Let $\lambda$ be an infinite cardinal. Recall that Weak diamond $\Phi_S$ on $S\subseteq\lambda^+$ is the following principle:

For every function $F:2^{<\lambda^+}\rightarrow 2$, there exists $g\in 2^{\lambda^+}$ such that for all functions $f:\lambda^+\rightarrow 2$, the set $\{\alpha\in S:~F(f\restriction\alpha)=g(\alpha)\}$ is stationary.

The following theorem is well known:

Theorem: (Devlin-Shelah) For every infinite $\lambda$, $\Phi_{\lambda^+}\Leftrightarrow 2^\lambda<2^{\lambda^+}$.

Unlike the above situation, failure of the weak diamond at a stationary set is not enough to get $2^\lambda=2^{\lambda^+}$. In fact, we have the following.

Theorem: (Shelah) The theory ${\rm ZFC}+\neg\Phi_{S^{\omega_2}_\omega}+2^{\omega_1}<2^{\omega_2}$ is consistent.

where $S^{\omega_2}_{\omega}=\{\alpha<\omega_2: {\rm cof}(\alpha)=\omega\}$, and similarly for $S^{\omega_2}_{\omega_1}=\{\alpha<\omega_2:{\rm cof}(\alpha)=\omega_1\}$ we have:

Theorem:(Shelah) The theory ${\rm ZFC}+\neg\Phi_{S^{\omega_2}_{\omega_1}}+{\rm GCH}$ is consistent.

Now in the light of above theorems, I would like to know what is the answer to the following question.

Question: Does simultaneous failure of $\Phi_{S^{\omega_2}_{\omega }}$ and $\Phi_{S^{\omega_2}_{\omega_1}}$ imply $2^{\aleph_1}=2^{\aleph_2}$?


For a proof of 2nd theorem look at Shelah's paper "More on the weak diamond"-1985. A reference for the 3rd one was given in the comments by Golshani.


1 Answer 1


Yes, it does.

The collection $I^{\omega_2}_{WD}= \{ S \subset \omega_2: \neg \Phi^{\omega_2}_S \}$ of subsets of $\omega_2$ is an ideal (a proof this fact is provided in the proposition below; moreover I think this result is originally due to Shelah, however a reference escapes me at the moment.)

Moreover, $\Phi_{\omega_2}$ fails, precisely when $I^{\omega_2}_{WD}$ is not a proper ideal, which in this case is equivalent to some club subset of $\omega_2$ being a member of $I^{\omega_2}_{WD}$.

So if $\Phi^{\omega_2}_S$ fails for both $E^{\omega_2}_\omega$ and $E^{\omega_2}_{\omega_1}$. Then $E^{\omega_2}_{\omega}\cup E^{\omega_2}_{\omega_1} =\lim(\omega_2) \in I^{\omega_2}_{WD}$ and this implies the failure of $\Phi_{\omega_2}$, which is equivalent to $2^{\aleph_1} = 2^{\aleph_2}$.

Proposition: For any cardinal $\lambda$, the set $I^{\lambda^{+}}_{WD} = \{S \subset \lambda^{+}: \neg \Phi^{\lambda^{+}}_S \}$ is a $\lambda^{+}$-complete ideal extending the non-stationary ideal on $\lambda^{+}$.

Proof: That $I_{WD}^{\lambda^{+}}$ contains the non-stationary ideal on $\lambda^{+}$ follows by definition. As such, all that remains is to show $I^{\lambda^{+}}_{WD}$ is an appropriately complete ideal.

To this end suppose $S \in I^{\lambda^{+}}_{WD}$ as witnessed by the sequence $\langle F_\alpha: \alpha \in S \rangle$ where for each $\alpha \in S$, $F_{\alpha}:\,^{\alpha}2\rightarrow 2$. Then, for every $g:\lambda^{+} \rightarrow 2$, there is some $f:\lambda^{+} \rightarrow 2$ and club $C_g \subset \lambda^{+}$ such that

$$ \{ \alpha \in S \cap C_g: F_\alpha(f\vert_\alpha) = g(\alpha) \} = \emptyset $$

as such, for every $S_0 \subset S$ we have $ \{ \alpha \in S_0 \cap C_g: F_\alpha(f\vert_\alpha) = g(\alpha) \} = \emptyset $, hence the sub-sequence $\langle F_\alpha : \alpha \in S_0\rangle$ witnesses $S_0 \in I^{\lambda^{+}}_{WD}$.

Next, fix a bijection $\varphi:\lambda^{+}\times \lambda^{+}\rightarrow \lambda^{+}$ and let $C = \{ \alpha \in \lim(\lambda^{+}): \varphi[\alpha \times \alpha] = \alpha \}$ be the club subset of $\lambda^{+}$ consisting of closure points of $\varphi$. Now, assume $\{ S_\gamma: \gamma \in \delta \} \subset I_{WD}^{\lambda^{+}}$ (with $\delta \in \lambda^{+}$) are stationary, pairwise disjoint, and that for each $\gamma\in \delta$, the sequence $\langle F^{\gamma}_\alpha : \alpha \in S_\gamma \rangle$ witnesses the failure of $\Phi^{\lambda^{+}}_{S_\gamma}$.

Letting $E = \bigcup \{ S_\gamma : \gamma \in \delta \}$, define the sequence $\langle F^{\ast}_\alpha : \alpha \in E\rangle$ of functions $F^{\ast}_\alpha : \,^{\alpha}2\rightarrow 2$ as follows, given $\alpha \in E$ and $h \in \,^{\alpha}2$:

  • if $\alpha \in C\cap S_\gamma$, set $F^{\ast}_{\alpha}(h) = F^{\gamma}_{\alpha}([h]_\gamma)$, where $[h]_\gamma$ denotes the function defined by $[h]_\gamma(\xi) = h(\varphi(\gamma,\xi))$, and if $\alpha \not\in C$, simply take $F^{\ast}_{\alpha}(h) = 0$.

Now, for every $g:\lambda^{+} \rightarrow 2$, there are functions $\{ h_\gamma : \gamma \in \delta \} \subset \,^{\lambda^{+}}2$ and clubs $\{ C_\gamma: \gamma \in \delta \} \subset \mathcal{P}(\lambda^{+})$ such that, for each $\gamma \in \delta$,

$$ \{ \alpha \in S_\gamma \cap C_\gamma : F^{\gamma}_{\alpha}(h_\gamma\vert_\alpha) = g(\alpha) \} = \emptyset. $$

So, letting $D=\cap \{ C_\gamma : \gamma \in \delta \}$ and defining $h:\lambda^{+}\rightarrow 2$ by $h(\varphi(\gamma,\xi))=h_\gamma(\xi)$ for $\gamma \in \delta$, and $h(\varphi(\gamma, \xi)) = 0$ otherwise; we have $\alpha \in S_\gamma \cap C \cap D\implies $

$$ F^{\ast}_{\alpha}(h\vert_\alpha) = F^{\gamma}_{\alpha}([h]_{\gamma}\vert_\alpha)=F^{\gamma}_{\alpha}(h_{\gamma}\vert_\alpha)=1 - g(\alpha).$$ Therefore, $ \{ \alpha \in E\cap C \cap D: F^{\ast}_{\alpha}(h\vert_\alpha) = g(\alpha) \} = \emptyset $, and it immediately follows that $$E=\cup \{ S_\gamma: \gamma \in \delta \} \in I^{\lambda^{+}}_{WD}$$ $\square$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.