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 $V[G]$).