\kappa$ is regular uncountable cardinal):

Definition: a sequence $\mathcal{C} = \langle \mathcal{C}_\alpha | \alpha < \kappa,\,\alpha \text{ limit ordinal} \rangle$ is a $\square (\kappa, <\lambda )$-sequence when for every $\alpha$, $|\mathcal{C}_\alpha| < \lambda$, every $C \in \mathcal{C}_\alpha$ is a club in $\alpha$, if $\gamma \in \text{acc }C$ then $C\cap \gamma \in \mathcal{C}_\gamma$ and most importantly - there is no thread i.e. there is no club $D\subset \kappa$ such that for every $\alpha \in \text{acc } D$, $D\cap \alpha \in \mathcal{C}_\alpha$.

We say that a club $D$ in $\kappa$ is a weak thread for $\mathcal{C}$ when for every $\alpha \in \text{acc }D$ there is $C \in \mathcal{C}_\alpha$ such that $D\cap \alpha \subset C$. (I don't know if this is the standard terminology)

Clearly every thread is a weak thread.

If $\lambda < \kappa$, there is no weak thread for $\square (\kappa, <\lambda)$ sequence, since it will imply the existence of a thread (we can prove it by showing that the candidates for initial segment of a thread that contain the weak thread, ordered by end extension, forming a tree of height $\kappa$ and width $<\lambda$, so by Kurepa it has a cofinal branch). On the other hand, if we have a $\square(\kappa, <\kappa)$ sequence, we can always build another $\square(\kappa, <\kappa)$ sequence that has a weak thread (but doesn't have a thread).

Question: Assuming the existence of a $\square (\kappa, <\kappa)$ sequence. Can we prove that there is a $\square (\kappa, <\kappa)$ sequence with no weak threads?

Edit: I'll try to explain why this question is interesting.

Todorcevic, in Partitioning pairs of countable ordinals present his method of minimal walks and showed that the existence of $\square(\kappa)$ implies the existence of Aronszajn tree on $\kappa$. His proof easily generalizes to show that $\square(\kappa,<\kappa)$ implies the existence of Aronszajn tree on $\kappa$. On the other hand, from Aronszajn tree it's easy to produce a $\square(\kappa, <\kappa)$ sequence (the elements of the square sequence will be some appropriate closures of the bounded branches of the tree on some club) so is seems to be equivalence. The square sequences derived this way always have weak threads.

Now, if the answer to the above question is positive we conclude that c.c.c. forcing can't destroy the tree property: every $\square(\kappa, <\kappa)$ sequence in $V[G]$ agrees on many elements with coherent sequence of width $<\kappa$ in $V$ (by the c.c.c.), and this sequence has a thread - which will be a weak thread for the original sequence in $V[G]$. So in $V[G]$ every $\square(\kappa,<\kappa)$ sequence has a weak thread and my question translate to "do there are any (thread-less) $\square(\kappa,<\kappa)$ sequences in $V[G]$" (i.e. does the tree property still holds in $V[G]$).

  • $\begingroup$ Is there a known answer to the converse question, that is given a $\square(\kappa,<\kappa)$ sequence, can we build one which is weakly threaded? $\endgroup$ – Asaf Karagila Nov 9 '13 at 20:17
  • $\begingroup$ Yes. Starting with a $\square(\kappa,<\kappa)$ sequence, you can build an Aronszajn tree on $\kappa$. From partial branches of this tree, you can build the weakly threaded square sequence. Assume the tree elements are the successors ordinals in $\kappa$. On a club $\forall x \in T_\alpha,\,\{b\leq_T x | b \in T\}\text{ is unbounded in }\alpha$. This sets will never have thread since it will be a branch in $T$. Making it a real square sequence (taking closures, ensuring coherence and complete the elements that are outside the club) is standard (and straight-forward). $\endgroup$ – Yair Hayut Nov 10 '13 at 6:30

The answer is negative (at least at double successors of regular cardinals): it's possible that every square sequence is weakly threaded and there are Aronszajn trees. For simplicity let work with $\omega_2$.

We start with supercompact cardinal. Using Unger construction in Fragility and indestructibility of the tree property we collapse it to be $\omega_2$ while forcing the tree property at $\omega_2$ to be indestructible under any $\omega_2$-directed closed forcing.

Now, we add a generic normal $\omega_2$-tree by the forcing $\mathbb{Q}$ (the conditions of $\mathbb{Q}$ are the normal trees of successor ordinal height $<\omega_2$ and width $\omega_1$. $\leq_{\mathbb{Q}}$ is end extension). Let $V$ be the universe at this point. $V$ has Aronszajn trees and therefore also $\square(\omega_2,\omega_1)$ sequences.

The generic tree that was added by $\mathbb{Q}$ is Suslin, and it's easy to see that the forcing that adds a branch through this generic Suslin tree, $\mathbb{T}$, satisfies that $\mathbb{Q\ast T}$ has a $\omega_2$(-directed)-closed subset. Since the tree is Suslin, $\mathbb{T}$ is $\omega_2$.c.c..

Let $\mathcal{C}$ be a $\square(\omega_2,\omega_1)$ sequence in $V$. Since $\mathbb{T}$ restore the tree property - it adds a thread to $\mathcal{C}$, which is a club $E$ such that for every $\alpha \in \text{acc }E$, $E\cap \alpha\in\mathcal{C}_\alpha$. But since $\mathbb{T}$ is $\omega_2$.c.c., this thread contains some club from $V$, and this club weakly threads $\mathcal{C}$.

  • $\begingroup$ Using similar arguments we can obtain the same separation result at every successor cardinal. $\endgroup$ – Yair Hayut Feb 7 '14 at 14:29
  • $\begingroup$ Very nice answer. Does it work for successors of singulars? $\endgroup$ – Mohammad Golshani Feb 24 '14 at 4:33
  • $\begingroup$ Not exactly - I don't know how to force indestructible tree property at successor of singular. However, I can achieve similar situation (that adding a branch to certain generic Suslin tree forces the tree property), using some variant of Sinapova's argument. I can send you the details, if you want. $\endgroup$ – Yair Hayut Feb 24 '14 at 6:05
  • $\begingroup$ Yes, that seems interesting, please send it to me. $\endgroup$ – Mohammad Golshani Feb 25 '14 at 3:49

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