A Weakening of the Tree Property

Question

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.

Answer 1

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

Answer 2

Douglas: here's Yair's argument. Assume that $P$ is $\sigma$-closed and adds a $b$ as described. First, if $p$ forces that $b|\alpha=\beta$, then there are extsnions $p'$ and $p''$ of $p$ which force $b|\beta=g',g''$ such that $g'$ and $g''$ differ. (otherwise $b$ is in $V$.) Notice that you can make $\beta$ arbitrarily large. Now build the decreasing sequences $p_0\geq p_1\geq\cdots$ and $q_0\geq q_1\geq\cdots$ and increasing sequence of ordinals $\alpha_0<\alpha_1<\cdots$ such that $p_i$ forces that $b$ restricted to $[\alpha_i,\alpha_{i+1})$ is $g_i$, $q_i$ forces that $b$ restricted to $[\alpha_i,\alpha_{i+1})$ is $g'_i$ and $g_i\neq g'_i$. Let $p,q$ be the extensions of $p_i$, $q_i$, resp. Then they force that $b_\alpha$ is almost equal to two functions, which differ at infinitely many places, a contradiction.