If $S \subset \omega_1$ is stationary, then the weak diamond principle $\Phi(S)$ states that for any $F: 2^{<\omega_1} \to 2$, there is a $g: \omega_1 \to 2$ such that for all $f: \omega_1 \to 2$, the set $\{\alpha \in S: F(f \restriction_\alpha) = g(\alpha)\}$ is stationary. Let $\Phi^*$ be the statement: for all stationary $S \subseteq \omega_1$, $\Phi(S)$.

Let $MA^*$ be the following weakening of Martin's Axiom: $\mathbb{R}$ is not the union of $\omega_1$ nowhere dense sets. (Equivalently, $MA^* = MA_{\omega_1}($countable POs$)$; or $MA^* =``\mbox{cov}(\mathcal{B}) \geq \omega_2."$)

I would like to know the consistency of $\Phi^* \land MA^*$. (If consistent this would have applications in model theory.)

Remark 1. It is consistent that $\Phi(\omega_1)$ and $MA^*$ holds; for example, start with $\mathbb{V} = \mathbb{L}$ and force over $\mathsf{Fn}(\omega_{\omega_1}, \omega, \omega)$, or over $\mathsf{Fn}(\omega_2, \omega, \omega) \times \mathsf{Fn}(\omega_3, \omega_1, \omega_1)$. For all I know, both of these forcing extensions witness the consistency of $\Phi^* \land MA^*$. (Here $\mathsf{Fn}(I, J, \lambda)$ is the set of all partial functions from $I$ into $J$ of cardinality less than $\lambda$.)

Remark 2. A related, and probably easier problem is: suppose we start with $\mathbb{V} = \mathbb{L}$ and force over $\mathsf{Fn}(\omega_1, \omega, \omega)$ to get $\mathbb{V}[G]$. Then $\mathbb{V}[G] \models \diamondsuit$. But does $\mathbb{V}[G] \models \forall \mbox{ stationary } S \subseteq \omega_1, \diamondsuit(S)$?

  • $\begingroup$ Earlier I had referenced to "mathoverflow.net/questions/125308/forcing-diamond", but I had misunderstood the answer there. There it is shown that Fn(omega_1, omega, omega_1) forces for all stationary S, daimond(S). $\endgroup$ Oct 12 '14 at 2:36
  • $\begingroup$ On Remark 2: $\diamondsuit_S$ is preserved under ccc forcings of size $\leq \omega_1$. $\endgroup$
    – Ashutosh
    Oct 12 '14 at 8:10
  • $\begingroup$ Do you have an example for a model of $\neg CH$ and $\Phi^{*}$? $\endgroup$
    – Yair Hayut
    Oct 12 '14 at 17:50
  • $\begingroup$ @Yair Hayut: No, actually. In fact the only example I have of a model of $\Phi^*$ is $\mathbb{V}=\mathbb{L}$. $\endgroup$ Oct 12 '14 at 17:59
  • $\begingroup$ (It occurs to me--$\Phi^*$ holds also after forcing over $\mathsf{Fn}(\omega_1, \omega, \omega_1)$ But of course here CH holds.) $\endgroup$ Oct 12 '14 at 18:02

So I was looking through related questions on this site and the book "Proper and Improper Forcing" by Shelah kept popping up. So I checked it out and the appendix actually resolves the question. I rephrase the proof there because I think this way it's simpler and also because it suggests an interesting question (see the end).

Let $P_0 := \mathsf{Fn}(\omega_2, 2, \omega)$ and let $P_1 := \mathsf{Fn}(\omega_3, 2, \omega_1)$. The following lemma is basic:

Lemma $P_0 \Vdash MA^*$.

Proof. Let $G$ be $P_0$-generic over $\mathbb{V}$. Let $f: \omega_1 \to 2$ be the generic function.

Suppose $X \subseteq \omega_1$ codes an $\omega_1$-sequence of open dense sets $(\mathcal{O}_\alpha: \alpha < \omega_1)$ of $(\,^\omega \omega)^{\mathbb{V}[G]}$. Then for some $I \subset \omega_2$ with $|I| = \aleph_1$, we have $X \in \mathbb{V}[G \cap \mathsf{Fn}(I, 2, \omega)]$. But let $\alpha = (\sup I) + 1$ and let $x \in \,^\omega \omega$ be defined by $x(n) = f(\alpha + n)$. Then $x \in \bigcap_{\alpha < \omega_1} \mathcal{O}_\alpha$.

End proof of lemma.

We aim to show that $P_0 \times P_1 \Vdash MA^* \land \Phi^*$ provided we start with a model of $CH$. This proof is gleaned from the proof of Theorem 2.11 in the Appendix of Shelah's "Proper and Improper forcing."

Customarily we view this iterated forcing as starting with $P_1$ and following it by $P_0$, since $(P_0)^{\mathbb{V}^{P_1}} = P_0$. However to get $\Phi^*$ we have to view it the other way around. So we need to look at $P_1$ in $\mathbb{V}^{P_0}$.

Lemma. Suppose $G_0$ is $P_0$-generic over $\mathbb{V} \models CH$. Then in $\mathbb{V}[G_0]$:

(a) $\mathsf{Fn}(\omega_3, 2, \omega) \subset P_1 \subset \mathsf{Fn}(\omega_3, 2, \omega_1)$, and for any two elements $p, q \in P_1$, if $p$ and $q$ are compatible then $p \cup q \in P_1$,

(b) $P_1$ has the $\omega_2$ c.c.,

(c) Forcing with $P_1$ does not add any reals,

(d) The set $\mathcal{B} := \{\mbox{dom}(p): p \in P_1\}$ is closed under intersections, unions, and relative complements, and if $p \in P_1$ and $X \in \mathcal{B}$, then there is some $q \, || \, p$ with $\mbox{dom}(q) = X$,

(e) If $X \subseteq \omega_3$ is countable there is some (countable) $Y \in \mathcal{B}$ containing $X$.


(a) Obvious.

(b) Showing that $(Q \mbox{ has the $\omega_2$ c.c.})^{\mathbb{V}[G_0]}$ is the same as showing it in $\mathbb{V}$ (just go through the proof of the $\Delta$-system lemma and see that it works for any antichain in $Q$).

(c) Let $G_1$ be $Q$-generic over $\mathbb{V}[G_0]$. Then also $G_1$ is $Q$-generic over $\mathbb{V}$, and $G_0$ is $P$-generic over $\mathbb{V}[G_1]$ and $G_0 \times G_1$ is $P \times Q$-generic over $\mathbb{V}$.

Now it suffices to show that if $x \in \mathcal{P}(\omega) \cap \mathbb{V}[G_0 \times G_1]$ then $x \in \mathcal{P}(\omega) \cap \mathbb{V}[G_0]$. To see this, consider a $P$-nice name $\sigma$ for $x$ in $\mathbb{V}[G_1]$; so $\sigma = \{(\hat{n}, A_n): n \in \omega\}$ where $A_n \in \mathbb{V}[G_1]$ is an antichain in $P$. So $A_n \subseteq P$ is countable, so $A_n \in \mathbb{V}$ already (since $(Q \mbox{ is $\omega$-closed})^{\mathbb{V}}$). Hence $\sigma \in V$ so $x \in \mathbb{V}[G_0]$.

(d) Obvious.

(e) The existence of $Y$ follows from the c.c.c. of $P_0$.

End proof of lemma.

Lemma. Suppose $2^{\aleph_0} \leq \aleph_2$ and $Q$ is a forcing notion satisfying $(a)$ through $(e)$ above. Then $Q \Vdash \Phi^*$.

Proof. Suppose towards a contradiction that $1_Q \Vdash \lnot \Phi^*$. Then (by the maximal principle) there are names $\dot{F}, \dot{S}$ such that

\begin{eqnarray*} 1_Q &\Vdash& ``\mbox{ $\dot{F}: \,^{<\omega_1} 2 \to 2$ and $\dot{S} \subseteq \omega_1$ is stationary and } \\ && \forall g: \omega_1 \to 2 \,\,\, \exists f: \omega_1 \to 2 \mbox{ such that } \{\alpha \in \dot{S}: \dot{F}(f \restriction_\alpha)= g(\alpha)\} \\ && \mbox{ is nonstationary}." \end{eqnarray*} We can choose $\dot{F}$ and $\dot{S}$ to be nice names; that is $\dot{F} = \{(\hat{(\eta, i)}, A_{\eta, i}): \eta \in \,^{<\omega_1} 2, i \in 2\}$ and $\dot{S} = \{(\hat{\alpha}, A_\alpha) : \alpha < \omega_1\}$, where each $A_{\eta, i}$ and each $A_\alpha$ is an antichain in $Q$. (This uses that $(\,^{<\omega_1} 2)^{\mathbb{V}} = (\,^{<\omega_1} 2)^{\mathbb{V}^Q}$.

Let $\mathcal{A} = \bigcup_{\eta, i} A_{\eta, i} \cup \bigcup_{\alpha} A_\alpha$; so by the $\omega_2$-c.c., $|\mathcal{A}| \leq \aleph_2$. Hence $D:= \bigcup_{A \in \mathcal{A}} \mbox{dom}(A)$ has cardinality at most $\aleph_2$. Hence after relabeling we can suppose that $D \cap \omega_1 = \emptyset$.

Let $\dot{c}$ be the nice $Q$-name for the generic $\omega_3$-sequence: $\dot{c} = \{(\hat{(\alpha, i)}, \{(\alpha, i)\}): \alpha < \omega_3, i \in 2\}$. Let $\dot{g} = \dot{c} \restriction_{\omega_1}$, i.e. $\dot{g} = \{(\hat{(\alpha, i)}, \{(\alpha, i)\}): \alpha < \omega_3, i \in 2\}$.

Then $1_Q \Vdash ``\exists f: \omega_1 \to 2$ such that $\{\alpha \in \dot{S}: \dot{F}(f \restriction_\alpha) = g(\alpha)\}$ is nonstationary." So for some nice $Q$-name $\dot{f} = \{(\hat{(\alpha, i)}, B_{\alpha, i}): \alpha < \omega_1, i \in 2\}$, we have $1_Q \Vdash ``\dot{f}: \omega_1 \to 2$ and $\{\alpha \in \dot{S}: \dot{F}(\dot{f} \restriction_\alpha) = \dot{g}(\alpha)\}$ is nonstationary." Let $\dot{C}$ be a nice name for a club subset of $\omega_1$ such that $1_Q \Vdash ``\forall \alpha \in \dot{S} \cap \dot{C}: \dot{F}(\dot{f} \restriction_\alpha) \not= \dot{g}(\alpha)$. Say $\dot{C} = \{(\hat{\alpha}, B_\alpha): \alpha < \omega_1\}$.

Let $G$ be $Q$-generic over $\mathbb{V}$; we work in $\mathbb{V}[G]$. Let $F = \dot{F}_G$, let $S = \dot{S}_G$, let $g = \dot{g}_G$, let $f = \dot{f}_G$ and let $C = \dot{C}_G$. For each $\alpha < \omega_1$, let $p_\alpha$ be the unique element of $G \cap (B_{\alpha, 0} \cup B_{\alpha, 1})$ and let $q_\alpha$ be the unique element of $G \cap B_\alpha$ if it exists, or $q_\alpha = 1_Q$ if $G \cap B_\alpha = \emptyset$. Let $D_\alpha = \mbox{dom}(p_\alpha) \cup \mbox{dom}(q_\alpha)$ and let $X_\alpha = \bigcup_{\beta < \alpha} D_\beta$. Then the set $C^* = \{\alpha < \omega_1: X_\alpha \cap \omega_1 \subseteq \alpha\}$ is club. So there is some $\delta \in S \cap \mbox{acc}(C) \cap C^*$.

(In the following we use the properties $(a), (d)$ and $(e)$ heavily.)

Let $I \subset \omega_3$ be countable such that $I \supseteq \delta \cup \{X_\alpha \backslash \omega_1\}$ and $I \in \mathcal{B}$ (where $\mathcal{B}$ is as in the lemma). Let $J = I \backslash \{\delta\}$ (possibly $I = J$); so $J \in \mathcal{B}$.

Let $p_0$ be the unique element of $G \cap A_\delta$, so $p_0 \Vdash \delta \in \dot{S}$. Let $p_1 = g \restriction_J$, so $p_1 \Vdash \dot{f} \restriction_\delta = f \restriction_\delta$ and $p_1 \Vdash \dot{C} \cap \delta = C \cap \delta$. Since $\delta \in \mbox{acc}(C)$, $p_1 \Vdash \delta \in \dot{C}$.

Let $p = p_0 \cup p_1$; note that $\delta \not \in \mbox{dom}(p)$.

Back in $\mathbb{V}$, $p \Vdash ``\hat{\delta} \in \dot{S} \cap \dot{C} \land \dot{F}(\dot{f} \restriction_\delta) = i"$, for some $i \in 2$. But let $q = p \cup \{(\delta, i)\}$; then $q \vDash \dot{F}(\dot{f} \restriction_\delta) = \dot{g}(\delta)$, contrary to definition of $\dot{C}$.

End proof of lemma.

Theorem. Suppose $\mathbb{V} \models CH$. Then $P_0 \times P_1 \Vdash MA^* \land \Phi^*$.


Let $G_0 \times G_1$ be $P_0 \times P_1$-generic over $\mathbb{V}$. By Lemma 1, $\mathbb{V}[G_0 \times G_1] = \mathbb{V}[G_1][G_0] \models MA^*$. By Lemma 2, $\mathbb{V}[G_0 \times G_1] = \mathbb{V}[G_0][G_1] \models \Phi^*$.

End proof of theorem.

Question. Under what conditions can we find a $Q$ satisfying conditions (a) through (e)?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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