If $f$ and $g$ are two functions, define $f \sim g$ if they differ only finitely often on their common domain.


The following property of a large cardinal arose from a problem in model theory. I am interested in its strength. Say that a cardinal $\kappa$ has the weak tree property if the following holds: 


Suppose $(b\_\alpha: \alpha < \kappa)$ is a sequence such that:


* Each $b\_\alpha \in 2^\alpha$,
* For each $\alpha < \beta < \kappa$, $b\_\alpha \sim b\_\beta$.


Then there is some $b \in 2^\kappa$ such that for all $\alpha < \kappa$, $b \sim b\_\alpha$.


The following summarizes what I know about the weak tree property:


* If $\kappa$ has the tree property then it has the weak tree property. (A sequence $(b\_\alpha)$ such as above can also be viewed as defining an Aronszajn tree.)
* $\aleph\_1$ does not have the weak tree property. (One of the standard constructions of an $\aleph\_1$ Aronszajn tree yields this.)
* If $\kappa$ has countable cofinality then $\kappa$ has the weak tree property.


I am interested in knowing more. For example, might $\aleph\_2$ not have the weak tree property? (It is possible that $\aleph\_2$ has the weak tree property since it might have the full tree property.)


Any thoughts would be appreciated.


PS: The example question above has been resolved in the comments.



A Weakening of the Tree Property
I give here some upper bounds for the consistency of the weak tree property. Namely - for every successor of regular (including double successors of singulars) we can get the weak tree property by collapsing a weakly compact, and we can get the weak tree property everywhere by collapsing a strongly compact cardinal to be $\aleph\_2$.


Let start with some definitions:


**Definition:** We call a sequence $\mathcal{B} = \langle b\_\alpha | \alpha < \kappa \rangle$ *coherent* if $\forall \alpha , \beta < \kappa\,b\_\alpha\sim b\_\beta$.
A element $b\in 2^{\kappa}$ is a *thread* for $\mathcal{B}$ if $b\sim b\_\alpha$ for every $\alpha < \kappa$. 


**Lemma:** Let $\mathcal{B}$ be a coherent sequence with no thread. Let $\mathbb{P}$ be $\sigma$-closed forcing notion. Then after forcing with $\mathbb{P}$, $\mathcal{B}$ still has no thread. 


**Proof:** [Peter Komjath already gave a proof in his answer above. I give here a slightly more detailed one]


Assume otherwise, and let $\dot{b}$ be a name for a thread. Note that since $\Vdash\dot{b} \notin V$, for every condition $p\in \mathbb{P}$ and ordinal $\alpha < \kappa$ there is ordinal $\alpha < \beta < \kappa$ and two conditions $q, q^\prime \leq p$ such that $q\Vdash \dot{b}(\beta) = 0$ and $q^\prime \Vdash \dot{b}(\beta) = 1$. 


Now, choose by induction ordinals $\alpha\_n < \kappa$ and conditions $p\_\eta \in \mathbb{P}$ for every $\eta \in 2^{<\omega}$ such that $\eta \trianglelefteq \eta^\prime \implies p\_\eta \geq p\_{\eta^\prime}$ and for all $\eta \in 2^n$ there is $\alpha\_n < \beta\_\eta$, such that $p\_{\eta \frown \langle 0 \rangle} \Vdash \dot{b}(\beta\_\eta )= 0$ and $p\_{\eta \frown \langle 1 \rangle} \Vdash \dot{b}(\beta\_\eta )= 1$. Let $\alpha\_{n+1} = \max\_{\eta \in 2^n} {\beta\_\eta} + 1$.


Let $\gamma = \sup \alpha\_n$. 


Now, using the $\sigma$-closure of $\mathbb{P}$ pick for every $f\in 2^\omega$ a condition $p\_f$ such that $p\_f \leq p\_{f\restriction n}$ for every $n$. By extending $p\_f$, if necessary, we may assume that the finite set of difference between $b\_\gamma$ and $\dot{b}\restriction \gamma$ is decides by $p\_f$, and let denote it by $s\_f$. Since $s\_f$ is bounded for every $f$, there is $n\_0 < \omega$ such that for infinite many $f\in 2^\omega$, $s\_f \subset \alpha\_{n\_0}$. But this is impossible since for every $\eta \in 2^n$ there is only one extension $\eta^\prime \in 2^m$, $m>n$ that force that the difference between $\dot{b}$ and $b\_\gamma\restriction \alpha\_m$ will be all below $\alpha\_n$, so there at most $2^n$ many such $f$ - a contradiction. **QED**


**Theorem:** Let $\mu$ be uncountable regular cardinal and $\kappa > \mu$. $\mathbb{C} = Col(\mu,<\kappa)$. Then if $\kappa$ is weakly compact, $\mathbb{C}$ forces the weak tree property at $\mu^+$, and if $\kappa$ is $\lambda$-strongly compact $\mathbb{C}$ forces the weak tree property at $\lambda$.


**Proof:** We start with the weakly compact case. Let $\dot{\mathcal{B}}$ be a $\mathbb{C}$-name for a coherent sequence of length $\kappa$. The elementary embedding $j: (V\_\kappa,\dot{\mathcal{B}},\mathbb{C},\dots) \rightarrow M$ extends to elementary embedding $\tilde{j}:V\_\kappa[G] \rightarrow M[G][H]$ in $V[G][H]$ where $H$ is a generic filter for $Col(\mu,[\kappa,j(\kappa) )$ which is $\mu$-closed.


$j(\mathcal{B}) = \langle b^{j}\_\alpha | \alpha < j(\kappa)\rangle$ is a coherent sequence so in particular, $b^{j}\_\kappa$ is a thread. This object was added by $Col(\mu,[\kappa,j(\kappa) )$ but this is a $\sigma$-closed forcing, so by the lemma, there was some thread already in $V[G]$. 


The proof for the strongly compact case is the same. We extend the strongly compact embedding $j$ to $\tilde{j}:V[G]\rightarrow M[G][H]$ be forcing with $Col(\mu,[\kappa,j(\kappa) )$. Now, if $\mathcal{B}$ was a coherent sequence of length $\lambda$ in $V[G]$, (and $\sup j^{\prime\prime} \lambda < j(\lambda)$) then for $j(\mathcal{B} = \langle b^j\_\alpha | \alpha < j(\lambda)\rangle$, we can pick $\delta = \sup j^{\prime\prime} \lambda$ and choose $b(\alpha) = b^j\_\delta (j(\alpha)$ for every $\alpha < \lambda$. It's clear that $b\in V[G][H]$ is a thread for $\mathcal{B}$, and by the lemma, there is a thread already in $V[G]$. **QED**


