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 $\mathbb{V}[G] \models \forall \mbox{ stationary } S \subseteq \omega\_1, \diamondsuit(S)$?



Consistency of Weak Diamond with a Weak Version of Martin's Axiom
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 $X$.


*Proof.*


(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 $P\_0 \times P\_1 \Vdash MA^\* \land \Phi^\*$.


*Proof.*


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)?


